Christina Battista
MA 515 – Analysis I
Course Notes
August 17, 2011
Introduction to Functional Analysis
⋆ Functional Analysis provides the necessary background for any further study in any area of analysis
⋆ Functional Analysis shows the unifying power of the abstract linear-space point of view in surveying
problems in many branches of math
Functional Analysis is the study of infinite dimensional spaces over R and C and the linear maps between
them. We will talk about this in terms of topological considerations.
Topology is finding a good abstract framework to talk about continuity.
f ∶ R → R is continuous at x ∈ R if ∀ε > 0, ∃δ > 0 s.t. ∣x − y∣ < δ implies ∣f (x) − f (y)∣ < ε.
Note: The most important way to build a topology is through a metric.
Page 1 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
August 19, 2011
Metric Spaces (play the same role as R in calculus)
A metric on a set X is a function d s.t.:
1. d ∶ X × X → [0, ∞) (i.e. d is a real-valued, finite, and non-negative)
2. d(x, y) = 0 ⇔ x = y
3. d(x, y) = d(y, x)
(Symmetry)
4. d(x, y) ≤ d(x, z) + d(z, y), ∀x, y, z ∈ X
(Triangle Inequality)
In words, a metric associates to any pair of points/elements in X a real number (called distance).
(X, d): metric space (equipped with a metric d).
Examples of metric spaces:
⋆ R, d(x, y) = ∣x − y∣ (this is called the Euclidean metric)
√
⋆ Rn , d(x, y) = ∑ni=1 (xi − yi )2 (Euclidean distance metric on Rn )
Note: the metric is not unique. For example,
d1 (x, y) = max{∣xi − yi ∣}
n
d2 (x, y) = ∑ ∣xi − yi ∣ taxicab metric
i=1
n
are also metrics on R .
⋆ Is d(x, y) = (x − y)2 a metric on R?
No.
We can give a counterexample: (x, y, z) = (5, 3, 4) and the Triangle Inequality does not hold.
⋆ Discrete Metric on X: d(x, y) = {
1
0
∶x≠y
∶x=y
⋆ Is X = C[0, 1] (the space of continuous functions on [0, 1]) with metric
d(f, g) = ∫
1
0
∣f (x) − g(x)∣ dx ≤ M
a metric?
Yes.
1
∫
0
∣f (x) − g(x)∣ dx ≤ ∫
1
0
∣f (x)∣ dx + ∫
Page 2 of 75
0
1
∣g(x)∣ dx ≤ M
Christina Battista
MA 515 – Analysis I
Course Notes
⋆ X = B(A), the set of bounded functions on A.
d(f, g) = sup ∣f (t) − g(t)∣
t∈A
∞
p
⋆ X = lp , p ≥ 1: space of sequences x = {xi }∞
i=1 s.t. ∑i=1 ∣xi ∣ < ∞.
1/p
⎛∞
⎞
d(x, y) = ∑ ∣xi − yi ∣p
⎝i=1
⎠
Note: l2 : Hilbert sequence space
Page 3 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
August 22, 2011
Classical Inequalities: Holder, Cauchy-Schwarz, and Minkowski
For p > 1, let q be its conjugate, i.e.
1
p
+
1
q
= 1.
1. u ⋅ v ≤ p1 up + 1q v q , ∀u, v > 0
1/p
2. ∑i ∣xi yi ∣ ≤ ( ∑j ∣xj ∣p )
( ∑k ∣yk ∣q )
1/q
, for x = {xi } ∈ lp and y = {yi } ∈ lq
Holder Inequality for Sums
Cauchy-Schwarz Inequality when p = q = 2
3. ( ∑i ∣xi + yi ∣p )
1/p
1/p
≤ ( ∑j ∣xj ∣p )
1/p
+ ( ∑k ∣yk ∣p )
, for x = {xi } ∈ lp and y = {yi } ∈ lp
Minkowski Inequality for Sums
Proof of Cauchy-Schwarz Inequality: (use convexity of e)
f ∶R→R
f is convex if ∀a, b, ∈ R and 0 ≤ α ≤ 1, we have f (αa + (1 − α)b) ≤ αf (a) + (1 − α)f (b). These quantities are
equal if and only if a = b or α = 0 or α = 1.
For ex :
eαa+(1−α)b ≤ αea + (1 − α)eb
(ea )α (eb )1−α ≤ αea + (1 − α)eb
Now we let ea = up and eb = v q . The Cauchy-Schwarz Inequality follows directly.
Special case: p=q=2
u⋅v ≤
u2 v 2
+
2
2
2uv ≤ u2 + v 2 ⇒ 0 ≤ (u − v)2
Proof of Holder Inequality for Sums:
Let {x̃i }i ∈ lp and {y˜i }i ∈ lq s.t. ∑i ∣x˜i ∣p = 1 and ∑i ∣y˜i ∣q = 1.
Let u = ∣x̃i ∣, v = ∣y˜i ∣. Apply Cauchy-Schwarz Inequality.
∣x̃i y˜i ∣ = ∣x̃i ∣∣y˜i ∣ ≤
∑ ∣x̃i y˜i ∣ ≤ ∑
i
i
∣x̃i ∣p ∣y˜i ∣q
+
p
q
∣x̃i ∣p
∣y˜i ∣q 1 1
+∑
= + =1
p
q
p q
i
⇒ ∑ ∣x̃i y˜i ∣ ≤ 1 ⋆
i
Page 4 of 75
Christina Battista
Let x̃i =
xi
1/p
(∑i ∣xi ∣p )
MA 515 – Analysis I
and y˜i =
yi
1/q .
(∑i ∣yi ∣q )
Course Notes
Then we have
xi yi
1/p
x̃i y˜i =
(∑i ∣xi ∣p )
1/q
(∑i ∣yi ∣q )
From ⋆ , and the above expression, we have
⎛
p⎞
∑ ∣xi yi ∣ ≤ ∑ ∣xi ∣
⎝i
⎠
i
1/p
⎛
q⎞
∑ ∣yi ∣
⎝i
⎠
1/q
Proof for Minkowski’s Inequality for Sums:
∣xi + yi ∣p = ∣xi + yi ∣∣xi + yi ∣p−1 ≤ (∣xi ∣ + ∣yi ∣)∣xi + yi ∣p−1 ≤ ∣xi ∣∣xi + yi ∣p−1 + ∣yi ∣∣xi + yi ∣p−1
p
p−1
+ ∑ ∣yi ∣∣xi + yi ∣p−1
∑ ∣xi + yi ∣ ≤ ∑ ∣xi ∣∣xi + yi ∣
i
i
i
We can then apply the Holder Inequality for Sums:
1/p
p
p
∑ ∣xi + yi ∣ ≤ ( ∑ ∣xi ∣ )
i
1/q
( ∑ ∣xi + yi ∣qp−q )
1/p
+ ( ∑ ∣yi ∣p )
∑i ∣xi + yi ∣p
⇒
i
i
i
i
1/p
( ∑i ∣xi + yi ∣p )
≤ ( ∑ ∣xi ∣p )
1/q
+ ( ∑ ∣yi ∣p )
≤ ( ∑ ∣xi ∣p )
i
1/p
1/p
+ ( ∑ ∣yi ∣p )
i
⇒ ( ∑ ∣xi + yi ∣p )
1/p
i
1/p
≤ ( ∑ ∣xi ∣p )
i
Page 5 of 75
1/p
i
i
1−1/q
⇒ ( ∑ ∣xi + yi ∣p )
i
1/q
( ∑ ∣xi + yi ∣qp−q )
1/p
+ ( ∑ ∣yi ∣p )
i
Christina Battista
MA 515 – Analysis I
Course Notes
August 24, 2011
1/p
Show that d(x, y) = ( ∑i ∣xi − yi ∣p )
is a metric on lp , ∀x, y ∈ lp . (This works for both real and complex
sequences; we only deal with real sequences in this course.)
1/p
Since x, y ∈ lp ∶ ( ∑i ∣xi ∣p )
1/p
< ∞ and ( ∑i ∣yi ∣p )
1/p
(M1): By Minkowski, ( ∑i ∣xi − yi ∣p )
< ∞.
≤ ( ∑i ∣xi ∣p )
1/p
1/p
+ ( ∑i ∣yi ∣p )
. Since both of these series are finite,
we know the original series must also be finite. It is obvious that it is also real-valued and non-negative.
(M2):
⋆If x = y ⇒ xi = yi , ∀i ⇒ d(x, y) = 0.
⋆If d(x, y) = 0 ⇒ ∣xi − yi ∣ = 0, ∀i ⇒ xi = yi , ∀i ⇒ x = y.
(M3): Obvious.
(M4): ∀x, y, z ∈ lp ∶ d(x, y) ≤ d(x, z) + d(z, y) (use Minkowski’s)
1/p
( ∑ ∣xi − yi ∣p )
i
= ( ∑ ∣xi − zi + zi − yi ∣p )
1/p
1/p
≤ ( ∑ ∣xi − zi ∣p )
+ ( ∑ ∣zi − yi ∣p )
i
i
i
Show that if x1 , x2 ∈ lp then x1 ± x2 ∈ lp , p ≥ 1.
Want to show: ∑ ∣x1i + x2i ∣p < ∞
i
( ∑ ∣x1i + x2i ∣p )
1/p
1/p
≤ ( ∑ ∣x1i ∣p )
i
1/p
+ ( ∑ ∣x2i ∣p )
<∞
i
i
For all x, y > 0, p ≥ 1
(x + y)p ≤ 2p (xp + y p )
Proof:
⋆x ≥ y ∶ (x + y)p = xp (1 + xy )p ≤ 2p xp
⋆y ≥ x ∶ (x + y)p = y p (1 + xy )p ≤ 2p y p
Putting the above statements together, we obtain: (x + y)p ≤ 2p (xp + y p ).
Jensen’s Inequality for Sums
p∗
1/p∗
( ∑ ∣xi ∣ )
i
1/p
≤ ( ∑ ∣xi ∣p )
for 0 < p < p∗
i
Prove that lp ⊂ lp , 1 < p < p∗ < ∞. This follows directly from Jensen’s Inequality.
∗
Page 6 of 75
1/p
≤ d(x, z) + d(z, y)
Christina Battista
MA 515 – Analysis I
l∞ is the space of (real) bounded sequences.
x ∈ l∞ ∶ ∣xi ∣ ≤ Cx , ∀i (the constant depends upon x.)
d(x, y) = sup ∣xi − yi ∣
i
lp ⊂ l∞ , ∀p ≥ 1
If x ∈ lp and x ∈ l∞ , then x is bounded.
∑i ∣xi ∣p < ∞ ⇒ xi → 0 as i → ∞ (by Test for Divergence) so xi is convergent and x is bounded.
Page 7 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
August 26, 2011
Maps on Metric Spaces
f ∶ (X, ρ) → (Y, σ)
What does it mean for f to be continuous at x ∈ X?
Recall: f ∶ R → R, x ∈ R ∶
∀ε > 0, ∃δ > 0 s.t. ∣x − y∣ < δ ⇒ ∣f (x) − f (y)∣ < ε
Generalizing this in terms of the Euclidean metric, d, on R we obtain:
∀ε > 0, ∃δ > 0 s.t. d(x, y) < δ ⇒ d(f (x), f (y)) < ε
In the general sense for any space with a given metric, we have that f ∶ (X, ρ) → (Y, σ) is continuous at
x ∈ X if ∀ε > 0, ∃δ > 0 s.t. ρ(x, y) < δ ⇒ σ(f (x), f (y)) < ε.
f ∶ (X, ρ) → (Y, σ) is continuous on X if f is continuous at every x ∈ X.
Example: X = any set.
ρ(x, y) = {
0
1
∶x=y
∶x≠y
Let x ∈ X. When is f ∶ (X, ρ) → (Y, σ) continuous on X?
Let ε > 0. Find δ > 0.
If δ = 1 or δ = 1/2, ρ(x, y) < 1 ⇒ x = y ⇒ f (x) = f (y) ⇒ σ(f (x), f (y)) = 0 < ε.
⇒ f is continuous on all x.
Note: Continuity does NOT depend on metric of the range.
Example: f ∶ R → R
√
Euclidean metric: ρ1 (x, y) = (x1 − y1 )2 + (x2 − y2 )2
Taxicab metric: ρ2 (x, y) = ∣x1 − y1 ∣ + ∣x2 − y2 ∣
Assume f is continuous on (R2 , ρ1 ). Show X is continuous on (R2 , ρ2 ).
Because f is continuous on (R2 , ρ1 ), we know
x ∈ X ∶ ∀ε > 0, ∃δ > 0 s.t. ρ1 (x, y) < δ ⇒ ∣f (x) − f (y)∣ < ε.
Want to show ∀ε > 0, ∃δ̃ > 0 s.t. ρ2 (x, y) < δ̃ ⇒ ∣f (x) − f (y)∣ < ε.
However, we know that ∣f (x) − f (y)∣ < ε when δ̃ = δ. Thus, we have found our δ̃.
Note: Continuity doesn’t depend on metric, it depends on open sets.
Page 8 of 75
Christina Battista
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
MA 515 – Analysis I
(X, ρ), x ∈ X, ε > 0
∅ ≠ B(x, ε) = {x′ ∈ X ∶ ρ(x, x′ ) < ε}
This is an open ball centered at x of radius ε.
(Also known as the ε−neighborhood of x.)
Closed ball: ∅ ≠ B̄(x, ε) = {x′ ∈ X ∶ ρ(x, x′ ) ≤ ε}.
Sphere: S(x, ε) = {x′ ∈ X ∶ ρ(x, x′ ) = ε} = B̄(x, ε) ∖ B(x, ε).
(The names for these all come from R3 with the Euclidean metric.)
Example: (X, ρ); X is the discrete metric space. x ∈ X ∶
⋆ B(x, 1) = {x}
⋆ B̄(x, 1) = X
⋆ S(x, 1) = X − {x}
⋆ B(x, 1/2) = {x}
⋆ B̄(x, 1/2) = {x}
⋆ S(x, 1/2) = ∅
A set U ⊂ X is open if ∀u ∈ U, ∃ε > 0 s.t. B(u, ε) ⊂ U .
A set K ⊂ X is closed if K c is open. (K c = X ∖ K)
Page 9 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
August 29, 2011
The definition of f being continuous at a point in a metric space using open balls:
∀ε > 0, ∃δ > 0 s.t. y ∈ Bρ (x, δ) ⇒ f (y) ∈ Bσ (f (x), ε).
Example: For X the discrete metric, all subsets are open.
Let U ⊂ X, u ∈ U. Choose ε = 1/2 ⇒ B(u, 1/2) = {u} ⊂ U .
Claim: Let (X, ρ) be a metric space. Let x ∈ X and ε > 0. Then B(x, ε) is open.
Proof:
B(x, ε) = {y ∈ X ∶ ρ(x, y) < ε}.
B(x0 , ε0 ) = {y ∈ X ∶ ρ(x0 , y) < ε0 }.
By the triangle inequality, we have
ρ(x, y) ≤ ρ(x, x0 ) + ρ(x0 , y) ≤ ρ(x, x0 ) + ε0 < ε.
Let ε0 = ε − ρ(x, x0 ).
THEOREM: ∅, X are both open and closed.
THEOREM: If Uα is a collection of open sets then U = ⋃α Uα is open.
Proof: Let u ∈ U . Then u ∈ Uα for some α. We know Uα is open so ∃ε > 0 s.t. B(u, ε) ⊂ Uα ⊂ U .
THEOREM: If U and V are open, then U ∩ V is open.
Proof: Let w ∈ U ∩ V ⇒ w ∈ U, w ∈ V . Because both are open,
∃ε1 > 0 s.t. B1 (w, ε1 ) ⊂ U
∃ε2 > 0 s.t. B2 (w, ε2 ) ⊂ V
Take ε = min{ε1 , ε2 } ⇒ B(w, ε) ⊂ (U ∩ V ). Therefore, U ∩ V is open.
The last two theorems are generally stated as
“Arbitrary unions of open sets are open. Finite intersections of open sets are open.”
PROPOSITION: Let (X, ρ) be a metric space. The collection of open sets for the metric ρ is a topology
for X. Hence, a metric space (X, ρ) is a topological space.
A topological space (X, T ) is a pair of a set X and T = collection of subsets of X (called topology/open
sets) satisfying:
Page 10 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
1. ∅, X ∈ T
2. If Uα ∈ T , then ⋃α Uα ∈ T
3. If U and V ∈ T , then U ∩ V ∈ T
THEOREM: [Equivalent Definition for Continuiuty]
f ∶ (X, ρ) → (Y, σ) is continuous if whenever U ⊂ Y is open (for σ), then f −1 (U ) = {x ∈ X ∶ f (x) ∈ U } is open
(for ρ). (In words, inverse images of open sets are open.)
Proof:
⇒
f is continuous (X, ρ) → (Y, σ).
Want to show that if U ⊂ Y open then f −1 (u) is open in X.
Let ũ ∈ f −1 (u) ⇒ f (ũ) ∈ U . Because U is open:
∃ε > 0 ∶ B(f (ũ), ε) ⊂ U.
f −1 (B(f (ũ), ε) ⊂ f −1 (u)
∃δ > 0 ∶ f −1 (B(f (ũ), ε)) ⊃ B(ũ, δ) (By definition of continuous)
Therefore, f −1 (u) is open.
⇐
f −1 (u) is open in X whenever U ⊂ Y .
Want to show that f is continuous on X (for each x ∈ X).
Let ε > 0. Want to find δ > 0 s.t.
Bρ (x, δ) ⊂ fσ−1 (B(f (x), ε))
B(f (x), ε) is open in Y ⇒ fσ−1 (B(f (x), ε)) is open.
⇒ x ∈ fσ−1 (B(f (x), ε)) is open at x
⇒ ∃δ s.t. Bρ (x, δ) ⊂ fσ−1 (B(f (x), ε))
If E ⊂ X, the union of all open sets U ⊂ E is the largest open set in E. It is called the interior of E. The
interior is denoted by E̊, Int(E).
The intersection of all closed sets F ⊃ E is the smallest closed set containing E. It is called the closure of
E. The closure is denoted by Ē.
E is said to be dense in X if Ē = X.
X is called separable if it has a countable dense subset.
Questions:
⋆ Let M ⊂ X be open. What is M̊ ?
⋆ Let K ⊂ X be closed. What is K̄?
⇒M
⇒K
Page 11 of 75
Christina Battista
MA 515 – Analysis I
⋆ Give an example of a dense set in R.
Course Notes
⇒Q
⋆ Show that R and C are separable.
⇒ Q ⊂ R, Q ⊂ C (both contain the subset Q which is dense.)
Sequences in a Metric (X, ρ)
We say {xn } is convergent to x ∈ X if ρ(xn , x) → 0 as n → ∞.
We write limn→∞ xn = x or xn → x, as n → ∞.
∀ε > 0, ∃N ∈ N s.t. ρ(xn , x) < ε, ∀n > N
Otherwise, {xn } is divergent.
Example: Let X = (0, 1), ρ(x, y) = ∣x − y∣, xn = 1/n.
Is xn convergent in X? ⇒ No. 0 ∉ X.
Note: limit MUST be in X for sequence to converge in X.
THEOREM: (X, ρ), E ⊂ X, and let x ∈ X. The following statements are equivalent:
1. x ∈ Ē
2. B(x, r) ∩ E ≠ ∅, ∀r > 0
3. ∃{xn } in E s.t. xn → x as n → ∞
Proof: (1)⇔(2)
(1)⇒(2)
x ∈ Ē ⇒ ∃r > 0 s.t. B(x, r) ∩ E = ∅ ⇒ E ⊂ B c (x, r) (B c (x, r) must be closed)
Because x is in closure(Ē), then x ∈ B c (x, r).
**THIS IS A CONTRADICTION** → x is NOT in the complement of B(x, r)
We have (1)⇒(2).
(2)⇒(1)
B(x, r) ∩ E ≠ ∅, ∀r > 0. Assume x ∉ Ē ⇒ x ∈ E c . (E c is open)
∃ε > 0 s.t. B(x, ε) ⊂ Ē c ⊂ E c .
⇒ B(x, ε) ∩ E = ∅ **THIS IS A CONTRADICTION**
Proof: (2)⇔(3)
(2)⇒(3)
B(x, r) ∩ E ≠ ∅, ∀r > 0. Take rn = 1/n.
B(x, 1/n) ∩ E ≠ ∅ ⇒ ∃xn ∈ [B(x, 1/n) ∩ E] ⇒ xn → x as n → ∞.
(3)⇒(2)
{xn } s.t. xn → x as n → ∞.
Assume ∃r > 0 s.t. B(x, r) ∩ E = ∅, then ρ(x, y) ≥ r, ∀y ∈ E. Then no sequence in E can converge to x.
**THIS IS A CONTRADICTION**
Page 12 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 2, 2011
If (X, ρ) is the discrete metric space, prove no proper subset of X can be dense in X.
Proof: If A ⊂ X, want to show Ā ≠ X.
Suppose x ∈ Ā ⇔ B(x, r) ∩ A ≠ ∅, ∀r > 0.
→ {x}∩A ≠ ∅ which means {x}∩A = {x} ⇒ x ∈ A ⇒ Ā = A, ∀A ⊂ X. If dense, it means that Ā = X ⇔ A = X.
So the subset of X that can be dense in X is itself. No proper subset exists that is dense in X.
Facts:
⋆ Any countable set is separable because the whole space is a countable dense subset of itself.
⋆ R is an uncountable separable space since Q̄ = R.
Q is countable and dense in R
⋆ For every n, Rn is separable since the set {(x1 , ⋯, xn ) ∈ Rn ∶ xi ∈ Q, ∀i = 1, ⋯, n} is a countable dense
subset of Rn .
⋆ The discrete metric space (X, d) is separable if and only if X is countable.
⋆ No proper subset of X can be dense in X; the only dense set in X is X (where X is the discrete metric
space).
⋆ lp , 1 ≤ p < ∞ is separable.
⋆ l∞ is NOT separable.
Page 13 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 7, 2011
THEOREM: Every convergent sequence {xn } in (X, ρ) is Cauchy.
Example: X = (0, 1] with Euclidean metric. Let {xn } = 1/n.
Is {xn } Cauchy on X?
0 ≤ ρ(xn , xm ) = ∣ n1 −
1
∣
m
≤
1
n
+
1
m
→ 0.
Is {xn } convergent on X?
0 ∉ X so {xn } is NOT convergent on X.
X is complete if every Cauchy sequence is convergent. Otherwise, X is incomplete.
THEOREM:
⋆ Any closed subset of a complete metric space is complete.
⋆ A complete subset of an arbitrary metric space is closed.
Proof: Any closed subspace of a complete metric space is complete
We have a complete metric space X, which means all Cauchy sequences in it converge.
E ⊂ X so all Cauchy sequences in E are subsets of all Cauchy sequences in X.
E = Ē (because E is closed)
Want to show E is complete.
Let {xn } be a Cauchy sequence in E. So {xn } is a sequence in X. ⇒ {xn } → x ⇒ x ∈ Ē ⇒ x ∈ E. Therefore,
{xn } is convergent in E.
Proof: A complete subset of an arbitrary metric space is closed.
(X, ρ), E ⊂ X, E is complete.
Want to show E is closed (E = Ē)
E ⊂ Ē is always true. So we just need to show containment in the other direction.
Let x ∈ Ē ⇒ ∃{xn } ∈ E s.t. xn → x, as n → ∞. Therefore, x ∈ E (because E is complete.) We have that
E = Ē and thus, E is closed.
THEOREM: f ∶ (X, ρ) → (Y, σ) is continuous at x0 ∈ X if and only if [xn → x0 implies f (xn ) → f (x0 )]
Page 14 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 9, 2011
{(x, y) ∈ R2 ∶ x2 + y 2 ≤ 1} ⊂ R2 ...is this complete?
⇒ Yes. It is a closed subset in a complete metric space so it is complete.
{(x, y) ∈ R2 ∶ x2 + y 2 < 1} ⊂ R2 ...is this complete?
⇒ No. Cauchy sequences could converge to the boundary of the space.
E ⊂ X (discrete metric space)... is this complete?
⇒ Yes. E is closed subset of X and X is complete so E is complete
R is complete (with Euclidean metric.)
Proof:
Want to show: any Cauchy sequence {xn } is convergent in R.
{xn } is Cauchy: ∀ε > 0, ∃N s.t. ∣xn − xm ∣ < ε, ∀m, n > N . Choose ε = 1.
∃N s.t. ∣xm − xN ∣ < 1, ∀m ≥ N
∣xm ∣ − ∣xN ∣ ≤ ∣xm − xN ∣ < 1 ⇒ ∣xm ∣ ≤ ∣xm − xN ∣ + ∣xN ∣, ∀m ≥ N .
We have shown xm is bounded by max{∣x1 ∣, ∣x2 ∣, ⋯, ∣xN −1 ∣, 1 + ∣xN ∣}. Because xm is bounded, ∃{xnk } subsequence that converges to a ∈ R.
∀ε > 0, ∃N2 ∈ N s.t. ∣xnk − a∣ < ε, ∀k > N2 .
Want to show: {xn } → a as n → ∞: ∀ε > 0, ∃N ∈ N s.t. ∣xn − a∣ < ε, ∀n > N .
Choose k > N2 s.t. nk > N1 . Then xn → a ⇒ every Cauchy sequence is convergent.
Rn is complete with metric d(x, y) = ( ∑ni=1 (xi − yi )2 )
1/2
l∞ is complete (in R)
Note: lp is complete (p ≥ 1), C[a, b] is complete with d(f, g) = maxt∈[a,b] ∣f (t) − g(t)∣.
Note: Q with the Euclidean metric is incomplete.
1
Note: C[0, 1] with metric d(x, y) = ∫0 ∣x(t) − y(t)∣dt is incomplete.
Page 15 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 12, 2011
Normed Spaces: Every normed space is a metric space. If you have a norm, you can construct a metric.
A vector space over a field K(R or C) is a set X ≠ ∅ whose elements are called vectors with 2 algebraic
operations:
1. Addition: X × X → X
⋆ x+y =y+x
⋆ (x + y) + z = x + (y + z)
⋆ ∃0x s.t. x + 0x = x
⋆ x + (−x) = 0x
2. Multiplication by scalar: K × X → X
⋆ α(βx) = (αβ)x
⋆ 1x = x
3. Combination of Addition and Multiplication:
⋆ α(x + y) = αx + αy
⋆ (α + β)x = αx + βx
Examples:
⋆ R is NOT a vector space over C
⋆ C is a vector space over R
⋆ R is a vector space over R
⋆ C is a vector space over C
⋆ C[a, b] is a vector space over R
⋆ Rn is a vector space over R.
⋆ lp is a vector space over R with 1 ≤ p ≤ ∞ (can show with Minkowski’s)
Y ⊂ X is a subspace if αy1 + βy2 ∈ Y, ∀α, β ∈ K, ∀y1 , y2 ∈ Y.
If Y ≠ ∅, Y ≠ X, Y is called a proper subspace. Otherwise, it is an improper subspace.
For x1 , ⋯, xm ∈ X, ∑m
i=1 αi xi is a linear combination in X, αi ∈ K.
Page 16 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
{x1 , ⋯, xm } is linearly independent if and only if
m
∑ αi xi = 0 ⇔ ∀αi = 0, ∀i.
i=1
Otherwise, {x1 , ⋯, xm } is linearly dependent.
Example: Is the set {cosx, lnx, sin2x} on C(0, ∞) linearly independent over R?
Yes. The only way for their sum to equal 0 is if all the coefficients in front of each term are identically 0.
Example: Is the set {0, lnx, ex , x3 + 9} linearly independent?
No. They are linearly dependent: 13(0) + 0(lnx) + 0(ex ) + 0(x3 + 9) = 0.
All coefficients are not identically 0 but their sum is still 0.
As soon as 0 is in the set, it becomes linearly dependent.
We say that M ⊂ X is linearly independent if every nonempty finite subset of M is linearly independent.
Otherwise, M is linearly dependent.
X is finite-dimensional if ∃n ≥ 0 s.t. {e1 , ⋯, en } ⊂ X is linearly independent but any set of (n + 1) or more
elements is linearly dependent. ⇒ dim(X) = n
If dim(X) = n, a linearly independent set {e1 , ⋯, en } is called a basis for X. Any element x ∈ X can be
constructed by a unique combination of the basis vectors.
Examples:
⋆ basis of R ∶ 1 ⇒ dim(R) = 1
⋆ basis of Rn is the canonical basis (think of columns of identity matrix) ⇒ dim(Rn ) = n
X is any vector space. B is linearly independent in X. B spans X (span(B) = X). B is called a Hamel
Basis for X. (It’s the set of all linear combinations of vectors in B.)
THEOREM: Every nonempty vector space has a Hamel basis.
Page 17 of 75
Christina Battista
MA 515 – Analysis I
Example: M = {(1, 0), (0, 7)} ⊂ R2
α(1, 0) + β(0, 7) = (α, 7β), α, β ∈ R2
⇒ span(M ) = R2
Page 18 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
September 14, 2011
Normed Spaces – Banach Spaces
A normed space X is a vector space over K(R or C) with a norm defined on it. The norm is a function
∥ ⋅ ∥ ∶ X → R that satisfies the following:
1. ∥x∥ ≥ 0, ∀x ∈ X
2. ∥x∥ = 0 ⇔ x = 0
3. ∥αx∥ = ∣α∣∥x∥, ∀α ∈ K
4. ∥x + y∥ ≤ ∥x∥ + ∥y∥, ∀x, y ∈ X (the triangle inequality)
PROPOSITION: Any norm on X defines a metric on X by d(x, y) = ∥x − y∥, i.e. every normed space is a
metric space.
Proof: (for the case K = R)
We want to show that d(x, y) = ∥x − y∥ is a metric.
1. d(x, y) ∈ [0, ∞)
This is obvious because vector spaces are closed under subtraction. x − y = z ∈ X
2. d(x, y) = 0
⇒x−y =0⇒x=y
3. d(x, y) = d(y, x)
∥x − y∥ = ∣ − 1∣∥y − x∥ = ∥y − x∥
4. d(x, y) ≤ d(x, z) + d(z, y)
∥x − y∥ = ∥x − z + z − y∥ ≤ ∥x − z∥ + ∥z − y∥ = d(x, z) + d(z, y)
Note: The converse is NOT true!! To see this, we can show that the discrete metric space on a vector space
cannot be obtained from a norm:
Let V be a vector space over R. Assume d(x, y) = ∥x−y∥ (any norm on X, the discrete metric space.) We know
∥⋅∥={
1
0
∶x≠y
∶x=y
Pick an α ∈ R, then d(αx, αy) = ∥αx − αy∥ = ∣α∣∥x − y∥ = ∣α∣d(x, y). This would have to be equivalent to the
norm defined above. Take α = 1/2.
d(αx, αy) = {
1/2
0
∶x≠y
∶x=y
**THIS IS A CONTRADICTION** Therefore, the discrete metric cannot be obtained from a norm.
Page 19 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
A Banach space is a complete normed space (complete with respect to the metric induced by the norm.)
Examples of Banach spaces:
⋆ (R, ∣ ⋅ ∣)
⋆ (Rn , ∥x∥ =
√
∑i x2i )
⋆ C
⋆ (lp , ∥x∥ = ( ∑i ∣xi ∣p )
1/p
)
⋆ (C[a, b], maxt∈[a,b] ∣x(t)∣)
⋆ (l∞ , supi ∣xi ∣)
Example of non-Banach space:
1
⋆ (C[0, 1], ∫0 ∣x(t)∣dt)
Property: Let (X, ∥ ⋅ ∥) be a normed space. Then we have the following inequality
∣∥x∥ − ∥y∥∣ ≤ ∥x − y∥
Proof:
∥x∥ = ∥x − y + y∥ ≤ ∥x − y∥ + ∥y∥
→ ∥x∥ − ∥y∥ ≤ ∥x − y∥
∥y∥ = ∥y − x + x∥ ≤ ∥x∥ + ∥y − x∥
→ ∥y∥ − ∥x∥ ≤ ∥x − y∥
⇒ ∣∥x∥ − ∥y∥∣ ≤ ∥x − y∥
PROPOSITION: The norm is a continuous function from X → R.
Proof: (use the above property)
Want to show ∀ε > 0, ∃δ s.t. d(x, y) < δ ⇒ d(∥x∥, ∥y∥) < ε.
∥x − y∥ < δ ⇒ ∣∥x∥ − ∥y∥∣ < ε
∣∥x∥ − ∥y∥∣ < ∥x − y∥ < δ = ε
Page 20 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 16, 2011
Partial order is a relation on X s.t.
1. x ≤ x, ∀x ∈ X (reflexive)
2. x ≤ y & y ≤ x ⇒ x = y, ∀x, y ∈ X (anti-symmetric)
3. x ≤ y, y ≤ z ⇒ x ≤ z (transitive)
Let M be a subset of the poset S. Then u ∈ S is an upper bound of M if m ≤ u, ∀m ∈ M .
A maximal element of a poset S is an element m ∈ S s.t. m ≤ x ⇒ m = x (there is nothing bigger than m).
A chain C ⊆ S is a totally ordered set.
Zorn’s Lemma
Let S be a nonempty poset. If every nonempty chain in S has an upper bound, then S has at least one
maximal element.
THEOREM: Every non-trivial vector space has a basis.
Let X be the set of all linearly independent subsets = c of V and order V by inclusion. Let C be a nonempty
chain in X.
Claim: ⋃z∈c z is an upper bound of C. Let a ∈ c for some c ∈ C. By construction, a ∈ ⋃z∈c x.
Let {b1 , ⋯, bn } be a finite subset of ⋃z∈c z, where each bi ∈ zi for some zi ∈ c. Then {b1 , ⋯, bn } is linearly
independent since C ⊆ X. Hence, C has an upper bound and from Zorn’s Lemma, C has a maximal element
M.
Let P = span{M }. Recall that P ⊆ V is a subspace. Suppose P ⊂ V (proper subset). Then fix 0 ≠ v ∈ V ∖ P .
So v ∈/ span{M }, which means v is linearly independent in M . Thus, M ∪ {V } is linearly independent but
M is the maximal element. **THIS IS A CONTRADICTION** Thus P = V so span{M } = V ⇒ M is a
basis for V.
Page 21 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 19, 2011
Normed Spaces – Banach Spaces (continued)
Let (X, ∥ ⋅ ∥) be a normed space.
Subspace:
1. Y ⊂ X is a called a subspace if Y is a subspace as a vector space, with the norm obtained by restricting
the norm on X to Y .
2. If (X, ∥ ⋅ ∥) is a Banach space, “Y ⊂ X is a subspace” should be understood as a subspace of the normed
space (X, ∥ ⋅ ∥). i.e. Y is NOT required to be complete.
THEOREM: Let (X, ∥ ⋅ ∥) be a Banach space. A subspace Y ⊂ X is complete if and only if Y is a closed
subspace of X.
Proof:
⇒
It is enough to show that sn is Cauchy in X because X is Banach. ∥sn − sm ∥ WLOG m < n.
n
0 ≤ ∥sn − sm ∥ = ∥ ∑ xi ∥ ≤ ∥xm+1 ∥ + ⋯ + ∥xn ∥ = ∥s˜n − s˜m ∥ → 0 as m, n → ∞
i=m+1
⇒ ∥sn − sm ∥ → 0 as m, n → ∞
Therefore, sn is Cauchy in X and convergent in X.
⇐
Want to show that X is complete.
Let {xn } be Cauchy in X and then show it converges in X. It is enough to prove that the Cauchy sequence
has a convergent subsequence.
PROPOSITION: If a Cauchy sequence {xn } has a convergent subsequence {xnk }, then it is convergent.
Proof: Since {xn } is convergent,
ε
ε
∀ > 0, ∃N1 ∈ N s.t. d(xn , xm ) < , ∀m, n > N1
2
2
ε
ε
∀ > 0, ∃N2 ∈ N s.t. d(xnk , x) < , ∀k > N2
2
2
∀ε > 0, ∃N ∈ N s.t. d(xn , x) < ε, ∀n > N
d(xn , x) ≤ d(xn , xnk ) + d(xnk , x) < ε
Fix k > N2 s.t. nk > N1 . For N1 , d(xn , x) ≤
ε
2
+
ε
2
= ε.
Sequences
Let {xn } be a sequence in (X, ∥ ⋅ ∥). Recall that the metric induced by the norm is d(x, y) = ∥x − y∥.
{xn } is convergent if ∀ε > 0, ∃N ∈ N s.t. ∥xn − x∥ < ε, ∀n > N (or ∃x ∈ X s.t. ∥xn − x∥ → 0 as n → ∞)
Page 22 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
{xn } is Cauchy is ∀ε > 0, ∃N ∈ N s.t. ∥xn − xm ∥ < ε, ∀m, n > N .
Series
∞
∑ xn = x1 + x2 + ⋯
n=1
∞
∞
n
n=1
n=1
i=1
∑ xn is convergent and the sum of the series is s = ∑ xn if the sequence of partial sums, sn = ∑ xi → s
as n → ∞.
What other types of convergence are there?
∞
∞
n=1
n=1
Absolute convergence: ∑ xn is absolutely convergent in X if ∑ ∥xn ∥ is convergent in R.
Note: In general, absolute convergence does NOT imply convergence.
THEOREM: A normed space is complete if and only if every absolutely convergent series is convergent.
Proof:
⇒
X is Banach. Let ∑∞
n=1 xn be an absolutely convergent series in X. This means that ∑ ∥xn ∥ is convergent in
R ⇒ s˜N = ∑N
n=1 ∥xn ∥ is convergent in R.
∞
Want to show sN = ∑N
n=1 xn is convergent in R which would imply ∑n=1 xn is convergent in X. It is enough
to show sN is Cauchy in X. WLOG, assume M < N .
∥sN − sM ∥ = ∥xM +1 + xM +2 + ⋯ + xN ∥ ≤ ∥xM +1 ∥ + ⋯ + ∥xN ∥ = ∣s˜N − s˜M ∣ → 0
So sN is Cauchy in X and since X is Banach, we have sN converges in X.
⇐ Want to show X is complete. Let {xn } be Cauchy in X. It is enough to prove that a Cauchy sequence
has a convergent subsequence (see proof above).
Every absolutely convergent series converges.
Let {xn } be Cauchy in X. Want to show that this sequence (or its subsequence) is convergent in X.
Because we have a Cauchy sequence: ∀ε > 0, ∃N ∈ N s.t. ∥xn − xm ∥ < ε, ∀m, n > N . Take:
1
ε1 = 1/2 ∶ ∃N1 s.t. ∥xn − xm ∥ < , ∀m, n > N
2
1
ε2 = 1/4 ∶ ∃N2 s.t. ∥xn − xm ∥ < 2 , ∀m, n > N
2
1
ε3 = 1/8 ∶ ∃N3 s.t. ∥xn − xm ∥ < 3 , ∀m, n > N
2
Take y1 = xN1 , y2 = xN2 − xN1 , y3 = xN3 − xN2 , ⋯, yi = xNi − xNi−1 . Therefore, we have xNk = ∑ki=1 yi . We want
to show this converges as k → ∞.
Claim: ∑ yi is absolutely convergent in X which means ∑ ∥yi ∥ is convergent in R.
1 j
∑ ∥yi ∥ ≤ ∥y1 ∥ + ∑( ) = ∥y1 ∥ + 1 < ∞
2
∑ yi is absolutely convergent ⇒ ∑ yi is convergent
∑ yi = x ∈ X ⇒ lim xNk = x ∈ X
k→∞
Page 23 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 21, 2011
Every incomplete metric space/normed space can be completed.
Completion of Normed Spaces/Metric Spaces
ˆ complete metric space s.t. X is isometric (≃) to W ⊂ X̂ s.t. W = X̂.
THEOREM: For every (X, d), ∃(X̂, d)
ˆ
Completion (X̂, d) is unique up to isometries.
ˆ complete normed space s.t. X is isometric (≃) to W ⊂ X̂ s.t. W = X̂.
THEOREM: For every (X, d), ∃(X̂, d)
T ∶ (X1 , d1 ) → (X2 , d2 ) is an isometry if d2 (T (x), T (y)) = d1 (x, y), ∀x, y ∈ X1 (preserves distances).
X1 and X2 are isometric if there exists a bijection isometry T ∶ X1 → X2 (X1 ≃ X2 )
Note: Isometry is ALWAYS injective.
Proof:
Let x, y ∈ X s.t. T (x) = T (y). We want to show x = y.
T (x) = T (y) ⇒ d2 (T (x), T (y)) = 0 ⇒ d1 (x, y) = 0 ⇒ x = y
Example: R is the completion of Q.
Example: X = discrete metric space is its own completion because X is complete.
Example: X = set of all continuous, real-valued functions on [a, b] with
∥x∥ = ( ∫
b
a
1/p
∣x(t)∣p dt)
, ∀x ∈ X
is NOT complete. (See [a, b] = [0, 1] with p = 2).
The completion of X is Lp [a, b]. Lp [a, b] is a Banach space, similar to lp but not with sequences. Lp deals
with functions.
Let (X, ∥ ⋅ ∥) be a normed space. We define the Schauder Basis {e1 , ⋯, en , ⋯} if
1. linearly independent
2. ∀x ∈ X, x = ∑∞
i=1 αi ei , αi ∈ R
∞
∥x − ∑ αi ei ∥ → 0 as n → ∞
i=1
Page 24 of 75
Christina Battista
MA 515 – Analysis I
Note: Schauder basis may not exist.
Example: lp , ei = {0, 0, ⋯, 1, 0, 0, ⋯} where the 1 is in the ith position.
THEOREM: If a normed space has a Schauder basis, then it is separable.
Note: The converse is NOT true.
Example: l∞ does NOT have a Schauder basis because we know it is NOT separable.
Page 25 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
September 23, 2011
Finite Dimensional Normed Spaces
THEOREM: Let X be a vector space of dimX = n. Then any proper subspace Y ⊂ X has dimY < n.
Proof:
Y ⊂ X is a proper subspace. Clearly dim(Y ) ≤ n. Assume dim(Y ) = n. Then there exists a basis for
Y ∶ {e1 , ⋯, en }. This must also be the basis for X ⇒ X = Y
**THIS IS A CONTRADICTION** So dim(Y ) < n.
THEOREM: Every finite dimensional normed space is complete. Every finite dimensional subspace Y of
a normed space is complete.
Proof:
Y is finite dimensional with dim(Y ) = n. Then there exists a basis for Y ∶ {e1 , ⋯, en }. Let {ym } be Cauchy
in Y:
∀ε > 0, ∃N ∈ N s.t. ∥ym − yr ∥ < ε, ∀m, r > N
We know that ∀ym ∶ ym = ∑n αmi ei , αmi ∈ R. This means that our new criteria for Cauchy can be written as
n
n
∀ε > 0, ∃N ∈ N s.t. ∥ ∑ αmi ei − ∑ αri ei ∥ < ε, ∀m, r > N
n
⇒ ∥ ∑(αmi − αri )ei ∥ < ε
LEMMA: Let x1 , ⋯, xn be linearly independent in normed space X. Then ∃c > 0 s.t. ∀αi , i = 1, ⋯, n we
have ∥α1 e1 + ⋯ + αn en ∥ ≥ c(∣α1 ∣ + ⋯ + ∣αn ∣)
So we know ∃c > 0 ∶ c ∑n ∣αmi − αri ∣ ≤ ∥ ∑n (αmi − αri )ei ∥.
For any i: ∣αmi − αri ∣ < ε/c, (∀m, r > N ). Therefore αmi is Cauchy in R ∶ αmi → αi as m → ∞ in R.
Let y = α1 e1 + ⋯ + αn en ∈ Y :
n
n
∥ym − y∥ = ∥ ∑(αmi − αi )ei ∥ ≤ ∣ ∑(αmi − αi )∣∥ei ∥ → 0.
Therefore, ym is convergent in Y .
COROLLARY: Any finite dimensional subspace of a normed space is closed.
Page 26 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
September 26, 2011
Finite Dimensional Normed Spaces (continued...)
THEOREM: One a finite dimensional space, ALL norms are equivalent.
∥ ⋅ ∥1 and ∥ ⋅ ∥2 are equivalent on X if ∃a, b > 0 s.t. a∥x∥1 ≤ ∥x∥2 ≤ b∥x∥1 .
Note: Equivalent norms define the same topology for X (i.e. open sets are the same!)
Proof of Theorem:
Let {e1 , ⋯, en } be a basis in X. So for all x ∈ X, x can be written as x = ∑ni αi ei . Let k = maxi ∥ei ∥1 .
n
n
n
i
i
i
∥x∥1 = ∥ ∑ αi ei ∥ ≤ ∑ ∣αi ∣∥ei ∥1 ≤ k ∑ ∣αi ∣
n
n
i
i
∥x∥2 = ∥ ∑ αi ei ∥2 ≥ c( ∑ ∣ai ∣), for some c > 0
Putting these two statements together, we obtain
∥x∥1 ≤
k
k
∥x∥2 , where > 0.
c
c
We could use the same method to show this in the other direction.
LEMMA: Let {e1 , ⋯, en } be a linearly independent set in a normed space X. Then ∃c > 0 s.t. ∀αi
(i = 1, ⋯, n), we have
∥ ∑ni αi xi ∥ ≥ c ∑ni ∣αi ∥
Proof:
Let ∑ni ∣αi ∣ = s ∈ R. If s = 0, we are done.
Assume s > 0. ⋆ ⇔ 1s ∥ ∑ni αi xi ∥ ≥ c.
⋆
n
αi
xi ∥ ≥ c
i=1 s
∥∑
Let αsi = βi .
Want to show ∃c > 0 ∶ ∥ ∑ni=1 βi xi ∥ ≥ c. Prove this by contradiction.
Note: ∑ni=1 ∣βi ∣ = 1.
(m)
Assume ∀c > 0 ∶ ∥ ∑n βi xi ∥ < c for ∑ni=1 ∣βi ∣ = 1. So ∃{y (m) } ∈ X s.t. {∥y (m) ∥ → 0 in X as m → ∞, ∑ ∣βi ∣ =
m
m
m
1, ∀m} ∶ y = β1 x1 + ⋯ + βn xn .
(m)
⇒ ∣βi ∣ ≤ 1, ∀i = 1, ⋯, n.
(m)
By Bolzano-Weierstrass on R, we have that {βi } has a convergent subsequence in R (for any fixed i).
⎧
⎪
⎪ {γ1(m) } ⊂ {β1(m) }
⎨
⎪
{y 1,m }
⎪
⎩
s.t. β˜1 is the limit of the convergent subsequence
is the corresponding subsequence of y (m)
Page 27 of 75
Christina Battista
MA 515 – Analysis I
n
(m)
y n,m = ∑ γi
n
(m)
xi and ∑ ∣γi
Course Notes
(m)
∣ = 1 with γi
n
⇒ y n,m → y = ∑ β̃i xi
Because xi are linearly independent, and ∑n ∣β̃i ∣ = 1, y ≠ 0.
***THIS IS A CONTRADICTION*** because we defined y → 0.
Page 28 of 75
→ β̃i as m → ∞
Christina Battista
MA 515 – Analysis I
Course Notes
September 30, 2011
Finite Dimensional Normed Spaces (continued...)
Recall:
⋆ Every finite-dimensional space is complete and every subspace is also complete.
⋆ Every finite-dimensional subspace of a normed space is closed.
⋆ All norms are equivalent.
⋆ Equivalent norms define the same topology for X.
Compactness
A metric space X is sequentially compact if every sequence in X has a convergent subsequence.
A subset M ⊂ X is compact considered as a subspace of X.
Is R compact?
⇒ No. Take {1, 2, ⋯} which has no convergent subsequence.
Is [0, 1] compact?
⇒ Yes.
Compact sets are “well-behaved.”
THEOREM: [Continuous Map] Let T ∶ X → Y be a continuous map between two metric spaces. Then
T (M ) is compact, whenever M is a compact subset of X.
Proof:
Let {yn } ∈ T (M ). Then ∃{xn } ∈ M s.t. yn = T (xn ), ∀n.
Because M is compact, ∃{xnk } s.t. xnk → x ∈ M as k → ∞.
Because T is continuous, T (xnk ) → T (x) ∈ T (M ) as k → ∞.
Therefore, yn has a convergent subsequence in T (M ) and T (M ) is compact.
COROLLARY: [Maximum and Minimum] Let M be a compact subset of a metric space X. T ∶ M → R be
a continuous map. Then T attains a maximum and minimum at some points of M .
LEMMA: A compact subset M of a metric space X is closed and bounded.
Page 29 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
Proof:
1. Want to show: M = M . We already know M ⊂ M . Show M ⊂ M .
Let m ∈ M ⇔ ∃{mn } ∈ M s.t. mn → m as n → ∞.
Because M is compact, {mnk } ∈ M s.t. mnk → m̃ ∈ M as k → ∞. By uniqueness of a limit, m̃ = m, m ∈ M .
∴M ⊂ M ⇒ M = M
2. Want to show M is bounded.
Proof by contradiction: Assume M is unbounded.
Let m ∈ M . Then there exists an unbounded sequence {yn } ∈ M s.t. d(yn , m) > n, ∀n.
yn cannot have a convergent subsequence. If it did, the subsequence would be bounded.
***THIS IS A CONTRADICTION*** M must be bounded.
Q: Is the converse true?
⇒ No.
Page 30 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 3, 2011
THEOREM: Let X be a finite dimensional normed space. A subset M ⊂ X is compact if and only if M is
closed and bounded.
Proof:
We just need to prove M is closed and bounded which then guarantees it is compact (because M ⊂ X, we
know M is finite dimensional.)
Let {x(m) } be a sequence in M . Let {e1 , ⋯, en } be a basis for X.
(m)
xj
n
(m)
= ∑ αi
ei , ∀m
i=1
M bounded ⇒ ∥x(m) ∥ ≤ k, ∀m.
n
(m)
k ≥ ∥x(m) ∥ = ∥ ∑ αi
n
i=1
∣ (for some c > 0) ⇒
i=1
(m)
So for any fixed i, ∣αi ∣ is bounded in R. By
(m)
(m)
subsequence {γi } ∶ γi → αi as m → ∞.
(m)
(m)
(m)
⇒ {z }= subsequence of {x
thus z ∈ M = M .
(m)
ei ∥ ≥ c ∑ ∣αi
} s.t. z
k n (m)
≥ ∑ ∣α ∣
c i=1 i
(m)
the Bolzano-Weierstrauss Theorem, {αi
} has a convergent
→ ∑ni=1 αi ei = z. Because M is closed, we know M = M and
Q: What are the compact subsets of R?
⇒ Closed intervals.
THEOREM: If the closed unit ball M = {x ∈ X∣∥x∥ ≤ 1} is compact in a normed space X, then X is finite
dimensional.
The proof is based on Riesz’s Lemma.
Riesz’s Lemma: Let Y and Z be subspaces of a normed space X. Assume that Y is closed and is a proper
subset of Z. Then ∀θ ∈ (0, 1), ∃z ∈ Z s.t.
∥z∥ = 1,
∥z − y∥ ≥ θ,
∀y ∈ Y
Proof of Riesz’s Lemma:
Let v ∈ Z ∖ Y . Let a = inf y∈Y ∥v − y∥ > 0 (this is true because Y is closed.) Let θ ∈ (0, 1). By definition of an
v−y0
infimum, ∃y0 ∈ Y s.t. a ≤ ∥v − y0 ∥ ≤ aθ . Let z = ∥v−y
∈ Z and ∥z∥ = 1.
0∥
v − y0
1
1
− y∥ =
∥v − y0 − y∥v − y0 ∥∥ =
∥v − (y0 + y∥v − y0 ∥)∥
∥v − y0 ∥
∥v − y0 ∥
∥v − y0 ∥
where y0 + y∥v − y0 ∥ ∈ Y .
a
a
≥
⇒ ∥z − y∥ ≥
≥ θ.
∥v − y0 ∥
∥v − y0 ∥
∥z − y∥ = ∥
Proof of Theorem:
Assume dimX = ∞. Choose x1 ∈ X with ∥x1 ∥ = 1.
Page 31 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
X1 = span{x1 } ⊂ X ⇒ dimX1 = 1 ⇒ X1 is a closed and complete proper subset of X.
Using Riesz’s Lemma, ∃x2 ∈ X with ∥x2 ∥ = 1 s.t. ∥x2 − x1 ∥ ≥ θ. Choose θ = 1/2.
∥x2 − x1 ∥ ≥
1
2
X2 = span{x1 , x2 } ⊂ X ⇒ dimX2 = 2 ⇒ X2 is a closed and complete proper subset of X.
⇒ ∃x3 ∈ X with ∥x3 ∥ = 1 s.t. ∥x3 − x1 ∥ ≥ 1/2 and ∥x3 − x2 ∥ ≥ 1/2.
1
2
1
∥x2 − x2 ∥ ≥
2
∥x3 − x1 ∥ ≥
Continue this process...
{xn } ∈ X and {xn } ∈ M s.t. ∥xn − xm ∥ ≥ 1/2, m ≠ n. But this means M is NOT compact. ***THIS IS A
CONTRADICTION***
We know M is compact ⇒ dimX ≠ ∞ ⇒ X is finite dimensional.
Page 32 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 5, 2011
Linear Operators
Let X and Y be vector spaces over K (where K is either R or C).
A ∶ D(A) ⊂ X → R(A) ⊂ Y
where D(A) is the domain of A and R(A) is the range of A.
Function A is a linear operator if D(A) is a subspace of X and
A(x1 + x2 ) = A(x1 ) + A(x2 )
and
A(αx) = αA(x)
for all x1 , x2 ∈ D(A) where α ∈ K.
Note: A(x) = Ax (just a difference in notation, but they mean the same thing)
⋆ A( ∑ni=1 xi ) = ∑ni=1 A(xi )
⋆ A(0) = 0
⋆ R(A) is a subspace of Y
Nullspace (kernel)
N (A) = {x ∈ D(A) ∶ Ax = 0} ≠ ∅
0 ∈ N (A) ALWAYS!!
N (A) is a subspace
Inverse (A−1 )
A ∶ D(A) → R(A) is automatically surjective. A is invertible if A is also injective
(i.e. if A(x1 ) = A(x2 ) ⇔ x1 = x2 )
THEOREM: Let A ∶ D(A) ⊂ X → R(A) ⊂ Y . Then A−1 exists if and only if N (A) = {0}. (When A−1
exists, it is also a linear operator.)
Proof:
⇒ A−1 exists (so we know A is bijective).
Assume N (A) = {0, x}.
A(0) = A(x) ⇒ x = 0
⇒ N (A) = {0}
Page 33 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
⇐ N (A) = {0}
Suppose A(x1 ) = A(x2 ) ⇒ A(x1 − x2 ) = 0 ⇒ x1 − x2 = 0 ⇒ x1 = x2 .
Thus, A is bijective⇒ A−1 exists.
Show A−1 is a linear operator.
A−1 ∶ R(A) → D(A)
By A being bijective, surjectivity and injectivity follow.
Let y1 , y2 ∈ R(A) ∶ x1 , x2 ∈ D(A) ∶ Ax1 = y1 , Ax2 = y2 .
A−1 (y1 + y2 ) = A−1 (Ax1 + Ax2 ) = A−1 (A(x1 + x2 )) = x1 + x2 = A−1 y1 + A−1 y2
A−1 (αy1 ) = A−1 (A(αx1 )) = αx1 = αA−1 y1
Thus, A−1 is a linear operator.
THEOREM: Let A ∶ D(A) = X → R(A) ⊂ Y be injective. Then R(A) is of the same (possibly infinite)
dimension as X.
Proof:
Take {x1 , ⋯, xn } be linearly independent in X (dimX = n).
n
∑ αi xi = 0 ⇒ αi = 0, ∀i = 1, ⋯, n
i=1
{Ax1 , Ax2 , ⋯, Axn } ∈ R(A) (WTS linearly independent)
n
n
i=1
i=1
n
∑ αAxi = 0 ⇒ A( ∑ αi xi ) = 0 + (A is injective) ⇒ N (A) = {0} ⇒ ∑ αi xi = 0 ⇒ αi = 0, ∀i
i=1
Now we have dimR(A) ≥ dimX. Do the same procedure with A
−1
to obtain dimX ≥ dimR(A).
⇒ dimR(A) = dimX
If A ∶ D(A) = X → R(A) = Y is injective, then A is called an isomorphism of X into Y .
→X ≃Y
→ dimX = dimY
Page 34 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 10, 2011
Applications of Linear Operators
A ∶ D(A) ⊂ X → R(A) ⊂ Y
Ax = y
Existence: For which y, ∃x s.t. Ax = y? (What is R(A)?)
Uniqueness: Is x unique s.t. Ax = y, ∀y ∈ R(A)? (Does A−1 exist? Is N (A) = {0}?)
Bounded Linear Operators
Let A ∶ D(A) = X → Y be a linear operator. We say that A is bounded if ∃c > 0 s.t. ∥Ax∥Y ≤ c∥x∥X , ∀x ∈ X.
Smallest possible c = sup
0≠x∈X
∥Ax∥Y
∥x∥X
Example: I ∶ X → X is the identity operator. (Ix = x)
⇒ Bounded; c = 1
Example: Differentiation operator: A ∶ C[0, 1] → C[0, 1] s.t. Ax(t) = x′ (t).
Consider xn (t) = tn , ∀n:
∥xn (t)∥ = max ∣tn ∣ = 1
t∈[0,1]
∥x′n (t)∥
= max ∣ntn−1 ∣ = n
t∈[0,1]
⇒ Unbounded.
1
Example: Integral operator: A ∶ C[0, 1] → C[0, 1] s.t. Ax = y = ∫0 k(t, τ )x(τ )dτ where k(t, τ ) is the kernel
of a given function.
∥Ax∥ = max ∫
t∈[0,1]
1
0
k0 ∥x(t)∥dτ = k0 ∥x(t)∥(1 − 0) = k0 ∥x(t)∥
⇒ Bounded; c = k0
THEOREM: X is a normed space with dimX = n. Then every linear operator on X is bounded.
Page 35 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
Proof:
X has a basis {e1 , ⋯, en }. For any x ∈ X, we can write it as x = ∑ni=1 αi ei
n
n
n
i=1
i=1
i=1
n
∥Ax∥ = ∥A ∑ αi ei ∥ = ∥ ∑ αi Aei ∥ ≤ ∑ ∣αi ∣∥Aei ∥ ≤ max ∥Aei ∥ ∑ ∣αi ∣
≤ max ∥Aei ∥
i=1,⋯,n
i=1,⋯,n
∥x∥
∥ ∑ni=1 αi ei ∥
= max ∥Aei ∥
i=1,⋯,n
c
c
⇒ A is bounded.
B(X, Y ) is the set of all bounded linear operators from X to Y .
⋆ B(X, Y ) is a vector space
(T1 + T2 )(x) = T1 x + T2 x
(αT )x = α(T x)
⋆ B(X, Y ) is a normed space.
∥A∥ = sup
0≠x∈X
∥Ax∥Y
= sup ∥Ax∥Y
∥x∥X
x∈X,∥x∥=1
THEOREM: If Y is Banach, then B(X, Y ) is Banach.
Page 36 of 75
i=1
Christina Battista
MA 515 – Analysis I
Course Notes
October 12, 2011
Bounded Linear Operators (continued...)
A∶X →Y
where X, Y are normed spaces.
A is continuous at x0 ∈ X if ∀ε > 0, ∃δ > 0 s.t. ∥Ax − Ax0 ∥Y < ε whenever ∥x − x0 ∥X < δ.
THEOREM: A is continuous on X (or D(A)) if and only if A is bounded on X (or D(A)).
Proof:
⇒ A is continuous on X. Let x0 ∈ X s.t. ∀ε > 0, ∃δ > 0 s.t. ∥Ax − Ax0 ∥ < ε ⇒ ∥x − x0 ∥ < δ. Rewrite x as
x = x0 +
∥A(x − x0 )∥ < ε ⇒ ∥A(x − x0 )∥ = ∥A(
δ
y,
∥y∥
y∈X
δ
ε
δ
y)∥ =
∥Ay∥ < ε ⇒ ∥Ay∥ < ∥y∥ = c∥y∥
∥y∥
∥y∥
δ
⇒ A is bounded.
⇐ A is bounded. (∥Ax∥ ≤ c∥x∥)
Let x0 ∈ X. Let ε > 0.
Choose δ <
ε
c
∥A(x − x0 )∥ ≤ c∥x − x0 ∥
∥A(x − x0 )∥ ≤ c∥x − x0 ∥ < ε
⇒ A is continuous.
THEOREM: If A is continuous at one point in D(A), then A is continuous on D(A).
Proof: (follows directly from previous theorem’s proof)
COROLLARY: A is a bounded linear operator X → Y
⋆ xn → x in X then Axn → Ax in Y
⋆ N (A) is closed
Proof:
We know N (A) ⊂ N (A) is always true. Must show containment in the other direction.
Let x ∈ N (A) ⇔ ∃{xn } ∈ N (A) s.t. xn → x ⇒ Axn → Ax
⇒ Axn = 0 because xn ∈ N (A)
⇒ Axn → Ax = 0 ⇒ x ∈ N (A)
Page 37 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
Note: R(A) is NOT necessarily closed.
Linear Functionals
Special case of linear operators s.t. X → R (or C).
All definitions and theorems from linear operators apply here.
Notation: f, g, h (lowercase letters)
∥f ∥ = sup
0≠x∈X
∣f x∣
= sup ∣f (x)∣
∥x∥ x∈X,∥x∥=1
Example: ∥ ⋅ ∥ ∶ X → R is a functional but NOT linear.
∥a + b∥ ≠ ∥a∥ + ∥b∥
Example: Integral operator f ∶ C[a, b] → R
f (x) = ∫
b
x(t)dt is a functional AND linear
a
Q: Is this linear functional bounded?
∥f (x)∥ = ∣ ∫
b
a
x(t)dt∣ ≤ ∫
b
a
∣x(t)∣dt ≤ max ∣x(t)∣ ∫
t∈[a,b]
b
a
dt = max ∣x(t)∣(b − a) = ∥x∥C[a,b] (b − a)
t∈[a,b]
⇒ This is bounded; ∥f ∥ = (b − a)
Page 38 of 75
Christina Battista
MA 515 – Analysis I
October 14, 2011
Linear Functionals (continued...)
Example: Dot product: f ∶ R2 → R defined by f (x1 , x2 ) = x1 a + x2 a2 for a fixed (a1 , a2 ) ∈ R2 .
Questions:
⋆ Is f linear?
⇒ Yes.
⋆ Is f bounded?
∣a1 x1 + a2 x2 ∣ ≤ ∣a1 x1 ∣ + ∣a2 x2 ∣ ≤ max{∣a1 ∣, ∣a2 ∣}(∣x1 ∣ + ∣x2 ∣) taxicab metric
⇒ Yes.
⋆ Find ∥f ∥.
∥f ∥ ≤ ∥a∥ and ∥f ∥ ≥
∣f (a)∣ ∥a∥2
=
= ∥a∥
∥a∥
∥a∥
⇒ ∥f ∥ = ∥a∥
Dual Spaces
X ∗ (or X f ): set of all linear functionals on X. It is called the algebraic dual space of X.
X ′ : set of all bounded linear functionals on X. It is called the dual space of X.
Clearly we have X ′ ⊂ X ∗ .
Question: Can you add an assumption on X that will guarantee that X ′ = X ∗ ?
If X is finite dimensional, we have that X ′ = X ∗ .
Properties of Dual Spaces:
⋆ X ∗ and X ′ are vector spaces with the following operations:
(f1 + f2 )(x) = f1 (x) + f2 (x)
(αf )(x) = αf (x)
⋆ X ∗ and X ′ are normed spaces with the norm defined as before:
∥f ∥ = sup ∣f (x)∣
∥x∥=1
⋆ X ′ is Banach.
Page 39 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
THEOREM: If Y is Banach, then B(X, Y ) is Banach.
We can go one step further and find the dual of the dual:
X ∗∗ = (X ∗ )∗ : set of all linear functionals on X ∗ . It is called the 2nd algebraic dual space of X.
One way to obtain a g ∈ X ∗∗ is by choosing a fixed x ∈ X and setting
g(f ) = gx (f ) = f (x) where x is fixed in X and f is the variable in X ∗ .
So we have a map C ∶ X → X ∗∗ defined by C(x) = gx . This is called the canonical map or the embedding
map.
Let’s explain the “embedding map.”
Step 1: Show that C is linear.
C(x) = gx ;
C ∶ X → X ∗∗
g(f ) = gx (f ) = f (x) where f is linear.
C(αx + y) = gαx+y (f ) = f (αx + y) = αf (x) + f (y) = αgx + gy = αC(x) + C(y)
Now recall the definition of an isomorphism between two vector spaces:
T ∶ X → Y is an isomorphism if it is linear (thus preserves the operations) and bijective.
For two metric spaces X and Y , isomorphism T ∶ X → Y means that T is a bijective isometry (i.e. it
preserves distance).
For two normed spaces X and Y , isomorphism T ∶ X → Y means that T is linear, bijective, and preserves
norms (i.e. ∥T x∥Y = ∥x∥X ).
Step 2: C is injective, which means that X is isomorphic to R(C) ⊂ X ∗∗ as vector spaces. (R(C) is a
subspace of X ∗∗ since C is linear.) That is why we say that X embeds in X ∗∗ .
Explain why C is injective.
Let x ≠ 0, x ∈ X. Then ∃f ∈ X ∗ s.t. f (x) ≠ 0. Then C(x) = g(f ) = gx (f ) = f (x) ≠ 0. So x ∈/ N (C) ⇒ C is
injective.
Step 3: If C is also surjective, i.e. R(C) = X ∗∗ , then X is isomorphic to X ∗∗ as vector spaces and we say
that X is algebraically reflexive.
Page 40 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 17, 2011
LEMMA: If dimX = n < ∞, x0 ∈ X s.t. f (x0 ) = 0, ∀f ∈ X ∗ then x0 = 0. (This is the same as saying
C ∶ X → X ∗∗ C(x0 ) = 0 ⇒ x0 = 0).
Proof:
Let {e1 , ⋯, en } be a basis of X. For any x0 ∈ X, we can write it x0 = ∑ni=1 αi ei .
n
n
i=1
i=1
f (x0 ) = f (∑ αi ei ) = ∑ αi f (ei ) = 0, ∀f ∈ X ∗
n
⇒ ∑ αi βi = 0, for any βi , i = 1, ⋯, n
i=1
⇒ αi = 0, ∀i = 1, ⋯, n
⇒ x0 = 0
This says that C ∶ X → R(C) ⊂ X ∗∗ is injective (for finite dimensional X).
Linear Operators and Functionals in Finite Dimensional Spaces
A ∶ X → Y is linear with dimX = n and dimY = m. Let E = {e1 , ⋯, en } be a basis for X and F = {f1 , ⋯, fm }
be a basis for Y .
Note: A can be represented by a m × n matrix.
n
x ∈ X ⇒ x = ∑ αi ei
i=1
n
n
i=1
i=1
Ax = A ∑ αi ei = ∑ αi Aei where Aei ∈ Y
A is entirely determined by the values of A at the basis vectors.
For each i, Aei = ∑m
j=1 βji fj
⎡e ⎤
⎡f ⎤
⎢ 1⎥
⎢ 1⎥
⎢ ⎥
⎢ ⎥
e=⎢ ⋮ ⎥
f =⎢ ⋮ ⎥
⎢ ⎥
⎢ ⎥
⎢en ⎥
⎢fm ⎥
⎣ ⎦
⎣ ⎦
MEF = matrix(βji ) which represents our linear operator
T
Ae = MEF
f
m
m
n
m
m
n
j=1
j=1
i=1
j=1
j=1
i=1
Ax = ∑ γj fj ⇒ ∑ γj fj = ∑ αi ∑ βji fj = ∑ ( ∑ βji αi )fj
For each j, γj =
m
∑i=1 βji αi .
⇒ γ = MEF α
Note: Matrix representation is NOT unique–depends on the basis you are working with.
Page 41 of 75
Christina Battista
Example: T ∶ R3 → R3
⋆ Find matrix M .
MA 515 – Analysis I
T (x1 , x2 , x3 ) = (x1 , x2 , −x1 − x2 )
⎡1
⎢
⎢
M = ⎢0
⎢
⎢0
⎣
0
1
0
T
⋆ Find R(T ).
Course Notes
⎡1
−1⎤⎥
⎢
⎥
⎢
−1⎥ ⇒ M = ⎢ 0
⎥
⎢
⎢−1
0 ⎥⎦
⎣
0 0⎤⎥
⎥
1 0⎥
⎥
−1 0⎥⎦
R(T ) = span{α, β, −α − β} for α, β ∈ R
⋆ Find N (T ).
N (T ) = span{0, 0, 1}
Linear Functionals
f ∶ X → R where f ∈ X ∗ and dimX = n with basis {e1 , ⋯, en }.
n
x ∈ X ⇒ x = ∑ αi ei
i=1
n
n
i=1
i=1
f (x) = f (∑ αi ei ) = ∑ αi f (ei )
where the f (ei )’s are arbitrary and do NOT depend on x. Then f is uniquely defined by its values at the
basis vectors of X.
Given any scalars β1 , ⋯, βn , ∃!f ∈ X ∗ s.t. f (ei ) = βi , ∀i = 1, ⋯, n. Then f is given by f (x) = ∑ni=1 αi βi where
x = ∑ni=1 αi ei .
Special case:
Choice 1: Take β1 , ⋯, βn = 1, 0, ⋯, 0.
f1 (e1 ) = 1
f1 (ei ) = 0, ∀i ≠ 1
Choice 2: Take β1 , β2 , ⋯, βn = 0, 1, 0, ⋯, 0.
f2 (e2 ) = 1
f2 (ei ) = 0, ∀i ≠ 2
⋮
Choice n: Take β1 , β2 , ⋯, βn = 0, 0, ⋯1.
fn (en ) = 1
fn (ei ) = 0, ∀i ≠ n
Through the above method, we have defined the Kronecker Delta function.
δij = {
fj (ei ) = 1
fj (ei ) = 0
Page 42 of 75
i=j
i≠j
Christina Battista
MA 515 – Analysis I
Course Notes
CLAIM: fj , j = 1, ⋯, n is the dual basis in X ∗ . (dimX ∗ = n)
Note: dimX = dimX ∗ in finite dimensional space and dimX ∗ = dimX ′ in finite dimensional space as well.
THEOREM: {f1 , ⋯, fn } are a basis for X ∗ . Thus dimX = dimX ∗ = n.
Proof:
1. Show f1 , ⋯, fn are linearly independent.
Let β1 f1 + ⋯ + βn fn = 0, ∀x ∈ X, ej ∈ X.
β1 f1 (e1 ) + ⋯ + βj fj (ej ) + ⋯ + βn fn (en ) = 0
⇒ βj fj (ej ) = 0 ⇒ βj (1) = 0 ⇒ βj = 0
We can do thus for any j = 1, ⋯, n. Therefore, fi , i = 1, ⋯, n are linearly independent.
2. Show that for all f ∈ X ∗ , there exists scalars ωk s.t. f = ∑nk=1 ωk fk .
n
n
f (x) = ∑ ωk fk (x) ⇒ ∑ αi f (ei ) = ∑ ωk fk (x)
i=1
k=1
k=1
n
n
i=1
i=1
fi (x) = fi (∑ αi ei ) = ∑ αi fi (ei ) = αi
n
n
⇒ ∑ fi (x)f (ei ) ⇒ f (ei ) = ωi ⇒ ∑ ωi fi (x)
i=1
i=1
n
⇒ ∀f ∈ X ∗ , f = ∑ ωk fk (x)
k=1
Question: {e1 , e2 , e3 } in R3 . What is the dual basis?
f1 = [1
0
0]
f2 = [0
1
0]
f3 = [0
0
1]
Page 43 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 21, 2011
Examples of Dual Spaces
⋆ The dual space of Rn is Rn (i.e. (Rn )∗ is isomorphic to Rn .)
⋆ The dual space of l1 is l∞ .
⋆ The dual space of lp is lq where 1 < p < ∞ and
1
p
+
1
q
= 1.
Proof: ((Rn )∗ ≃ Rn )
ÐÐÐ→
Let T ∶ (Rn )∗ → Rn defined by T f = f (ei ).
We have to show that T is an isomorphism (linear, bijective, and norm-preserving).
1. Linear
T (αf + g) = (αf + g)(Ð
e→i ) = αf (Ð
e→i ) + g(Ð
e→i ) = αT f + T g
2. Bijective
n
n
n
i=1
i=1
i=1
T f = T g ⇒ f (x) = g(x), ∀x ∈ X; x = ∑ αi ei ⇒ ∑ αi f ei = ∑ αi gei
⇒ injective.
∀βi , i = 1, ⋯, n, βi ∈ Rn
Choose f s.t. f (ei ) = βi .
n
f (x) = ∑ αi f (ei )
i=1
where x =
n
∑i=1 αi ei .
⇒ surjective.
3. Norm-preserving f ∈ (Rn )∗
n
f (x) = ∑ αi f (ei ), f (ei ) = γi
i=1
n
n
i=1
i=1
⇒ ∣f (x)∣ ≤ ∑ ∣αi ∣∣f (ei )∣ ≤ ( ∑ ∣αi ∣2 )
1/2
n
( ∑ ∣f (ei )∣2 )
1/2
i=1
n
= ∥x∥( ∑ ∣f (ei )∣2 )
i=1
(by Cauchy-Schwarz inequality)
Take the supremum over ∥x∥ = 1:
n
∥f ∥ ≤ ( ∑ ∣f (ei )∣2 )
1/2
= ∥f (Ð
e→i )∥
i=1
n
Now recall ∥f ∥ = sup
x≠0
∣f (x)∣
= sup
∥x∥
x≠0
∣ ∑ αi f (ei )∣
i=1
n
.
(∑ ∣αi ∣2 )1/2
i=1
Choose αi = f (ei ):
∥f ∥ ≥
n
n
∑i=1 f (ei )2
=
(
f (ei )2 )1/2 = ∥f (Ð
e→i )∥
∑
n
(∑i=1 f (ei )2 )1/2
i=1
Page 44 of 75
1/2
Christina Battista
MA 515 – Analysis I
Course Notes
⇒ ∥f ∥ = ∥f (Ð
e→i )∥
THEOREM: A finite dimensional vector space X is algebraically reflexive, i.e. X ≃ X ∗∗ .
Proof:
Let {e1 , ⋯, en } be a basis for X.
C ∶ X → X ∗∗
We know this is linear and we know this is injective if X is finite dimensional.
We know C ∶ X → R(C) is bijective (an isomorphism).
⇒ dimX = dimR(C)
We also know dimX = dimX ∗ in finite dimensional space, which also means dimX ∗ = dimX ∗∗ in finite
dimensional space.
⇒ dimX = dimX ∗∗ in finite dimensional space
R(C) ⊂ X ∗∗ but because they have the same dimension, R(C) = X ∗∗ ⇒ X ≃ X ∗∗
Page 45 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 24, 2011
Direct Sums
Let M, N ⊂ X be subspaces of X. The sum M + N is defined as
M + N = {m + n∣m ∈ M, n ∈ N } − − the smallest subspace in X containing M and N
When M ∩ N = {0}, we write M ⊕ N and we call it the direct sum of M and N .
THEOREM 1: Let M, N ⊂ X be subspaces of X. Then X = M ⊕ N if and only if each x ∈ X can be
written in the form x = m + n, with m ∈ M and n ∈ N , in one and only one way.
When X = M ⊕ N , we say that M and N are complementary subspaces (or that N is the complement
of M in X.
Remark: Any subspace M ⊂ X has at least one complement in X.
THEOREM 2: If X = M ⊕ N , then dimX = dimM + dimN .
Proof:
Case 1: If any one of M or N have dim = ∞, then dimX = ∞.
Case 2: Now assume dimM = m, dimN = n.
Claim: {e1 , ⋯, em , f1 , ⋯, fn } is a basis for X.
1. Show our claimed basis is linearly independent.
c1 e1 + ⋯ + cm em + cm+1 f1 + ⋯ + cm+n fn = 0
⇒ c1 e1 + ⋯ + cm em = −(cm+1 f1 + ⋯ + cm+n fn )
where c1 e1 + ⋯ + cm em ∈ M and cm+1 f1 + ⋯ + cm+n fn ∈ N . Because M ∩ N = {0}, each side must be zero.
c1 e1 + ⋯ + cm em = 0 ⇒ ci = 0, ∀i = 1, ⋯, m
cm+1 f1 + ⋯ + cm+n fn = 0 ⇒ cm+j = 0, ∀j = 1, ⋯, n
⇒ this basis is linearly independent!
2. Show that our proposed basis spans X.
x ∈ X ⇒ x = m + n if X = M ⊕ N (by Theorem 1)
m
n
i=1
j=1
x = m1 + n1 = ∑ ci ei + ∑ cm+j fj
Therefore, it spans X.
Page 46 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
THEOREM 3: Let X and Y be two vector spaces and let A ∶ X → Y be linear. If X = W ⊕ N (A), then
A∣W ∶ W → R(A) is an isomorphism.
Proof:
1. It is obvious that A∣W is linear.
2. Bijectivity:
⋆ Show injectivity:
Let w1 ∈ W. Assume A∣W = {0, w1 }. Because N (A) ∩ W = {0}, w1 = 0. Therefore, A∣W = {0} and thus
A∣W is injective. We now know that A∣W ∶ W → R(A∣W ) is an isomorphism.
⋆ Show R(A) = R(A∣W ):
We know R(A∣W ) ⊂ R(A).
Let y ∈ R(A). Then y = Ax for some x ∈ X. ⇒ x = w1 + n1 for some w1 ∈ W, n1 ∈ N (A). Then
y = Aw1 + An1 = Aw1 ⇒ y ∈ R(A∣W ) ⇒ R(A) = R(A∣W ).
THEOREM 4: Let X and Y be two vector spaces and let A ∶ X → Y be linear. Then
dim(X) = dim(N (A)) + dim(R(A)).
Proof:
From Theorem 2, dimX = dimW + dimN (A). By Theorem 3, dimW = dimR(A) (because W ≃ R(A).)
⇒ dimX = dimR(A) + dimN (A)
Quotient Spaces
Let M ⊂ X be a subspace of a vector space X.
We say that two elements x1 and x2 are equivalent modulo M if x1 − x2 ∈ M .
”Equivalence modulo M” has the usual properties of an equivalence relation:
⋆ reflexivity: x ∼ x
x − x ∈ M ⇒ 0 ∈ M (because M is a subspace)
⋆ symmetric: x ∼ y ⇔ y ∼ x
x1 − x2 ∈ M ⇒ x2 − x1 ∈ M (because M is a subspace)
⋆ transitivity: x ∼ y, y ∼ z → x ∼ z
x1 − x2 ∈ M, x2 − x3 ∈ M ⇒ x1 − x2 + x2 − x3 = x1 − x3 ∈ M
[x] denotes the equivalence class that contains x, i.e.
[x] = {u ∈ X∣x − u ∈ M }
where x ∈ X. Then [x] = [u] if and only if x − u ∈ M .
X/M is the set of all these equivalence classes. It is called the quotient space of X modulo M .
X/M is a vector space with addition and multiplication by scalars defined as follows:
Page 47 of 75
Christina Battista
MA 515 – Analysis I
⋆ [x + y] = [x] + [y]
⋆ [αx] = α[x]
Q: What is the zero element [0] of X/M ?
⇒ M.
[0] = {u ∈ X∣u − 0 ∈ M } = M
Page 48 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
October 28, 2011
Quotient Spaces
Example: X = R3 , M = {(x, 0, 0)∣x ∈ R} (the x−axis)
⋆ X/M = R2
⋆ X/X = {0}
⋆ X/{0} = X
THEOREM 1: Let M be a subspace of X, and let N be any complement of M . Then X/M is isomorphic
to N .
Proof:
X =M ⊕N
φ ∶ X → X/M
φ ∶ x ↦ [x]
φ is linear: αx + y ↦ [αx + y] = α[x] + [y] = αφ(x) + φ(y)
φ is surjective: x ↦ [x], ∀x ∈ X (all equivalence classes) R(φ) = X/M .
φ is injective: By Theorem 3 of direct sums, N (φ) ∶ φ(x) = [0] = M ⇒ N (φ) = M .
Let W = N . Then X = M ⊕ N where M = N (φ).
φ∣N ∶ N → R(φ) = X/M
is an isomorphism by Theorem 3 of direct sums and therefore N ≃ X/M .
THEOREM: Let A be a linear operator between two vector spaces X and Y . Then R(A) is isomorphic
to X/N (A).
Proof:
A∶X →Y
X = X/N (A) + N (A) ⇒ X = W + N (A)
A∣X/N (A) ∶ X/N (A) → R(A) is an isomorphism.
R(A) ≃ W where W ⊕ N (A) = X.
By Theorem 1, X/N (A) ≃ W where M = N (A).
⇒ R(A) ≃ X/N (A)
Note: dimX/M = dimX − dimM. for finite-dimensional X.
Page 49 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
Inner Product Spaces
X is an inner product space if X is a vector space with inner product ⟨⋅, ⋅⟩ ∶ X × X → K (where K is R or
C) s.t.
⋆ ⟨x, x⟩ ≥ 0 with equality if x = 0
⋆ ⟨x + y, z⟩ = ⟨x, z⟩ + ⟨y, z⟩
⋆ ⟨αx, y⟩ = α⟨x, y⟩
⋆ ⟨x, y⟩ = ⟨y, x⟩
If an inner product space is complete, it is called a Hilbert space.
Immediate consequences:
⋆ ⟨x, y + z⟩ = ⟨x, y⟩ + ⟨x, z⟩
⟨x, y + z⟩ = ⟨y + z, x⟩ = ⟨y, x⟩ + ⟨z, y⟩ = ⟨x, y⟩ + ⟨x, z⟩
⋆ ⟨x, αy⟩ = α⟨x, y⟩
⟨x, αy⟩ = ⟨αy, x⟩ = α⟨y, x⟩ = α⟨x, y⟩
⋆ ⟨x, 0⟩ = ⟨0, x⟩ = 0 (scalar α = 0)
Page 50 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
October 31, 2011
Inner Product Spaces. Hilbert Spaces
Examples:
⋆ Cn ,
⟨(a1 , a2 , ⋯, an ), (b1 , b2 , ⋯, bn )⟩ = ∑ni=1 ai bi
b
⟨f, g⟩ = ∫a f (t)g(t)dt
⋆ L2 [a, b] (the completion of C[a, b]),
Q: Why is ⟨f, g⟩ = ⟨g, f ⟩?
⟨g, f ⟩ = ∫
b
a
g(t)f (t)dt = ∫
b
a
g(t)f (t)dt = ∫
b
a
g(t)f (t)dt = ⟨f, g⟩
Let y, z ∈ X. Suppose ⟨x, y⟩ = ⟨x, z⟩, ∀x ∈ X. Then y = z.
Proof:
⟨x, y⟩ = ⟨x, z⟩
⟨x, y⟩ − ⟨x, z⟩ = 0
⟨x, y − z⟩ = 0, ∀x ∈ X
Choose x = y − z.
⟨y − z, y − z⟩ = 0 ⇒ y − z = 0 ⇒ y = z
Fix y ∈ X. The mapping f ∶ X → C defined by f (x) = ⟨x, y⟩ is a linear functional on X.
THEOREM: If X is an inner product space, then ∥x∥ = ⟨x, x⟩1/2 is a norm on X.
Proof:
⋆ ∥x∥ = ⟨x, x⟩1/2 ≥ 0 ⇔ ∥x∥ = 0 ⇒ 0
⋆ ∥αx∥ = ⟨αx, αx⟩1/2 = (αα⟨x, x⟩)1/2 = (∣α∣2 ⟨x, x⟩)1/2 = ∣α∣∥x∥
⋆ ∥x + y∥2 = ⟨x + y, x + y⟩ = ∥x∥2 + ∥y∥2 + ⟨x, y⟩ + ⟨y, x⟩ = ∥x∥2 + ∥y∥2 + 2Re(⟨x, y⟩)
Need to show 2Re(⟨x, y⟩) ≤ 2∥x∥∥y∥ = 2⟨x, x⟩1/2 ⟨y, y⟩1/2
– Case 1: ∥x∥ = ∥y∥ = 1
0 ≤ ⟨x − y, x − y⟩ = ∥x∥2 + ∥y∥2 − 2Re(⟨x, y⟩) = −2Re(⟨x, y⟩) + 2
⇒ Re(⟨x, y⟩) ≤ 1
Page 51 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
y
x
∥x∥, ∥y∥
∥y∥⟩)
– Case 2: Re(⟨x, y⟩) = Re(⟨ ∥x∥
Re(⟨
x
y
x
y
∥x∥,
∥y∥⟩) = ∥x∥∥y∥Re(⟨
,
⟩)
∥x∥
∥y∥
∥x∥ ∥y∥
By the previous case, we know this is less than one.
⇒ Re(⟨x, y⟩) ≤ ∥x∥∥y∥
Remark: Re(⟨x, y⟩) ≤ ∥x∥∥y∥. This leads directly to the Cauchy-Schwarz Inequality for Inner Products:
∣⟨x, y⟩∣ ≤ ∥x∥∥y∥
Proof:
∣⟨x, y⟩∣ =
where α =
∣⟨x,y⟩∣
⟨x,y⟩
∣⟨x, y⟩∣
⟨x, y⟩ = ⟨αx, y⟩ = Re(⟨αx, y⟩) ≤ ∣α∣∥x∥∥y∥ ≤ ∥x∥∥y∥
⟨x, y⟩
with ∣α∣ = 1.
Q: When do we have equality?
If either x = 0 or y = 0 or x = λy for any λ ∈ C.
Page 52 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 2, 2011
Hilbert Spaces
A Hilbert Space is a complete inner product space.
Q: When is a Banach space a Hilbert space?
THEOREM: If X is an inner product space, then ∀x, y ∈ X
∥x + y∥2 + ∥x − y∥2 = 2(∥x∥2 + ∥y∥2 ).
(1)
(In words, all elements in X satisfy the parallelogram identity ((1)).)
Proof:
⟨x + y, x + y⟩ + ⟨x − y, x − y⟩ = ⟨x, x⟩ + ⟨y, y⟩ + 2Re(⟨x, y⟩) + ⟨x, x⟩ + ⟨y, y⟩ − 2Re(⟨x, y⟩)
= 2(∥x∥2 + ∥y∥2 )
1
Re(⟨x, y⟩) = (∥x + y∥2 − ∥x − y∥2 )
4
1
Im(⟨x, y⟩) = (∥x + iy∥2 − ∥x − iy∥2 )
4
Putting together the above two equations, we obtain the polarization identity :
1
1
⟨x, y⟩ = (∥x + y∥2 − ∥x − y∥2 ) + i (∥x + iy∥2 − ∥x − iy∥2 )
4
4
Examples of Hilbert spaces:
⋆ Cn is complete and an inner product space ⇒ Cn is a Hilbert space
⋆ l2 is a Hilbert space: ⟨x, y⟩ = ∑∞ xi yi
⋆ L2 [a, b] (the completion of C 2 [a, b]) is a Hilbert space
Note: lp (p < ∞) is Banach, but NOT a Hilbert space.
To see this, we will provide a simple example.
Page 53 of 75
(2)
(3)
Christina Battista
MA 515 – Analysis I
x = (0, 1, 0, 0, 0, ⋯)
∥x∥ = 1
y = (1, 0, 0, 0, 0, ⋯)
∥y∥ = 1
x + y = (1, 1, 0, 0, ⋯)
x − y = (−1, 1, 0, 0, ⋯)
∥x + y∥ = 21/p
∥x − y∥ = 21/p
Plugging these values into the parallelogram identity...
2 ⋅ 22/p = 2 ⋅ (1 + 1) ⇒ 22/p = 2
This is only true when p = 2.
Page 54 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
November 4, 2011
Orthogonal Complements and Direct Sums
Recall: δ = inf y∈M d(x, y) (in a metric space) or δ = inf y∈M ∥x − y∥ (in a normed space)
Q: Does ∃y0 ∈ M s.t. δ = ∥x − y0 ∥? If yes, is y0 unique? (This is a question of existence and uniqueness)
x x δ The graph on the left has M as an open interval. Thus, the point y s.t. δ = d(x, y) is NOT in M .
No for existence.
The graph on the right has M as the x-axis and y-axis. However, given the point x, we can see that there
are three points which satisfy δ = d(x, y).
Yes for existence, but no for uniqueness.
M ⊂ X, M is convex if ∀x, y ∈ M , z = αx + (1 − α)y ∈ M where α ∈ [0, 1]. (z is called the segment joining x
and y.)
Remarks:
⋆ EVERY subspace is convex.
⋆ The intersection of convex sets is convex.
⋆ The union of convex sets is NOT necessarily convex.
Page 55 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
THEOREM: [Minimizing Vector] If X is an inner product space and M ⊂ X is a complete, convex
subset then ∀x ∈ X, ∃!y0 ∈ M s.t. δ = inf y∈M ∥x − y∥ = ∥x − y0 ∥.
Proof:
1. Existence:
Let x ∈ X. By definition of infimum, ∃{yn } ∈ M s.t. δn = ∥x − yn ∥ → δ.
Claim: yn → y0 ∈ M as n → ∞. Then by continuity of norms, ∥x − yn ∥ → ∥x − y0 ∥ → δ (by uniqueness of
limit). M is complete so it is enough to show yn is Cauchy.
∥yn − ym ∥2 = ∥vn − vm ∥2
(Let vn = yn − x.)
= −∥vn + vm ∥2 + 2∥vm ∥2 + 2∥vn ∥2
(by the parallelogram identity)
1
∥vn + vm ∥ = ∥yn + ym − 2x∥ = 2∥ (yn + ym ) − x∥ ≥ 2δ
2
because M is convex, 12 (yn + ym ) ∈ M .
∥yn − ym ∥2 ≤ −4δ 2 + 2∥vm ∥2 + 2∥vn ∥2 → −4δ 2 + 2δ 2 + 2δ 2 = 0
Thus, ∥yn − ym ∥ → 0 as m, n → ∞ ⇒ CAUCHY! So y0 ∈ M (by M being complete.)
2. Uniqueness:
By contradiction, assume ∃y0 , y1 s.t. δ = ∥x − y0 ∥ = ∥x − y1 ∥.
∥y1 − y0 ∥2 = ∥y1 − x + x − y0 ∥2 = −∥y1 − 2x + y0 ∥2 + 2∥x − y0 ∥2 + 2∥y1 − x∥2
1
= −4∥ (y1 + y0 ) − x∥2 + 4δ 2
2
≤ −4δ 2 + 4δ 2 = 0
⇒ ∥y1 − y0 ∥2 = 0 ⇒ y1 = y0
THEOREM: [Orthogonality] In the minimizing vector theorem, take M to be a complete subspace.
x ∈ X. Then z = x − y0 is orthogonal to M ⇒ ⟨z, m⟩ = 0, ∀m ∈ M . (y0 s.t. ∥x − y0 ∥ = δ)
Proof:
By contradiction, assume ∃y1 ∈ M s.t. ⟨z, y1 ⟩ = β ≠ 0. For all scalars α ∶
∥z − αy1 ∥2 = ⟨z − αy1 , z − αy1 ⟩ = ∥z∥2 − α⟨z, y1 ⟩ − α⟨y1 , z⟩ + ∣α∣2 ∥y1 ∥2
= ∥z∥2 − αβ − αβ + ∣α∣2 ∥y1 ∥2
= ∥z∥2 − αβ − α(β − α∥y1 ∥2 )
Choose α =
β
.
∥y1 ∥2
∥x − (y0 + αy1 )∥2 = ∥z − αy1 ∥2 = ∥z∥2 −
= ∥z∥2 −
ββ
∥y1 ∥2
∣β∣2
< ∥z∥2 = δ 2
∥y1 ∥2
Because M is a subspace, y0 + αy1 ∈ M .
∥x − (y0 + αy1 )∥2 ≥ δ 2
but we found ∥x − (y0 + αy1 )∥2 < δ 2 . **THIS IS A CONTRADICTION**
⇒ ⟨z, m⟩ = 0, ∀m ∈ M.
Page 56 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 7, 2011
M ⊂ X (subset). M ⊥ = {x ∈ X ∶ ⟨x, m⟩ = 0, ∀m ∈ M } is called the orthogonal complement of M .
⋆ M ⊥ ≠ ∅ ⇒ 0 ∈ M ALWAYS!
⋆ If M ⊂ X is complete subspace, then z = x − y0 ∈ M ⊥ .
Examples in R2 :
⋆ {(0, 0)}⊥ = R2
⋆ {(1, 1)}⊥ = span{(1, −1)}
⋆ [0, 1]⊥ = y-axis
THEOREM: M ⊥ is a closed subspace of X for any subset M ⊂ X.
Proof:
1. Show M ⊥ is a subspace.
Let x, y ∈ M ⊥ ⇒ ⟨x, m⟩ = 0, ⟨y, m⟩ = 0, ∀m ∈ M
⇒ ⟨x, m⟩ + ⟨y, m⟩ = 0 ⇒ ⟨x + y, m⟩ = 0 ⇒ x + y ∈ M ⊥ .
⟨αx, m⟩ = α⟨x, m⟩ = 0 ⇒ αx ∈ M ⊥
2. Show M ⊥ is closed.
Let y ∈ M . y ⊥ = {x ∈ X ∶ ⟨x, y⟩ = 0}. Because y is fixed, let F (x) = ⟨x, y⟩ linear functional and is continuous
(therefore, F is bounded.) So y ⊥ = N (F ). We know for bounded linear operators, the null space is closed
subspace. Thus, y ⊥ is closed and arbitrary intersections of closed sets is closed.
⇒ M ⊥ = ⋂ y ⊥ is closed
y∈M
THEOREM: [Projection Theorem] M ⊂ H (closed subspace of Hilbert space) then H = M ⊕ M ⊥ .
Proof:
1. Show M ∩ M ⊥ = {0}.
Assume ∃x ∈ M ∩ M ⊥ , x ≠ 0. ⟨x, x⟩ = 0 ⇔ x = 0 by properties of inner products.
2. Show x ∈ H can be written as y + z, y ∈ M, z ∈ M ⊥ in only one way.
From the minimizing vector theorem, ∃!y ∈ M s.t. ∥x − y∥ = δ.
z = x − y ⇒ ⟨z, m⟩ = 0 by orthogonality theorem
⇒ z ∈ M⊥ ⇒ x = y + z
Page 57 of 75
Christina Battista
MA 515 – Analysis I
Assume x = y1 + z1 .
y + z = y1 + z1
y − y1 = z1 − z with y − y1 ∈ M and z1 − z ∈ M ⊥
Because M ∩ M ⊥ = {0}, y − y1 = 0 and z1 − z = 0
⇒ y1 = y, z1 = z
P ∶ H → M , where P is called the projection map.
x ↦ y = P x =nearest point in M to x. (projection operator )
⟨x − P x, m⟩ = 0 ⇒ (x − P x) ⊥ M
Claim: P is a bounded linear operator.
1. Show P is bounded.
∥x∥2 = ⟨x, x⟩ = ⟨P x + z, P x + z⟩ = ∥P x∥2 + ∥z∥2 + ⟨P x, z⟩ + ⟨z, P x⟩ = ∥P x∥2 + ∥z∥2
∥P x∥2 ≤ ∥x∥2 ⇒ ∥P x∥ ≤ ∥x∥
2. On Assignment 7, we show P is linear.
Page 58 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
November 9, 2011
Given that H is a Hilbert space and M ⊂ H is closed subspace, what conclusions can we draw?
⋆ H = M ⊕ M ⊥ ⇒ ∀x ∈ H, x = P x + z where P x ∈ M and z ∈ M ⊥
⋆ ∃z = x + y0 s.t. d(x, M ) = δ = ∥x − y0 ∥ where x ∈ H and y0 ∈ M
⋆ M is complete and convex
⋆ M ⊥ is a closed subspace of H
⋆ x − Px ⊥ M
⋆ P ∶ H → M is a bounded linear operator (P ∶ x ↦ P x = y0 )
(y-axis)⊥ =x-axis
[0, 1]⊥ = y-axis
Note: In general, y ⊥⊥ ≠ y.
THEOREM: If M is a closed subspace of a Hilbert space then M ⊥⊥ = M .
Proof:
1. M ⊂ M ⊥⊥
Let x ∈ M ⇒ x ⊥ M ⊥ ⇒ x ∈ M ⊥⊥
2. M ⊥⊥ ⊂ M
Let x ∈ M ⊥⊥ ⇒ x ⊥ M ⊥ . By the projection theorem, x = P x + z where P x ∈ M ⇒ P x ∈ M ⊥⊥ and z ∈ M ⊥ .
Also, M ⊥ ∩ M ⊥⊥ = {0}.
Because M ⊥⊥ is a subspace, z ∈ M ⊥⊥ (because z = x − P x) but M ⊥ ∩ M ⊥⊥ = {0}.
⇒ z = 0 ⇒ x = Px ∈ M ⇒ x ∈ M.
M = M ⊥⊥ if M is a closed subspace.
THEOREM: [Riesz-Representation for Hilbert Spaces] L ∶ H → C is a continuous, linear functional.
Then ∃!y ∈ H s.t. L(x) = ⟨x, y⟩, ∀x ∈ H.
Proof: If L = 0, choose y = 0 and we are done.
1. Existence:
If L ≠ 0, then N (L) = {x ∈ H ∶ L(x) = 0} is a closed subspace of H. Let N (L) = M .
H = M ⊕ M ⊥.
Take z ∈ M ⊥ with ∥z∥ = 1. For all x ∈ H, (Lx)z − (Lz)x ∈ H.
L((Lx)z − (Lz)x) = L((Lx)z) − L((Lz)x) = (Lx)Lz − (Lz)Lx = 0
⇒ (Lx)z − (Lz)x ∈ N (L) = M
⟨(Lx)z − (Lz)x, z⟩ = 0 because one is in M and the other is in M ⊥ .
(Lx)⟨z, z⟩ − (Lz)⟨x, z⟩ = Lx − (Lz)⟨x, z⟩ = 0
Page 59 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
⇒ Lx = ⟨x, (Lz)z⟩ where (Lz)z ∈ H
Thus, y = (Lz)z. THIS GIVES THE FORMULA FOR y.
2. Uniqueness:
Assume Lx = ⟨x, y1 ⟩ = ⟨x, y2 ⟩
Choose x = y1 − y2 .
⇒ ⟨x, y1 − y2 ⟩ = 0, ∀x ∈ H
⟨y1 − y2 , y1 − y2 ⟩ = 0 ⇒ y1 − y2 = 0 ⇒ y1 = y2
A set {vα }α∈F in H is called orthonormal if
⋆ ⟨vα , vα ⟩ = 1, ∀α ∈ F
⋆ ⟨vα , vβ ⟩ = 0, α ≠ β
PROPOSITION: If S ⊆ H is orthonormal then S is linearly independent. (Also, every finite subset of S
is linearly independent.)
Proof:
Let {v1 , ⋯, vn } ⊆ S.
⟨c1 v1 + ⋯ + cn vn , vi ⟩ = ci ⟨vi , vi ⟩ = ci ∥vi ∥2 = ci = 0, ∀i
⇒ c1 v1 + ⋯ + cn vn = 0 implies ci = 0, ∀i
Page 60 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 11, 2011
Note: If H is finite-dimensional, then every orthonormal set is finite.
x ∈ H. x̂(n) = ⟨x, vn ⟩, ∀n where vn is in an orthonormal set. These are called the Fourier Coefficients of
x with respect to {vn }.
THEOREM: [Best Approximation Theorem] If {vi }ni=1 is a finite orthonormal set in H, then ∑n ⟨x, vi ⟩vi
is the unique best approximation to x in span{vi }.
(i.e. ∥x − ∑n ⟨x, vi ⟩vi ∥ ≤ ∥x − s∥, ∀s ∈ span{vi }).
Also, ∥x∥2 ≥ ∑n ∣⟨x, vi ⟩∣2 .
Proof:
1. Let M = span{vi }.
Claim: M is a closed subspace of H.
x ∈ H, P x =projection of x onto M , P x ∈ M .
n
⇒ P x = ∑ ci vi
i=1
∥x − P x∥ = δ ≤ ∥x − s∥, ∀s ∈ M
Show ci = ⟨x, vi ⟩, ∀i = 1, ⋯, n. We know x − P x ⊥ M ⇒ ⟨x − P x, m⟩ = 0, ∀m ∈ M ⇒ ⟨x − ∑n ci vi , vi ⟩ = 0, ∀i.
n
⇒ ⟨x, vi ⟩ − ⟨∑ ci vi , vi ⟩ = 0
⟨x, vi ⟩ = ci because ⟨vi , vi ⟩ = 1 and ⟨vi , vj ⟩ = 0.
2.
n
n
n
n
n
∥x∥2 ≥ ∥P x∥2 = ⟨∑ ci vi , ∑ ci vi ⟩ = ⟨∑⟨x, vi ⟩vi , ∑⟨x, vi ⟩vi ⟩ = ∑ ∣⟨x, vi ⟩∣2
Immediate Consequence: sn = ∑ni=1 ∣⟨x, vi ⟩∣2 is bounded by ∥x∥2 .
sn is increasing and thus, sn converges.
∞
∞
∑i=1 ∣⟨x, vi ⟩∣2 is convergent and ∑i=1 ∣⟨x, vi ⟩∣2 ≤ ∥x∥2 . This is known as Bessel’s inequality .
Q: Given a linearly independent set, can we find an orthonormal set?
⇒Yes.
This process is called the Gram-Schmidt Process.
Given a linearly independent set {xi }, we will transform it into an orthonormal set {ei }.
1.
x1
e1 =
∥x1 ∥
Page 61 of 75
Christina Battista
MA 515 – Analysis I
2.
ei+1 =
xi+1 − ∑ik=1 ⟨xi+1 , ek ⟩ek
∥xi+1 − ∑ik=1 ⟨xi+1 , ek ⟩ek ∥
Page 62 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
November 14, 2011
Total Orthonormal Sets and Parseval’s Equality
X is a normed space, M ⊂ X subset. Then M is total if spanM = X. We say that the span of M is dense
in X.
X is an inner product space, M is an orthonormal set in X. M is a total orthonormal set if M is total
in X. We call this an “orthonormal basis for X.”
Note: Every Hilbert space has a total orthonormal set.
**In general, this is NOT true for an inner product space.**
THEOREM: [Equivalent definition for total] H a Hilbert space, ∅ ≠ M is total in H if and only if
[x ⊥ M ⇒ x = 0] (i.e. M ⊥ = {0})
If M is a total orthonormal set it means M is a maximal orthonormal set (it cannot be properly contained
in an orthonormal set).
Proof:
⇒ Let x ∈ M ⊥ . WTS: x = 0.
x ∈ M ⊥ ⇒ x ∈ H = spanM
So there exists {xn } ∈ spanM s.t. xn → x ∈ M ⊥ .
⟨xn , x⟩ = 0 → ⟨x, x⟩ = 0 ⇒ x = 0
⇐ spanM = subspace in H.
spanM = closed subspace in H.
⊥
H = spanM ⊕ spanM .
⊥
Left to show: spanM = {0}.
M ⊂ spanM ⊂ spanM
⊥
⊥
spanM ⊂ spanM ⊥ ⊂ M ⊥ = {0} ⇒ spanM = {0}.
Thus, H = spanM + 0 = spanM .
THEOREM: [Parseval’s Equality] M = {en }∞ orthonormal set in H. M is total if and only if ∀x ∈ H,
x = ∑∞ ⟨x, en ⟩en and the following relation holds
∞
2
2
∑ ∣⟨x, en ⟩∣ = ∥x∥
n=1
Proof:
⇐ Assume M is NOT total. (spanM ≠ H and M ⊥ ≠ {0})
Page 63 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
Let 0 ≠ x ∈ M ⊥ where x ∈ H.
⟨x, en ⟩ = 0, ∀n ⇒ ∥x∥ = 0 (from Parseval’s) ⇒ x = 0 ***THIS IS A CONTRADICTION***
Thus M ⊥ = {0} ⇒ M is total.
⇒ We need to show three things:
1. ∑∞ ⟨x, en ⟩en is convergent.
sk = ∑k ⟨x, en ⟩en and show sk is convergent in H. (Since we are in a Hilbert space, Cauchy implies convergence.)
WLOG, assume k > m.
k
k
k
k
n=m+1
n=m+1
n=m+1
n=m+1
∥sk − sm ∥2 = ∥ ∑ ⟨x, en ⟩en ∥2 = ⟨ ∑ ⟨x, en ⟩en , ∑ ⟨x, en ⟩en ⟩ = ∑ ∣⟨x, en ⟩∣2
k
m
n=1
n=1
= ∑ ∣⟨x, en ⟩∣2 − ∑ ∣⟨x, en ⟩∣2 = s˜k − s˜m
as k, m → ∞ ∶
s˜k − s˜m = ∥x∥ − ∥x∥ = 0
Thus sk → s so it is convergent in H.
2. Show x = s.
Fix n.
k
⟨s, en ⟩ = ⟨ lim sk , en ⟩ = lim ⟨sk , en ⟩ = lim ⟨∑⟨x, ei ⟩ei , en ⟩ = ⟨x, en ⟩
k→∞
k→∞
k→∞
⇒ ⟨s, en ⟩ = ⟨x, en ⟩ ⇒ ⟨s − x, en ⟩ = 0, ∀n
s − x ∈ M ⊥ where M is total ⇒ s − x = 0 ⇒ x = s
3. Show Parseval’s Equality.
k
sk → x ⇒ ∑ ⟨x, en ⟩en → x
n=1
By continuity of norm, ∥sk ∥ → ∥x∥ as k → ∞.
2
2
k
∞
n=1
n=1
∥sk ∥2 = ∑ ∣⟨x, en ⟩∣2 → ∑ ∣⟨x, en ⟩∣2
2
By uniquness of limit, ∥x∥2 = ∑∞
n=1 ∣⟨x, en ⟩∣ .
Page 64 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 16, 2011
Separable Hilbert Spaces
THEOREM 1: If H is separable, then every orthonormal set is countable.
THEOREM 2: If H contains a total orthonormal sequence, then H is separable.
Proof of (1):
Let B be dense in H (B = H). Let M be an orthonormal set.
Want to show: M is countable.
√
√
√
Let x, y ∈ M s.t. ∥x − y∥ = 2. Pick Nx = {x0 ∈ M ∶ ∥x − x0 ∥ ≤ 32 } and Ny = {y0 ∈ M ∶ ∥y − y0 ∥ ≤ 32 }. Nx
and Ny are disjoint (Nx ∩ Ny = ∅).
∃b1 ∈ Nx , b2 ∈ Ny (because B is dense) b1 , b2 ∈ B.
Assume M is NOT countable. Then B is uncountable for any B. ***THIS IS A CONTRADICTION*** H
is separable so there must exist a B that is countable and dense in H.
Thus, M is countable.
Note: A maximal element of P need not be an upper bound.
THEOREM: Every Hilbert space (≠ {0}) has a total orthonormal basis.
x
Let M be a set of orthonormal sets (M ≠ ∅). If x ∈ H, ∥x∥
∈ M.
Define “ ≤ ” to be ⊂ (set inclusion) (i.e. A1 ⊂ A2 ⊂ ⋯)
Look at every chain in M (C ⊂ M )
Claim: Every C has an upper bound (= [union of subsets of M ] ∈ C). By Zorn’s Lemma, M has at least
one maximal element, F .
Claim: F is a total orthonormal set.
Assume F is NOT total.
∃0 ≠ x ⊥ F
x
F̃ = F ∪ { ∥x∥
} is orthonormal then F ⊂ F̃ is not maximal. ***THIS IS A CONTRADICTION***
Thus, F is total.
Page 65 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 18, 2011
Hahn-Banach Theorem (Extension of Linear Functionals)
Problem: Let H is a Hilbert space and let Z ⊂ H be a closed subspace of H (This means Z is complete.)
Let f be a bounded linear functional on Z. Can we find an extension f˜ of f to H s.t. ∥f˜∥H ′ = ∥f ∥Z ′ ? YES.
f ∶Z→K
By Riesz Representation Theorem, ∃z ∈ Z s.t. f (x) = ⟨x, z⟩, ∀x ∈ Z.
f˜ = ⟨x, z⟩, ∀x ∈ H
f˜∣Z = f
f˜ is obviously linear.
∥f ∥Z ′ = ∥z∥
∣f (x)∣ = ∣⟨x, z⟩∣ ≤ ∥x∥∥z∥ →
sup
∥x∥=1,x∈Z
∥x∥∥z∥ = ∥z∥
∥f ∥Z ′ ≤ ∥z∥
f (z) = ⟨z, z⟩ = ∥z∥ but ∣f (z)∣ ≤ ∥f ∥∥z∥ ⇒ ∥z∥ ≤ ∥f ∥
2
∥f ∥Z ′ = ∥z∥
We can do the same for ∥f˜∥H ′ :
∥f˜∥H ′ = ∥z∥ = ∥f ∥Z ′
Let X be a real vector space. p ∶ X → R is a sublinear functional on X if:
⋆ p is subadditive, i.e. p(x + y) ≤ p(x) + p(y), ∀x, y ∈ X
⋆ p is positive homogeneous, i.e. p(αx) = αp(x), ∀α ≥ 0, ∀x ∈ X
Obviously every linear functional is sublinear.
Examples:
⋆ All linear functionals are sublinear (i.e. inner product)
⋆ Norms are sublinear (but not linear)
Page 66 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
THEOREM: [Hahn-Banach for Real Vector Spaces] Let X be a real vector space, and let Z be a
subspace of X. Let p ∶ X → R be a sublinear functional on X, and f ∶ Z → R a linear functional satisfying
f (x) ≤ p(x), ∀x ∈ Z.
Then f has a linear extension f˜ ∶ X → R (i.e. f˜(x) = f (x), ∀x ∈ Z), s.t.
f˜(x) ≤ p(x), ∀x ∈ X.
Proof: (look in book) The proof uses Zorn’s Lemma and the following ideas:
⋆ Let P = {g ∶ D(g) ⊂ X∣D(g) ⊂ X subspace, g is linear , Z ⊂ D(g), g∣Z = f and g(x) ≤ p(x), ∀x ∈ D(g)}.
Note: P is not empty. f ∈ P
⋆ Define “ ≤ ” as follows: [g1 ≤ g2 ⇔ D(g1 ) ⊂ D(g2 ) and g2 ∣D(g1 ) = g1 .]
⋆ Consider C a chain in P and show that C has an upper bound (which by Zorn’s Lemma means that
P has a maximal element tildef )
⋆ Then show D(f˜) = X
THEOREM: [Hahn-Banach (complex/generalized version)] Let X be a complex vector space, and
let Z be a subspace of X. Let p ∶ X → R be a sublinear functional of n X, i.e.
⋆ p is subadditive, i.e. p(x + y) ≤ p(x) + p(y), ∀x, y ∈ X
⋆ p is homogeneous, i.e. p(αx) = ∣α∣p(x), ∀α ∈ C, ∀x ∈ X
Let f ∶ Z → C be a linear functional, satisfying
∣f (x)∣ ≤ p(x), ∀x ∈ Z.
Then f has a linear extension f˜ ∶ X → C (i.e. f˜(x) = f (x), ∀x ∈ Z), s.t.
∣f˜(x)∣ ≤ p(x), ∀x ∈ X.
Page 67 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 21, 2011
THEOREM: [Hahn-Banach (normed spaces)] **STANDARD VERSION** Let X be a normed space.
Let f be a bounded linear functional on Z ⊂ X subspace of X. Then there exists a bounded linear functional
f˜ on X s.t.
f˜∣Z = f, and ∥f˜∥X ′ = ∥f ∥Z ′
(and ∥f ∥Z = 0 for Z = {0}.)
Proof:
Need p(x) s.t. ∣f (x)∣ ≤ p(x), ∀x ∈ Z.
∣f (x)∣ ≤ ∥f ∥Z ′ ∥x∥
Let p(x) = ∥f ∥Z ′ ∥x∥ → R.
⋆ p(x + y) = ∥f ∥Z ′ ∥x + y∥ ≤ ∥f ∥Z ′ (∥x∥ + ∥y∥) = p(x) + p(y)
⋆ p(αx) = ∥f ∥Z ′ ∥αx∥ = ∥f ∥Z ′ ∣α∣∥x∥ = ∣α∣p(x)
Thus p is sublinear on X.
By the generalized version of Hahn-Banach, ∃f˜ ∶ X → C s.t. f˜∣Z = f and ∣f˜(x)∣ ≤ p(x), ∀x ∈ X.
∥f˜∥X ′ = sup ∣f˜(x)∣ ≤ ∥f ∥Z ′ ∥x∥ = ∥f ∥Z ′
∥x∥=1
∥f ∥Z ′ =
sup
∥x∥=1,x∈Z
∣f (x)∣ ≤
sup
∥x∥=1,x∈X
∣f˜(x)∣ = ∥f˜∥X ′
THEOREM: [Bounded Linear Functionals] Let X be a normed space and let x0 ≠ 0 (where x0 is a
fixed point) be any element of X. Then there exists a bounded linear functional f˜ on X s.t. ∥f˜∥ = 1 and
f˜(x0 ) = ∥x0 ∥.
Proof:
1. Define a subspace.
Z ⊂ X ∶ Z = {αx0 ∶ α ∈ C}
2. Define a bounded linear functional:
∀x ∈ Z, f (x) = f (αx0 ) = α∥x0 ∥
This is linear.
∣f (x)∣ = ∣α∣∥x0 ∥ = ∥αx0 ∥ = ∥x∥
This is bounded with c = 1.
3. By the standard version of Hahn-Banach, ∃f˜ ∶ X → C s.t. f˜∣Z = f and ∥f˜∥X ′ = ∥f ∥Z ′ but ∥f ∥Z ′ = 1 ⇒
∥f˜∥X ′ = 1.
f˜(x0 ) = f (x0 ) (because x0 ∈ Z) = ∥x0 ∥
COROLLARY: [Norm, Zero Vector] For all x ∈ X (X a normed space), we have
∥x∥X =
sup
f ∈X ′ ,f ≠0
∣f (x)∣
.
∥f ∥X ′
Page 68 of 75
Christina Battista
MA 515 – Analysis I
Hence, if x0 ∈ X s.t. f (x0 ) = 0, ∀f ∈ X ′ , then x0 = 0.
Proof:
∣f (x)∣ ≤ ∥f ∥∥x∥ ⇒ ∥x∥ ≥
∣f (x)∣
∣f (x)∣
, ∀f ∈ X ′ ⇒ ∥x∥ ≥ sup
∥f ∥
f ∈X ′ ,f ≠0 ∥f ∥
∣f (x)∣ ∣f˜(x)∣
≥
= ∣f˜(x)∣ = ∥x∥
∥f˜∥X ′
f ∈X ′ ,f ≠0 ∥f ∥X ′
sup
From the bounded linear functional theorem, we have
∥x∥X =
sup
f ∈X ′ ,f ≠0
∣f (x)∣
∥f ∥X ′
Page 69 of 75
Course Notes
Christina Battista
MA 515 – Analysis I
Course Notes
November 28, 2011
Reflexivity of Normed Spaces
Let X be a normed space.
⋆ X ′ = set of all bounded linear functionals on X, called the dual space of X.
⋆ X ′′ = set of all bounded linear functionals on X ′ , called the second dual space of X.
Q: What is the difference between X ′ and X ∗ ?
X ′ is all bounded linear functionals on X whereas X ∗ is all linear functionals on X.
Note: X ′ ⊂ X ∗ always but if X is finite-dimensional, X ′ = X ∗ .
Now fix x ∈ X, and define a functional on X ′ by: gx (f ) = f (x). In this case, f is the variable, f ∈ X ′ where
f ∶ X → K and g ∶ X ′ → K.
LEMMA: gx ∈ X ′′ , for every fixed x ∈ X, and ∥gx ∥ = ∥x∥.
Proof:
⋆ Linearity:
This follows directly from f being linear.
gx (αf + h) = (αf + h)(x) = αf (x) + h(x) = αgx (f ) + gx (h)
⋆ Boundedness:
∥gx ∥X ′′ =
Canonical Map
∣f (x)∣
∣gx (f )∣
= sup
= ∥x∥
f ∈X ′ ,f ≠0 ∥f ∥X ′
f ∈X ′ ,f ≠0 ∥f ∥X ′
sup
C ∶ X → X ′′
C(x) = gx
LEMMA: C ∶ X → R(C) is an isomorphism (i.e. X embeds in X ′′ .)
⋆ Linear: Let αx + y ∈ X.
gαx+y = f (αx + y) = αf (x) + f (y) = αgx (f ) + gy (f )
Page 70 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
⋆ Injective: Let x ∈ N (C).
C(x) = 0 ⇒ gx = 0 ⇒ gx (f ) = 0, ∀f ∈ X ′ ⇒ f (x) = 0, ∀f ∈ X ′ ⇒ x = 0 ⇒ N (C) = {0}
⋆ Isometric:
∣f (x)∣
∣gx (f )∣
= sup
= ∥x∥ from previous lemma
x∈X,x≠0 ∥f ∥
x∈X,x≠0 ∥x∥
∥gx ∥ = sup
⇒ ∥gx ∥ = ∥x∥
Note: In general, C will not be surjective.
A normed space X is called reflexive if R(C) = X ′′ . [Thus if X is reflexive, then X is isomorphic to X ′′ .]
THEOREM: If X is reflexive, then X is Banach (complete).
Proof: Because we know the dual and double dual spaces are Banach, then X is Banach.
THEOREM: If X is finite-dimensional, then X is reflexive.
Examples:
⋆ Reflexive: Rn , Cn , lp (1 < p < ∞), Lp [a, b](1 < p < ∞)
⋆ Non-reflexive: l1 , l∞ , L1 , L∞ , C[a, b]
THEOREM: Every Hilbert space is reflexive.
Proof:
Want to show: C ∶ H → H ′′ is an isomorphism. Need to show R(C) = H ′′ . (For g ∈ H ′′ , ∃x ∈ H s.t.C(x) = g.)
Let g ∈ H ′′ where g ∶ H ′ → K .
Want to show H ′ is an inner product space.
Let A ∶ H ′ → H. Af = z where z is s.t. f (x) = ⟨x, z⟩ with z given by Riesz Representation Theorem. Thus,
A is bijective.
Is A isometric? YES. By Riesz, ∥f ∥ = ∥z∥.
Is A linear?
A(f1 + f2 ) = z1 + z2 = A(f1 ) + A(f2 )
A(αf ) = αz = αA(f )
Therefore, A is conjugate linear.
On H ′ , define: ⟨f1 , f2 ⟩H ′ = ⟨Af2 , Af1 ⟩H . Check the properties of an inner product:
⋆ ⟨f, f ⟩ ≥ 0 ⇒ ⟨f, f ⟩ = ⟨Af, Af ⟩ ≥ 0 ⇒ 0 = ⟨Af, Af ⟩ = ∥z∥2 = 0 ⇒ Af = 0 ⇒ f = 0 by injectivity of A
⋆ ⟨αf, g⟩ = ⟨Ag, A(αf )⟩ = ⟨Ag, αAf ⟩ = α⟨Ag, Af ⟩ = α⟨f, g⟩
⋆ ⟨f, g⟩ = ⟨Ag, Af ⟩ = ⟨Af, Ag⟩ = ⟨g, f ⟩
Page 71 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
⋆ ⟨f + g, h⟩ = ⟨Ah, Af + Ag⟩ = ⟨Ah, Af ⟩ + ⟨Ah, Ag⟩ = ⟨f, h⟩ + ⟨g, h⟩
So now H ′ is an inner product space. Thus, H ′ is complete and H ′ is Hilbert. So we can use Riesz
Representation Theorem on g ∶ H ′ → K.
∃!f0 ∈ H ′ s.t. g(f ) = ⟨f, f0 ⟩H ′ = ⟨Af0 , Af ⟩H = ⟨x, z⟩H = f (x) because Af0 = x and Af = z.
Extension Theorem for Linear Operators in Vector Spaces
THEOREM: Let X and Y be vector spaces and M a proper subspace of X. Let A be a linear operator
A ∶ M → Y . Then there exists a linear operator à ∶ X → Y s.t. à is an extension of A.
THEOREM: Let M be a proper subspace of a vector space X and suppose that x0 ∈ X ∖ M . Then there
exists an element A ∈ X ∗ s.t. Ax = 0 if x ∈ M and Ax0 = 1.
THEOREM: Let X be a vector space. If x ∈ X s.t. Ax = 0, for all A ∈ X ∗ then x = 0.
Note: This theorem gives injectivity for the canonical map C ∶ X → X ∗∗ .
Page 72 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
November 30, 2011
THEOREM: If X ′ is separable then X is separable.
Note: In general, the converse is NOT true.
Example
⋆ l1 is separable but l∞ is NOT separable, thus l1 is NOT reflexive.
COROLLARY: If X is reflexive, X ≃ X ′′ . And if X is separable and reflexive, then X ′′ is separable and
then X ′ is separable.
Direct consequence of this: If X is separable and X ′ is NOT separable, then X CANNOT be reflexive.
Page 73 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
December 2, 2011
Category Theorem. Uniform Boundedness Theorem
Let X be a metric space. A subset M ⊂ X is
1. rare (nowhere dense) in X if M̄ has no interior points.
2. meager (of the 1st category) in X if M is the union of countably many rare sets in X.
3. nonmeager (of the 2nd category) in X if M is not meager in X.
Note: Recall that if A ⊂ X, the interior of A is the union of all open sets of X that are contained in A. If A
has empty interior, then A contains no open sets of X other than ∅. Or, A has empty interior if every point
of A is a limit point of Ac , i.e. Ac is dense in X.
Examples:
⋆ Q ⊂ R is meager (but not rare) (Q̄ = R)
⋆ Z ⊂ R is rare (and meager) (Z̄ = Z)
⋆ Rare sets in a discrete metric space X: ∅
THEOREM: [Baire’s Category Theorem] If a metric space X =≠ ∅ is complete, then it is nonmeager
in itself.
Therefore if X ≠ ∅ is complete and X = ⋃∞
k=1 Ak , then at least one Ak contains a nonempty open subset.
THEOREM: [Uniform Boundedness Theorem] Let X be a Banach space and Y a normed space. Let
{Tn } be a sequence in B(X, Y ) s.t. {∥Tn x∥} is bounded for every x ∈ X (i.e. ∥Tn x∥ ≤ cx , n = 1, 2, ⋯), then
the sequence {∥Tn ∥} is bounded, i.e.
∃c s.t. ∥Tn ∥ ≤ c, n = 1, 2, ⋯
(i.e. the sequence {∥Tn x∥} is uniformly bounded.)
One way to use the theorem: If you can construct a sequence of bounded linear operators {Tn } on X s.t.
{∥Tn x∥} is bounded for every x ∈ X but not uniformly bounded, then X can not be complete.
A sequence {xn } in a normed space X is weakly convergent if there is an x ∈ X s.t. for every f ∈ X ′ we
have
lim f (xn ) = f (x).
n→∞
Page 74 of 75
Christina Battista
MA 515 – Analysis I
Course Notes
We say that xn converges weakly to x and we write:
w
xn ⇀ x or xn Ð
→ x.
Rewrite this definition in a Hilbert space:
w
xn Ð
→ x in H iff [⟨xn , z⟩ → ⟨x, z⟩, ∀z ∈ H].
LEMMA: If xn ⇀ x, then
1. The weak limit is unique.
Proof:
Assume xn ⇀ y and xn ⇀ x. Then f (xn ) → f (x) and f (xn ) → f (y), ∀f ∈ X ′ . By uniqueness of limit,
f (x) = f (y), ∀f ∈ X ′ . And since f is a linear functional, f (x) = f (y) ⇒ f (x − y) = 0, ∀f ∈ X ′ ⇒ x − y =
0 ⇒ x = y.
2. Every subsequence of {xn } converges weakly to x.
3. The sequence {∥xn ∥} is bounded.
Proof:
{f (xn )} is convergent ⇒ ∣f (xn )∣ ≤ cf , ∀n.
C ∶ X → X ′′ where gn (f ) = f (xn ) ⇒ ∣gn (f )∣ = ∣f (xn )∣ ≤ cf
g ∶ X ′ → K where X ′ is complete so we can apply the Uniform Boundedness Theorem ⇒ ∥gn ∥ ≤
c Rightarrow∥xn ∥ = ∥gn ∥ ≤ c
THEOREM: [Strong and Weak Convergence] Let {xn } be a sequence in a normed space X.
1. Strong convergence implies weak convergence with the same limit.
Proof: (Given ∥xn − x∥ → 0 strong convergence)
0 ≤ ∣f (xn ) − f (x)∣ = ∣f (xn − x)∣ ≤ ∥f ∥∥xn − x∥ → 0
2. The converse is NOT generally true. For example, take an orthonormal sequence in a Hilbert space.
3. If dim(X) < ∞, then weak convergence implies strong convergence.
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