FUNCTIONAL ANALYSIS NOTES
The Baire Category Theorem and Related
Denition 1. A set
A
in a metric space is called
nowhere dense if the closure of A has empty interior. (Ā)◦ = ∅.
Example 2. The rationals are NOT nowhere dense. A nite number of points is nowhere dense. Cantor sets are
nowhere dense sets. Subsets of nowhere dense sets are nowhere dense.
Remark 3.
Finite unions of nowhere dense sets are still nowhere dense.
Denition 4. A set
Meagre.
A
◦
∪ni=1 Ai = ∪ni=1 Āi
◦
which can be written as a countable union of nowhere dense sets is called
Sets which cannot be written this way are called
◦
= ∪ni=1 Āi = ∅.
2nd Category or Nonmeagre.
1st Category
or
Example 5. The countable union of meagre sets is still meager. Any subset of a meager set is still meager.
Lemma 6.
Proof.
Let X be a complete metric space. If {Un } is a sequence of open dense sets, then ∩n Un is also dense.
V , the intersection V ∩ (∩n Un ) 6= ∅. This is a simple rescurisve
U1 is open and dense, ∃Br1 (x1 ) so that Br1 (x1 ) ⊂ V ∩ U1 . Since U2 is open and dense, ∃Br2 (x2 )
Br2 (x2 ) ⊂ Br1 (x) ∩ U2 ⊂ V ∩ (U1 ∩ U2 ). Continuing in this way, we get a nested sequence of balls. Since we
It is sucient to show that for any open set
argument. Since
so that
are in a complete metric space, by the nested sets property for complete spaces, there is a point in the intersection
of all the balls. (Another way to do this is to assume WOLOG that
rn → 0 and then the sequence xn
will be forced
to be Cauchy).
Theorem 7.
Proof.
A complete metric space is always second catergory or non-meager.
Suppose by contradiction that
c
An = E n .
X
This is a sequence of open sets. Moreover,
dense! But then by the lemma, the intersection
we have
∪n En 6= X
◦
X = ∪n En with En = ∅ for each n. Let
◦ c
= En
= ∅c = X . This means each An is
is 1st category and we can write
∩n An
An = En
c
is dense too, so
∩n An 6= ∅.
But then, by taking complements,
which is a contradiction.
Important Consequences:
Name
Banach-Schauder
Statement
Let
X, Y
Open Mapping
be Banach spaces and let
Suppose morevoer that
T
T ∈ B(X, Y )
is onto. Then
be a bounded linear map.
T
is an open map.
Theorem
T
k·k2 are two norms on a space X , and there is an m so that
k·k1 ≤ m k·k2 , then there exists M so that k·k1 ≥ M k·k2
Let X, Y be Banach spaces and let T : X → Y be linear. Let
Γ(T ) = {(x, T (x)) : x ∈ X} be the graph of T . Then T is continuous if and
only if Γ(T ) is closed.
Suppose X, Y are Banach spaces and (Tα )α∈Λ is a collection of bounded
linear maps. Let E = {x ∈ X : supα∈Λ kTα xk < ∞}. If E is 2nd category or
0
nonmenager, then supα∈Λ kTα xk < ∞. I.e. the Tα s are uniformly bounded.
By the Baire category theorem, it is enough to show E = X .
c
(Same setup as above) Let M = E = {x ∈ X : supα kTα xk = ∞}. Then
either M is empty or M is a dense Gδ set.
The Closed Graph
Theorem
Banach-Steinhaus
Uniform
Boundedness
Principle
(Slightly Stronger
version)
then
T −1
Corr
k·k1
to
Y
If
If
is a continuous linear bijection from
X
Corr
is continous too.
and
Theorem 8. (Banach-Steinhaus Uniform Boundedness Principle) Suppose X, Y are Banach spaces and (Tα )α∈Λ
is a collection of bounded linear maps. Let E = {x ∈ X : supα∈Λ kTα xk < ∞}. If E = X is all of X then
supα∈Λ kTα xk < ∞. I.e. the Tα0 s are uniformly bounded.
Proof.
−1
En = {x ∈ X : supα∈Λ kTα xk ≤ n} so that E = ∪n En . Notice also that En = ∩α kTα (·)k [0, n] is an
intersection of closed sets (because x → kTα xk is continuous), so En is closed. Since E is not 1st category, we know
Let
1
FUNCTIONAL ANALYSIS NOTES
that
E
2
cannot be written as a countable union of nowhere dense sets. Hence it must be the case that at least one
∃n0 so that En◦0 6= ∅. Hence ∃x0 , r so that Br (x0 ) ⊂ En0 .
For any x with kxk ≤ r now, notice that x0 + x ∈ Br (x0 ) ⊂ En0 . Hence for such x, we know by denition
that supα kTα (x0 + x)k ≤ n0 . Have then for any kxk ≤ r :
En
is not nowhere dense. In other words,
sup kTα xk
=
of
E n0
sup kTα (x0 + x) − Tα (x0 )k
α
α
≤ sup (kTα (x0 + x)k + kTα (x0 )k)
α
≤ n0 + n0 = 2n0
So by scaling, we conclude that for any
supα kTα k = supα supkxk=1 kTα xk ≤
2n0
r
x
with
kxk ≤ 1
that
supα kTα xk ≤
2n0
r .
Have nally then that
< ∞.
(The slightly stronger version) (Same setup as Banach-Steinhaus) Let M = E c = {x ∈ X : supα kTα xk < ∞}.
Then either M is empty or M is a dense Gδ set.
Theorem 9.
Let
Un = {x ∈ X : supα kTα xk > n}
so that
Un
=
M=
[
T
α
Un .
Notice that we can write:
{x ∈ X : kTα xk > n}
α
=
[
kTα (·)k
−1
(n, ∞)
α
−1
x → kTα xkTis continuous each set kTα (·)k (n, ∞) is open and we see from this that Un is a union
of open sets. Since M =
α Un , we see that M is a Gδ -set.
Claim: Either M is empty or Un is dense set for every n ∈ N.
Pf: It suces to show the following: if there is a single n0 for which Un0 is not dense, then M is empty. Suppose
c
Un0 is not dense. Then, by denition of dense, Un0 6= X . In other words this is Un0 6= ∅. Now, Un0 is a closed set,
c
so we know that Un0 is an open set. Hence, since this is a non-empty open set, we can nd x0 ∈ X and r > 0 so
c
that Br (x0 ) ⊂ Un0 .
c
/ Un0 . By denition of Un
Consider any x ∈ X with kxk ≤ r . Then x0 + x ∈ Br (x0 ) ⊂ Un0 . Hence x0 + x ∈
this means that supα kTα (x0 + x)k ≤ n. Using scaling and translation invariance, we have then that for any x with
kxk ≤ 1 that:
since the map
sup kTα xk
=
α
=
≤
≤
=
Finally then we see that the
Tα are
1
sup kTα (rx)k
r α
1
sup kTα (x0 + rx) − Tα (x0 )k
r α
1
sup (kTα (x0 + rx)k + kTα (x0 + 0)k)
r α
1
(n0 + n0 ) since krxk ≤ rand k0k ≤ r
r
2n0
r
uniformly bounded,
sup kTα k
=
α
α kxk=1
≤
This means that
M
sup sup kTα xk
2n0
<∞
r
is the empty set, because for every
x∈X
we have that
kTα xk ≤ supα kTα k kxk < ∞
x∈
/ M.
Combining the initial remarks and the claim we see that
and every
Un
M
is either empty or otherwise we have that
so
M = ∩n Un
is dense. Since the countable intersection of open dense sets is dense (this was the main lemma in
the pf of Baire's theorem), in the latter case we see that
M
is a dense
Gδ
set, as desired.
FUNCTIONAL ANALYSIS NOTES
3
The Hahn-Banach Theorem and Related
H-B = Hahn-Banach for the rest of this section.
In all the statements the set up is:
(X, k·k) ≡
A normed vector space over the eld
F
≡
The eld that
M
≡
A linear subspace of
p
≡
A sub-linear functional
q
≡
X is
F
over. Will either be
Ror C
X
p : X → R,
i.e.
psatises:
p(x + y) ≤ p(x) + p(y), p(ax) = ap(x) ∀a > 0.
A semi-norm
q : X → R,
i.e.
q satises:
q(x + y) ≤ q(x) + q(y), q(λx) = |λ|q(x) ∀λ ∈ C.
(Rmk: semi-norm is stricter than sub-linear functional)
`A
≡
A linear functional
`A ≤ p
≡
Shorthhand for:`A (x)
≤ p(x) ∀x ∈ A
`A ≤ q
≡
Shorthhand for:`A (x)
≤ q(x) ∀x ∈ A
ext.
`B `A
Name
Baby H-B
F?
R
Thm
≡ ”`B
extends
`A ",
`M : A → Fwhere Awill
` M ≤ p ; x0 ∈ X − M .
Dene M ⊕ x0 R =
{x + λx0 : x ∈ M, λ ∈ R}
X
A ⊂ B and `A (x) = `B (x)∀x ∈ A”
shorthand for "
Hypothesis
be some subspace of
Conclusion
Pf ext.
∃`M ⊕x0 R `M so
`M ⊕x0 R ≤ p
that
`M ⊕x0 R as long
`M ⊕x0 R (x0 ) to be in
Can nd the correct
we can dene
as
some particular interval. The fact
`M ≤ p
can be used to show that this
interval is non-empty.
Real H-B
R
ext.
`M ≤ p
∃`X `M so
`X ≤ p
Thm
that
Zorn's Lemma is used with the partial
ext.
ordering .
Every totally ordered
set has a maximal element by taking
the union of all the subspaces. By
Zorn's Lemma, there is a maximal
element. By the Baby H-B Thm, the
maximal element cannot be a proper
subspace of
Complex
C
|`M | ≤ q
H-B Thm
X.
ext.
∃`X `M so
|`X | ≤ q
that
Use the Real H-B Thm to prove this
one. Its a pretty unenlightening proof
involving manipulations with complex
numbers.
(Not
F
x0 ∈ X − {~0}
named)
?
∃` ∈ X s.t.
k`kX ? = 1,`(x0 ) = kxk
Apply the H-B thm on the space
M = {λx0 : λ ∈ F} with the functional
`M (λx0 ) := λ kx0 k and seminorm
q(x) = kxk.
ext.
is what we want. The
gives that
`X `M
ineq |`X | ≤ q
The extension
k`kX ? ≤ 1
by linearity. The
other inequality is clear by pluggin in
x0 .
Analytic
H-B Thm
F
`M ∈ M
?
?
ext.
∃`X ∈ X , `X `M
k`X kX ? = k`M kM ?
with
Let
q(x) = k`M kM ? kxk
be the
seminorm. Then apply H-B thm. The
|`X | ≤ q gives that
k`X kX ? ≤ k`M kM ? by linearity.
other ineq is clear since M ⊂ X
ineq
The
FUNCTIONAL ANALYSIS NOTES
Projection
∃` ∈ X ? s.t. `|M = 0 and
`(x0 ) = 1 and
k`kX ? = dist(x10 ,M )
x0 ∈ X − M̄ such that
dist(x0 , M ) > 0
F
H-B Thm
4
M1 = M ⊕ x0 F and dene
`M1 (x + λx0 ) = λ.
|λ|
k`M1 k = supx∈M,λ∈F kx+λx0 k =
1
1
supx∈M,λ∈F x +x
=
k λ 0 k dist(x0 ,M )
Let
Then use the Analytic H-B thm to
extend to
`X .
(Baby H-B Thm) Suppose`M ≤ p and that x0 ∈ X−M . Dene M ⊕x0 R = {x + λx0 : x ∈ M, λ ∈ R}.
ext.
Then ∃`M ⊕x0 R `M so that `M ⊕x0 R ≤ p.
Theorem 10.
Proof.
Suppose we found a value`(x0 ) that we liked a lot. Then we could dene:
`M ⊕x0 R (x + λx0 ) = `M (x) + λ`(x0 )
ext.
`M ⊕x0 R `M we want! Of course, since we want `M ⊕xR ≤ p,
`(x0 ) to obey the following inequalities:
`(x0 ), dene `M ⊕x0 R (x + λx0 ) = `M (x) + λ`(x0 ). Then:
and we would have found the functional
value of
`(x0 )
not just any
will do. We need
Claim 1: For a xed value
`M ⊕x0 R ≤ p ⇐⇒ ∀x ∈ M,
Pf: (⇒) Plug in
x ± x0
into
`M ⊕x0 R ≤ p,
get
`M (x) + `(x0 ) ≤ p(x + x0 )
`M (x) − `(x0 ) ≤ p(x − x0 )
`M ⊕x0 R (x ± x0 ) ≤ p(x ± x0 ).
and
By using the denition of
`M ⊕x0 R
on
the LHS, we get the desired inequalities.
(⇐)Let
x + λx0 ∈ M ⊕ x0 R be arbitrary. There are two cases, one where λ > 0 and one where λ ≤ 0.
±sign abusivly and writing λ = ±|λ|. Write:
We handle
both cases simultaneously by using the
`M ⊕x0 R (x + λx0 )
`M ⊕x0 R (x ± |λ| x0 )
x
= |λ| `M ⊕x0 R
± x0
|λ|
x
= |λ| `M
± ` (x0 ) by def 'n of `M ⊕x0 R
|λ|
x
± p(x0 ) by the hypothesis inequalities
≤ |λ| p
|λ|
= p (x ± |λ| x0 ) since pis a sublinear functional
=
=
So indeed,
`M ⊕x0 R ≤ p
To show that a value of
for all
p(x + λx0 )
`(x0 )
exists which satises the inequalities from Claim 1, we need the following to hold
x ∈ M:
`M (x) − p(x − x0 ) ≤ `(x0 ) ≤ p(x + x0 ) − `M (x)
It suces then to show that ∀x1 , x2 ∈ M that `M (x1 ) − p(x1 − x0 ) ≤ p(x2 + x0 ) − `M (x2 ).
consequence of `M ≤ p. Pluggin in x1 + x2 into `M ≤ p , we have:
`M (x1 ) + `M (x2 )
Indeed, this is a
= `M (x1 + x2 )
≤ p(x1 + x2 )
= p ((x1 − x0 ) + (x0 + x2 ))
≤ p(x1 − x0 ) + p(x0 + x2 )
Rearranging now gives the desired inequality.
Lemma 11. (Zorn's Lemma) A partial ordering on a set P is a relation ” ” that is reexive (a a), antisymetric
(a b,b a =⇒ a = b), and transitive (a b, b c =⇒ a c). Suppose that every totally ordered subset (i.e. a
set in which for every pair a, b either a b or b a), {aα }α∈Λ has an upper bound in P (i.e. an element a?,Λ ∈ P
so that aα a?,Λ for all α ∈ Λ). Then P contains at least one maximal element (i.e. an a? so that a a? for all
a ∈ P ).
Remark
.
12
This is equivalent to the axiom of choice, but the proof is non-trivial!
(Real H-B Theorem) Let X be a normed vector space over R, p a sublinear functional on X , M a
subspace, and `M : M → R a linear functional such that `M ≤ p (i.e.`M (x) ≤ p(x) ∀x ∈ M ). Then ∃`X a linear
ext.
functional that extends `X `M (i.e. `M (x) = `X (x)∀x ∈ M ) and `X ≤ p (i.e.`X (x) ≤ p(x) ∀x ∈ X )
Theorem 13.
FUNCTIONAL ANALYSIS NOTES
Proof.
Let
P =
n
o
ext.
`A : A → R : `A `M
5
be the space of all linear functions which are dened on subspaces
ext.
of
X.
Then
” ”
A
ext.
is a partial ordering on
ext.
Graph (`A )
⊃
(Rmk: one way to see this is to notice that
is inclusion of the
= {(x, f (x)), x ∈ Domain (f )}. Moreover,
{`Aα }α∈Λ is a totally ordered set, then dene
A?,Λ = ∪α∈Λ Aα and `A?,Λ (x) = `Aα (x) for x ∈ Aα . (This is well dened because {`Aα }α∈Λ is a totally ordered set).
Now by the conclusion of Zorn's lemma, there is a maximal element `A? for all of P . Now we claim that A? = X .
Indeed, if by contradiction, A? 6= X , then there is at least one element x0 ∈ X − A? . But now by the Baby
H-B Thm, we can get an extension `A? ⊕x0 R . But this contradicts the maximality of `A? in P ! So it must be that
A? = X .
graphs, that is
`A `B i
P
Graph
(`B )
where Graph(f )
every totally ordered subset has a maximum element in
P.
Namely, if
(Complex H-B Theorem) Let X be a normed vector space over C, q a seminorm on X , M a subspace,
and `M : M → C a linear functional such that |`M | ≤ q (i.e.|`M (x)| ≤ q(x) ∀x ∈ M ). Then ∃`X a linear functional
ext.
that extends `X `M (i.e. `M (x) = `X (x)∀x ∈ M ) and |`X | ≤ q (i.e.|`X (x)| ≤ q(x) ∀x ∈ X )
Theorem 14.
Proof.
(By manipulations using the Real H-B Thm) Let
`M = uM + ivM . uM and vM
R−linear actually!) Since `M is
are seen to be
actually
R−linear
C-linear,
uM (x) =
Re(`M (x)) and
functionals, because
`M
is
v(x) = Im(`M (x)) so
R-linear. (`M is more
that
than
we have that:
vM (x) = Im (`M (x)) = Re(−i`M (x)) = Re(`M (ix)) = uM (ix)
So then
`M (x) = uM (x) + iuM (ix)
ext.
we get a
uX uM
and
uX ≤ q(x).
uM . Now, q being a semi-norm, is also a
uM (x) ≤ |`M (x)| ≤ q(x). So applying the Real H-B Thm
can be entirely reconstructed from
sublinear map (which is a slightly looser condition), and
ext.
Now let
calculation early basically did this). Finally to
`X (x) = uX (x) + iuX (ix).
check that |`X | ≤ q have:
|`X (x)| = eiθ `X (x)
for some
One now veries that
`X `M
(our
θ
iθ
= `X (e x)
=
Re(`X (e
iθ
x))
since the LHS is real
iθ
= uX (e x)
≤ q(eiθ x)
iθ
since
|`X | ≤ q
= |e |q(x) = q(x)
since
q is
a seminorm.