Functional Analysis Lecture 4

[[lecture-data]]

← Lecture 3 | Lecture 5 →
Lecture Notes: Rodriguez, page 18

1. Normed and Banach Spaces

Recall from last time

concept

Baire Category Theorem

Let $M$ be a {as||complete} [[metric space]] and let ${C_{n}}_{n}$ be a {as||collection of closed subsets} of $M$ such that $M = ⋃_{n \in N} C_{n}$ Then {ha||at least one} of the $C_{n}$ {ha||contain an open ball}
{ha|| $B (x, r) = {y \in M : d (x, y) < r}$ }
(ie, {ha||at least one} $C_{n}$ has {ha||an interior point}).

[!NOTE]
A set that does not contain an interior point is called nowhere dense. Sometimes when we apply this theorem, we do not need $C_{n}$ to be closed. Then the result is that their closures must contain and open ball. ie, we cannot have all $C_{n}$ be all nowhere dense.

Note

We can use this theorem to prove that there exists a continuous function that is nowhere differentiable

Proof (by contradiction)

Suppose BWOC that there is some collection of closed subsets of $M$ such that $⋃_{n} C_{n} = M$ and all $C_{n}$ are nowhere dense.

Note

we'll show that there is a sequence in $M$ that does not converge

Since $M$ contains an open ball but $C_{1}$ does not, we have $M \neq C_{1}$ . Thus there is some $p_{1} \in M ∖ C_{1}$ . Since $C_{1}$ is closed but $M ∖ C_{1}$ is open, this implies that $\exists ϵ_{1} > 0$ s.t. $B (p_{1}, ϵ_{1}) \cap C_{1} = \emptyset$ .

Now, $B (p_{1}, \frac{ϵ_{1}}{3}) ⊄ C_{2}$ means there exists some $p_{2} \in M ∖ C_{2}$ . And since $C_{2}$ is closed, there exists $0 < ϵ_{2} < \frac{ϵ_{1}}{3}$ such that $B (p_{2}, ϵ_{2}) \cap C_{2} = \emptyset$

Now, suppose there we have found $k$ such points. Then for each $j = 1, \dots, k$ we have $p_{j} ⊄ C_{j}$ with $ϵ_{j}$ such that $B (p_{j}, ϵ_{j}) \cap C_{j} = \emptyset$

Since $B (p_{k}, \frac{ϵ_{k}}{3}) ⊄ C_{k + 1}$ , there exists some $p_{k + 1} \in B (p_{k}, \frac{ϵ_{k}}{3})$ such that $p_{k + 1} \notin C_{k + 1}$

Then there exists $ϵ_{k + 1} < \frac{ϵ_{k}}{3}$ such that $B (p_{k + 1}, ϵ_{k + 1}) \cap C_{k + 1} = \emptyset$

Thus, by indiction, we have found a sequence of points ${p_{k}}_{k} \subset M$ and $ϵ_{k} \in (0, ϵ_{1})$ such that for all $k$ ,

$p_{j} \in B (p_{j - 1}, \frac{ϵ_{j - 1}}{3})$
$B (p_{j}, ϵ_{j}) \cap C_{j} = \emptyset$

This sequence ${p_{k}}_{k}$ is Cauchy because for all $k, ℓ \in N$ , we have
$\begin{aligned} d (p_{k + 1}, p_{k + ℓ}) & \leq d (p_{k}, p_{k + 1}) + d (p_{k + 1}, p_{k + 2}) + \dots + d (p_{k + ℓ - 1}, p_{k + ℓ}) \\ < \frac{ϵ_{k}}{3} + \frac{ϵ_{k + 1}}{3} + \dots + \frac{ϵ_{k + ℓ - 1}}{3} \\ < \frac{ϵ_{1}}{3^{k}} + \frac{ϵ_{1}}{3^{k + 1}} + \dots + \frac{ϵ_{1}}{3^{k + ℓ}} \\ < ϵ_{1} \sum_{n = k}^{\infty} \frac{1}{3^{k}} \\ = \frac{ϵ_{1}}{3^{k}} (\frac{1}{1 - \frac{1}{3}}) \\ = \frac{ϵ_{1}}{2} 3^{- k + 1} \end{aligned}$
Since $M$ is complete, there exists some $p \in M$ such that $p_{k} \to p$ as $k \to \infty$ .

Now, for all $k \in N$ , we have
$\begin{aligned} d (p_{k + 1}, p_{k + ℓ + 1}) & < ϵ_{k + 1} [\frac{1}{3} + \frac{1}{3^{2}} + \dots + \frac{1}{3^{ℓ}}] \\ < ϵ_{k + 1} \sum_{n = 1}^{\infty} 3^{- n} \\ = \frac{ϵ_{k + 1}}{2} \\ ⟹ lim_{ℓ \to \infty} d (p_{k + 1}, p_{l + ℓ + 1}) & = d (p_{k + 1}, p) \\ \leq \frac{ϵ_{k + 1}}{2} \\ < \frac{ϵ_{k}}{6} \end{aligned}$
So as $ℓ \to \infty$ , we have
$d (p_{k + 1}, p)$

$⟹ d (p_{k}, p) \leq d (p_{k}, p_{k + 1}) + d (p_{k + 1}, p) \leq \frac{1}{3} ϵ + \frac{1}{6} ϵ_{k} < ϵ_{k}$
ie, $p \in B (p_{k}, ϵ_{k}) = B_{k}$ for all $k$ . And each of these balls has $B_{k} \cap C_{k} = \emptyset$ . Thus $p$ is not in any of the $C_{k}$ ie $p \notin ⋃_{k} C_{k} = M$
$⟹ ⟸$
$\tag*{$ \blacksquare $}$

References

#flashcards/math/functional

Mentions

File	Last Modified
Functional Analysis Lecture 3	2025-10-02
Functional Analysis Lecture 4	2025-10-02
open mapping theorem	2025-10-02
uniform boundedness theorem	2025-10-02

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Open Mapping Theorem

Let $B_{1}, B_{2}$ be two Banach spaces and let $T \in B (B_{1}, B_{2})$ be surjective. Then $T$ is an open map.

ie for all open subsets $U \subset B_{1}$ , we have $T (U) \subset B_{2}$ is open.

Proof

Note

We will first prove that if $B (0, 1) = {b \in B_{1} : | | b | | < 1}$ , then $T (B (0, 1))$ contains an open ball in $B_{2}$ centered at $0$ . Then, since $T$ is linear, we can scale everything properly to show that for any $b_{1} \in U \subset B_{1}$ open, there exists $ϵ > 0$ such that $B (T (b_{1}), ϵ) \subset T (U)$

Showing that the closure of the image

T (B (0, 1))

contains an open ball

B (0, r)

$T$ surjective means everything in $B_{2}$ gets mapped to by something in $B_{1}$ ie

B_{2} = ⋃_{n \in N} \overset{―}{T (B (0, n))}

So by the [[Baire Category Theorem]], there exists some $n_{0} \in N$ such that $\overset{―}{T (B (0, n_{0}))}$ contains an open ball.

Now, since $T$ is a linear operator, $n_{0} \overset{―}{T (B (0, 1))}$ contains an open ball. ie, $\overset{―}{T (B (0, 1))}$ contains an open ball. Equivalently, there exists some $v_{0} \in B_{2}$ and $r > 0$ such that $B (v_{0}, 4 r) \subset \overset{―}{T (B (0, 1))}$ .

ie, we can pick some $u_{1} \in B (0, 1)$ such that $v_{1} = T (u_{1}) \in T (B (0, 1))$ such that $| | v_{0} - v_{1} | | < 2 r$

Note

we choose $4$ here to make some calculations easier later

Then

B (v_{1}, 2 r) \subset B (v_{0}, 4 r) \subset \overset{―}{T (B (0, 1))}

If $| | v | | < r$ , then

\frac{1}{2} (2 v + v_{1}) \in \frac{1}{2} B (v_{1}, 2 r) \subset \frac{1}{2} \overset{―}{T (B (0, 1))} = \overset{―}{T (B (0, \frac{1}{2}))}

And this means

\begin{aligned} v = - T (\frac{u_{1}}{2}) + \frac{1}{2} (2 v + v_{1}) & \in - T (\frac{u_{1}}{2}) + \overset{―}{T (B (0, \frac{1}{2}))} \\ = \overset{―}{T \underset{\subset B (0, 1) since u_{1} \in B (0, 1)}{\underset{⏟}{(- \frac{u_{1}}{2} + B (0, \frac{1}{2}))}}} \\ \subset \overset{―}{T (B (0, 1))} \end{aligned}

ie, (in $B_{2}$ ), we have $B (0, r) \subset \overset{―}{T (B (0, 1))}$ . This implies that

B (0, 2^{- n} r) = 2^{- n} B (0, r) \subset 2^{- n} \overset{―}{T (B (0, 1))} = \overset{―}{T (B (0, 2^{- n}))}

for all $n \in N$

Now we prove that $B (0, \frac{r}{2}) \subset T (B (0, 1))$ .

Showing that the image

T (B (0, 1))

contains an open ball

B (0, \frac{r}{2})

Let $| | v | | < \frac{r}{2}$ . Then $v \in \overset{―}{T (B (0, \frac{1}{2}))}$ . This implies there exists some $b_{1} \in B (0, \frac{1}{2})$ such that $| | v - T b_{1} | | < \frac{r}{4}$

ie, $v - T b_{1} \in \overset{―}{T (B (0, \frac{1}{2}))} ⟹$ there is some $b_{2} \in B (0, \frac{1}{4})$ such that

| | v - T b_{1} - T b_{2} | | = | | v - T (b_{1} - b_{2}) | | < \frac{r}{4} \cdot \frac{1}{2} = \frac{r}{8}

continuing in this manner, we obtain a sequence ${b_{k}}_{k \in N} \subset B_{1}$ such that

$| | b_{k} | | < 2^{- k}$
$| | v - \sum_{k = 1}^{n} T b_{k} | | < 2^{- n - 1} r$

The series $\sum_{k = 1}^{\infty} b_{k}$ is absolutely summable (since the norm of the sums are bounded). And since $B_{1}$ is Banach, this means there exists some $b \in B_{1}$ such that $b = \sum_{k = 1}^{\infty} b_{k}$

Further, by the [[triangle inequality]], we have

| | b | | = lim_{n \to \infty} | | \sum_{k = 1}^{n} b_{k} | | \leq lim_{n \to \infty} \sum_{k = 1}^{\infty} | | b_{k} | |

And we can bound this

= \sum_{k = 1}^{\infty} | | b_{k} | | < \sum_{k = 1}^{\infty} 2^{- k} = 1

And since $T$ is a bounded linear operator (and thus continuous), we have

T b = lim_{n \to \infty} T (\sum_{k = 1}^{n} b_{k}) = lim_{n \to \infty} \sum_{k = 1}^{\infty} T b_{k} = v

ie, $v \in T (B (0, 1))$ . Thus $B (0, \frac{r}{2}) \subset T (B (0, 1))$ in $B_{2}$ .

Thus, if $0$ was an interior point of $U$ in the domain, then it is still an interior point in the image of $U$ .

Generalizing to any open set

Suppose $U \subset B_{1}$ is open and $b_{2} = T b_{1} \in T (U)$ . Then there exists some $ϵ > 0$ such that $b_{1} + B (0, ϵ) = B (b_{1}, ϵ) \subset U$ . And, from above, there exists some $δ > 0$ such that $B (0, δ) \subset T (B (0, 1))$ . ie

\begin{aligned} B (b_{2}, ϵ δ) & = b_{2} + ϵ B (0, δ) \\ \subset b_{2} + ϵ T (B (0, 1)) \\ = T (b_{1}) + ϵ T (B (0, 1)) \\ = T (b_{1} + ϵ B (0, 1)) \\ = T (b_{1} + B (0, ϵ)) \\ \subset T (U) \end{aligned}

Since $b_{1} + B (0, ϵ)$ is contained in $U$ . Thus we've found a ball around any $b_{2} \in T (U)$ .

\begin{matrix} ◼ \end{matrix}

(see [[open mapping theorem]])

Corollary

If $B_{1}, B_{2}$ are Banach spaces and $T : B_{1} \to B_{2}$ is a bijective bounded linear operator, then $T^{- 1} \in B (B_{2}, B_{1})$ is also.

Proof

$T^{- 1}$ is continuous $⟺$ for all $U \subset B_{1}$ open, $(T^{- 1})^{- 1} (U) = T (U)$ is open. And this is true by the [[open mapping theorem]]

see [[bijective bounded linear operators have bounded linear inverses]]

Corollary

If $B_{1}, B_{2}$ are Banach spaces, then $B_{1} \times B_{2}$ with [[norm]]

| | (b_{1}, b_{2}) | | := | | b_{1} | | + | | b_{2} | |

is a [[Banach space]].

Proof

Exercise

just need to check all the definitions for the norm
check that Cauchy sequences in $B_{1} \times B_{2}$ consist of Cauchy sequences in each of $B_{1}$ and $B_{2}$ (similar to completeness of $R^{2}$ )

see [[the cartesian product of banach spaces is banach]]

Closed Graph Theorem

If $B_{1}, B_{2}$ are Banach spaces and $T : B_{1} \to B_{2}$ is a linear operator, then

T \in B (B_{1}, B_{2}) ⟺ Γ (T) := {(u, T u) : u \in B_{1}} \subset B_{1} \times B_{2} is closed

Note

this may be easier to prove than proving something is a bounded linear operator. Normally, we need to show that sequences $u_{n} \to u$ have $T u_{n} \to T u$

Proof

(⟹)

Suppose $T \in B (B_{1}, B_{2})$ . Let ${(u_{n}, T u_{n})}_{n}$ be a sequence in $Γ (T)$ such that $u_{n} \to u$ and $T u_{n} \to v$ . Then

v = lim_{n \to \infty} T u_{n} = T (lim_{u \to \infty} u_{n}) = T u

since $T$ is continuous. Thus

(u, v) = (u, T u) \in Γ (T) $ $ i e $ Γ (T) $ i s c l o s e d .

Proof

(⟸)

Define

$π_{1} : Γ (T) \to B_{1}, π_{1} (u, T u) = u$
$π_{2} : Γ (T) \to B_{2}, π_{2} (u, T u) = T u$

Note that $Γ (T)$ is a Banach by the [[corrolary above]])

Further, we have $π_{1} \in B (Γ (T), B_{1})$ and $π_{2} \in B (Γ (T), B_{2})$

| | π_{2} (u, v) | | = | | v | | \leq | | u | | + | | v | | = | | (u, v) | |

(we can see it is bounded by the norm of $B_{1} \times B_{2}$ as we defined above, which must also be bounded)

$π_{1} : Γ (T) \to B_{1}$ is one-to-one and onto (bijective). Thus $S := π_{1}^{- 1} : B_{1} \to Γ (T)$ is also a bounded linear operator

And so $T = π_{2} \circ S$ is the composition of two bounded linear operators, and is thus itself a bounded linear operator.

ie $T \in B (B_{1}, B_{2})$

see [[closed graph theorem]]

Note

[[closed graph theorem]] $⟺$ [[open mapping theorem]]

Hahn Banach Theorem

See [[Hahn-Banach theorem]]

This theorem tries to answer the question

Question

Given a general, nontrivial, normed space $V$ , does $\underset{B (V, K)}{\underset{⏟}{V^{'}}} = {0}$ ?

(recall that we defined the [[dual space]] in the last lecture, and we define these specific duals to be "functionals")

Example

Spaces with nontrivial dual spaces

$(ℓ^{p})^{'} = ℓ^{p^{'}}$ where $\frac{1}{p} + \frac{1}{p^{'}} = 1$ for $1 \leq p < \infty$
$(c_{0})^{'} = ℓ^{1}$

In general, we want to know if we have elements in the dual space.

framework / axioms from set theory

Partial Order

A partial order on a set $E$ is a relation $⪯$ $E$ such that

for all $e \in E$ , $e ⪯ e$
for all $e, f \in E$ , if $e ⪯ f$ and $f ⪰ e$ then $e = f$
for all $e, f, g \in E$ , if $e ⪯ f$ and $f ⪯ g$ , then $e ⪯ g$

see [[partial order]]

Upper Bound

An upper bound of a set $D \subset E$ is an element $e \in E$ such that $d ⪯ e$ for all $d \in D$ .

see [[upper bound]]

Maximal Element

A maximal element of $E$ is an element $e$ such that if $f \in E$ and $e ⪯ f ⟹ e f$

see [[maximal element]]

Similar definitions for a minimal element

Chain

If $(E, ⪯)$ is a partially ordered set, a chain in $E$ is a set $C \subset E$ if for all $e, f \in C$ we have either $e ⪯ f$ or $f ⪯ e$

ie we can compare all the elements in the chain $C$

see [[chain]]

Hamel basis

A Hamel basis $H \subset V$ is a linearly independent set such that every element of $V$ is a finite linear combination of elements in $H$ .

see [[Hamel basis]]

Note

We know from linear algebra that we can find a basis and calculate the dimension of the space. Next time, we'll see how to apply these to infinite vector spaces.

Zorn's Lemma

If every partially ordered set $E$ has an [[upper bound]], then $E$ contains a [[maximal element]].

Without Proof

see [[Zorn's lemma]]

Note

We take this as an axiom of set theory and it can be used to prove the [[Axiom of Choice]]
We can also show that every [[vector space]] has a [[Hamel basis]]

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