MATH414 FUNCTIONAL ANALYSIS
HOMEWORK 3
Hw 3: 7/82 5/101 6/101 8/101 9/101
Problem (7/82)
If dimY < ∞ in Riesz’s lemma 2.5-4, show that one can even choose θ = 1.
Solution
Step1 If Y is a finite dimensional subspace of a normed space X then for
any x ∈ X there exists y0 ∈ Y such that d(x, y0 ) = d(x, Y ), that is,
||x − y0 || = inf ||x − y||.
y∈Y
Let a = d(x, Y ), then by definition for any n ∈ N there is an element yn in Y
∞
such that ||x − yn || ≤ a + n1 . So {yn }n=1 forms a bounded sequence in Y . Let
||yn || ≤ M for all n in N. Notice that A = {y ∈ Y : ||y|| ≤ M } is a compact
set since it is closed and bounded subset of a finite dimensional normed space
∞
Y (see theorem 2.5-3). Since {yn }n=1 is a sequence in A, it has a convergent
∞
subsequence, say {ynk }k=1 with limit y0 . We claim that d(x, y0 ) = ||x − y0 || = a.
Clearly it is ≥ a. On the other hand
||x − y0 || ≤ ||x − ynk || + ||ynk − y0 || ≤ a +
1
+ ||ynk − y0 || for all k ∈ N.
nk
So as k 7→ ∞ we find ||x − y0 || ≤ a. This proves the claim.
Step2 If dimY < ∞, then we can re-formulate Riesz’s lemma as:
If Y is a proper finite dimensional subspace of normed space X, then ∃ x in X
such that ||x − y|| ≥ 1 for all y ∈ Y .
Proof. Let v ∈ X − Y and a = inf y∈Y ||v − y||. Clearly a > 0 since Y is closed
(see theorem 2.4-2). We know from step1 that ∃ y0 ∈ Y such that a = ||v − y0 ||.
So let
v − y0
v − y0
=
.
x=
||v − y0 ||
a
Evidently ||x|| = 1 and for any y ∈ Y ,
||x − y|| = ||
v − y0
1
1
− y|| = ||v − y0 − ay || ≥ a = 1.
| {z }
a
a
a
in Y
1
Problem (5/101)
Show that the operator T : l∞ −→ l∞ ,
(ξj ) 7−→ (ξj /j) is linear and bounded.
Solution T is linear since if x = (ξj ), y = (ηj ) are vectors and α, β scalars then
T (αx + βy) = T (αξj + βηj ) = (
αξj + βηj
ξj
ηj
) = α( ) + β( ) = αT x + βT y.
j
j
j
And boundedness of T follows from
||T x|| = sup |
j
ξj
| ≤ sup |ξj | = ||x||
j
j
(c = 1).
Problem (6/101)
(Range) Show that the range R(T ) of a bounded linear operator T : X −→ Y
need not be closed in Y . Hint. Use T in Prob. 5.
Solution Consider the bounded
√
√ linear operator T in previous question. Since
for any n ∈ N, xn = (1, 2, ..., n, 0, 0, ...) is in l∞ , we obtain
1
1
T xn = yn = (1, √ , ..., √ , 0, 0, ...) is in R(T ) ∀ n ∈ N.
n
2
∞
∞
Clearly {yn }n=1 is a convergent sequence with limit y = ( √1n )n=1 . But y is not
∞
in R(T ). If not
√ ∃x = (ξn ) ∈ l such that T x = y, but the definition of T
requires ξn = n which is a contradiction.
Problem (8/101)
Show that the inverse T −1 : R(T ) −→ X of a bounded linear operator need
not be bounded. Hint. Use T in Prob. 5.
Solution
Consider again previous question. T is a bounded linear operator and its inverse
∞
−1
T −1 exits (from
is not bounded.
√ R(T )√→ l ) since T∞ is one-to-one. But T
Since xn = (1, 2, ..., n, 0, 0, ...) ∈ l for all n,
1
1
T xn = yn = (1, √ , ..., √ , 0, 0, ...) ∈ R(T ).
n
2
Clearly ||x|| =
gives us
√
n and ||y|| = 1. Given c ∈ R+ choosing n such that
||T −1 yn || = ||xn || =
2
√
n > c = c||yn ||.
√
n>c
Problem (9/101)
Let T : C[0, 1] −→ C[0, 1] be defined by
Z
t
x(τ ) dτ.
T x(t) = y(t) =
0
Find R(T ) and T −1 : R(T ) −→ C[0, 1]. Is T −1 linear and bounded?
Solution
Definition f : [0, 1] → R (or C) is said to be differentiable on [0,1], if f is
differentiable on (0,1) in usual sense and differentiable at t = 0 and t = 1 by
considering only one-sided limits.
Definition f : [0, 1] → R (or C) is said to be continuously differentiable on
[0,1], if f is differentiable on [0,1] and its derivative f 0 is continuous on [0,1].
(Notice that f (t) = t2 sin
1
t
for t 6= 0 and 0 for t = 0, is differentiable on [0,1] but its derivative
is continuous only on (0,1].)
R(T ) = continuously differentiable functions y on [0,1] with y(0) = 0.
Let y be a such function, so y 0 ∈ C[0, 1] and we have
T y 0 (t) =
Z
0
t
y 0 (τ ) dτ |{z}
= y(t) − y(0) = y(t) ⇒ y ∈ R(T ).
FTC
Conversely if y ∈ R(T ) then ∃x ∈ C[0, 1] such that
Z
T x(t) = y(t) =
t
x(τ ) dτ.
0
Again by FTC we obtain y 0 = x, so y is continuously differentiable on [0,1] and
clearly y(0) = 0.
It is easy to show T is one-to-one, so T −1 : R(T ) → C[0, 1] exists and linear (see theorem 2.6-10). Since for any y ∈ R(T ) T y 0 = y, then T −1 y = y 0 . But
T −1 is not bounded. Notice that yn (t) = tn is in R(T ) for any n, and we have
||yn || = 1, ||T −1 yn || = ||yn0 || = n. So given c ∈ R+ choosing n > c gives us
||T −1 yn || = ||yn0 || = n > c = c||yn ||.
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