Functional Analysis Lecture 4

[[lecture-data]]

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

1. Normed and Banach Spaces

Recall from last time

[[metabind-templates/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 $M={\bigcup }_{n\in N}{C}_n$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 ${\bigcup }_n{C}_n=M$ and all $C_n$ are nowhere dense.

$M$ that does not converge

we’ll show that there is a sequence in

Since $M$ contains an open ball but $C_1$ does not, we have $M\ne C_1$. Thus there is some $p_1\in M\setminus C_1$. Since $C_1$ is closed but $M\setminus C_1$ is open, this implies that $\exists {ϵ}_1>0$ s.t. $B(p_1,{ϵ}_1)\cap C_1=\varnothing$.

Now, $B\left(p_1,\frac{{ϵ}_1}{3}\right)\not\subset C_2$ means there exists some $p_2\in M\setminus C_2$. And since $C_2$ is closed, there exists $0<{ϵ}_2<\frac{{ϵ}_1}{3}$ such that $B(p_2,{ϵ}_2)\cap C_2=\varnothing$

Now, suppose there we have found $k$ such points. Then for each $j=1,\dots ,k$ we have $p_j\not\subset C_j$ with${ϵ}_j$ such that $B(p_j,{ϵ}_j)\cap C_j=\varnothing$

Since $B\left(p_k,\frac{{ϵ}_k}{3}\right)\not\subset C_{k+1}$, there exists some $p_{k+1}\in B\left(p_k,\frac{{ϵ}_k}{3}\right)$ such that $p_{k+1}\notin C_{k+1}$

Then there exists ${ϵ}_{k+1}<\frac{{ϵ}_k}{3}$ such that $B(p_{k+1},{ϵ}_{k+1})\cap C_{k+1}=\varnothing$

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 $k$,

1. $p_j\in B\left(p_{j-1},\frac{{ϵ}_{j-1}}{3}\right)$
2. $B(p_j,{ϵ}_j)\cap C_j=\varnothing$

This sequence $\{p_k{\}}_k$ is Cauchy because for all $k,\ell \in \mathbb{N}$, we have \begin{align} d(p_{k+1}, p_{k+\ell}) &\leq d(p_{k}, p_{k+1}) + d(p_{k+1}, p_{k+2}) + \dots + d(p_{k+\ell-1}, p_{k+\ell}) \\ &< \frac{\epsilon_{k}}{3} + \frac{\epsilon_{k+1}}{3} + \dots+\frac{\epsilon_{k+\ell-1}}{3} \\ &< \frac{\epsilon_{1}}{3^k}+\frac{\epsilon_{1}}{3^{k+1}}+\dots+\frac{\epsilon_{1}}{3^{k+\ell}} \\ &< \epsilon_{1} \sum_{n=k}^\infty \frac{1}{3^k} \\ &= \frac{\epsilon_{1}}{3^k}\left( \frac{1}{1-\frac{1}{3}} \right) \\ &= \frac{\epsilon_{1}}{2} 3^{-k+1} \end{align}$$ Since M$iscomplete,thereexistssome$p \in M$suchthat$p_{k} \to p$as$k \to \infty$.

Now, for all $k\in \mathbb{N}$, we have \begin{align} d(p_{k+1}, p_{k+\ell+1}) &< \epsilon_{k+1} \left[ \frac{1}{3} + \frac{1}{3^2} + \dots+\frac{1}{3^\ell} \right] \\ &< \epsilon_{k+1} \sum_{n=1}^\infty 3^{-n} \\ &= \frac{\epsilon_{k+1}}{2} \\ \implies \lim_{ \ell \to \infty } d(p_{k+1}, p_{l+\ell+1}) &=d(p_{k+1}, p) \\ &\leq \frac{\epsilon_{k+1}}{2} \\ &< \frac{\epsilon_{k}}{6} \\ \end{align}$$ So as \ell\to \infty$,wehave$d(p_{k+1}, p)$$

\implies d(p_{k}, p) \leq d(p_{k}, p_{k+1}) + d(p_{k+1}, p) \leq \frac{1}{3}\epsilon + \frac{1}{6}\epsilon_{k} < \epsilon_{k}$$ ie, p \in B(p_{k}, \epsilon_{k}) = B_{k}$\ast forall\ast$k$.Andeachoftheseballshas$B_{k} \cap C_{k} = \varnothing$.Thus$p$isnotinanyofthe$C_{k}$ie$p \not\in\bigcup_{k}C_{k} = M\implies \impliedby $\tag*{$\blacksquare$}

References

functional

Mentions

Mentions

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WHERE !direct_source
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Open Mapping Theorem

Let $B_1,B_2$ be two Banach spaces and let $T\in \mathcal{B}(B_1,B_2)$ 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

$B(0,1)=\{b\in B_1:||b||<1\}$, 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 $B(T(b_1),ϵ)\subset T(U)$

We will first prove that if

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 $B_2={\bigcup }_{n\in \mathbb{N}}\overline{T(B(0,n))}$

So by the Baire Category Theorem, there exists some $n_0\in \mathbb{N}$ such that $\overline{T(B(0,n_0))}$ contains an open ball.

Now, since $T$ is a linear operator, $n_0\overline{T(B(0,1))}$ contains an open ball. ie, $\overline{T(B(0,1))}$ contains an open ball. Equivalently, there exists some $v_0\in B_2$ and $r>0$ such that $B(v_0,4r)\subset \overline{T(B(0,1))}$.

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 $||v_0-v_1||<2r$

$4$ here to make some calculations easier later

we choose

Then $B(v_1,2r)\subset B(v_0,4r)\subset \overline{T(B(0,1))}$ If $||v||<r$, then $\frac{1}{2}(2v+v_1)\in \frac{1}{2}B(v_1,2r)\subset \frac{1}{2}\overline{T(B(0,1))}=\overline{T\left(B\left(0,\frac{1}{2}\right)\right)}$ And this means

v =-T\left( \frac{u_{1}}{2} \right) + \frac{1}{2}(2v + v_{1}) &\in -T\left( \frac{u_{1}}{2} \right) +\overline{T\left( B\left( 0, \frac{1}{2} \right) \right)} \\ &= \overline{T\underbrace{\left( -\frac{u_{1}}{2} + B\left( 0, \frac{1}{2} \right) \right)}_{\subset B(0,1) \text{ since } u_{1} \in B(0, 1)}}\\ &\subset \overline{T(B(0,1))} \\ \end{align}

ie, (in $B_2$), we have $B(0,r)\subset \overline{T(B(0,1))}$. This implies that $B(0,2^{-n}r)=2^{-n}B(0,r)\subset 2^{-n}\overline{T(B(0,1))}=\overline{T(B(0,2^{-n}))}$ for all $n\in \mathbb{N}$

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

Showing that the image $T(B(0,1))$ contains an open ball $B\left(0,\frac{r}{2}\right)$

Let $||v||<\frac{r}{2}$. Then $v\in \overline{T\left(B\left(0,\frac{1}{2}\right)\right)}$ . This implies there exists some $b_1\in B\left(0,\frac{1}{2}\right)$ such that $||v-T{b}_1||<\frac{r}{4}$

ie, $v-T{b}_1\in \overline{T\left(B\left(0,\frac{1}{2}\right)\right)}⟹$ there is some $b_2\in B\left(0,\frac{1}{4}\right)$ 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 $\{b_k{\}}_{k\in \mathbb{N}}\subset B_1$ such that

1. $||b_k||<2^{-k}$
2. $\left|\left|v-{\sum }_{k=1}^{n}T{b}_k\right|\right|<2^{-n-1}r$

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 $||b||={\lim }_{n\to \infty }\left|\left|{\sum }_{k=1}^n{b}_k\right|\right|\le {\lim }_{n\to \infty }{\sum }_{k=1}^{\infty }||b_k||$ 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 $Tb={\lim }_{n\to \infty }T\left({\sum }_{k=1}^n{b}_k\right)={\lim }_{n\to \infty }{\sum }_{k=1}^{\infty }T{b}_k=v$ ie, $v\in T(B(0,1))$. Thus $B\left(0,\frac{r}{2}\right)\subset T(B(0,1))$ 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 $U\subset B_1$ 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 $\delta >0$ such that $B(0,\delta )\subset T(B(0,1))$. ie

Suppose

B(b_{2}, \epsilon\delta) &= b_{2}+\epsilon B(0, \delta) \\ &\subset b_{2} + \epsilon T(B(0, 1)) \\ &= T(b_{1}) + \epsilon T(B(0, 1)) \\ &= T(b_{1} + \epsilon B(0, 1)) \\ &= T(b_{1} + B(0, \epsilon)) \\ &\subset T(U) \end{align}$$ Since $b_{1}+B(0, \epsilon)$ is contained in $U$. Thus we've found a ball around any $b_{2} \in T(U)$.

\tag*{$\blacksquare$}

(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 $T^{-1}\in \mathcal{B}(B_2,B_1)$ 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 ${\mathbb{R}}^2$)

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 $T\in \mathcal{B}(B_1,B_2)⟺\Gamma (T):=\{(u,Tu):u\in B_1\}\subset B_1\times B_2\text{ is closed}$

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 $T{u}_n\to Tu$

Proof

Proof $(⟹)$ Suppose $T\in \mathcal{B}(B_1,B_2)$. Let $\{(u_n,T{u}_n){\}}_n$ be a sequence in $\Gamma (T)$ such that $u_n\to u$ and $T{u}_n\to v$. Then $v={\lim }_{n\to \infty }T{u}_n=T({\lim }_{u\to \infty }{u}_n)=Tu$ since $T$ is continuous. Thus $(u,v)=(u,Tu)\in \Gamma (T)$ ie $\Gamma (T)$ is closed.

Proof $(⟸)$ Define

• ${\pi }_1:\Gamma (T)\to B_1,{\pi }_1(u,Tu)=u$
• ${\pi }_2:\Gamma (T)\to B_2,{\pi }_2(u,Tu)=Tu$

Note that $\Gamma (T)$ is a Banach space since $\Gamma (T)\subset B_1\times B_2$ is a closed subspace of $B_1\times B_2$ (which is Banach by the corrolary above)

Further, we have ${\pi }_1\in \mathcal{B}(\Gamma (T),B_1)$ and ${\pi }_2\in \mathcal{B}(\Gamma (T),B_2)$ $||{\pi }_2(u,v)||=||v||\le ||u||+||v||=||(u,v)||$ (we can see it is bounded by the norm of $B_1\times B_2$ as we defined above, which must also be bounded)

${\pi }_1:\Gamma (T)\to B_1$ is one-to-one and onto (bijective). Thus $S:={\pi }_1^{-1}:B_1\to \Gamma (T)$ is also a bounded linear operator

And so $T={\pi }_2\circ S$ is the composition of two bounded linear operators, and is thus itself a bounded linear operator.

ie $T\in \mathcal{B}(B_1,B_2)$

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 $\underset{\mathcal{B}(V,\mathbb{K})}{\underbrace{V^{\prime }}}=\{0\}$?

(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

• $({\ell }^p{)}^{\prime }={\ell }^{p^{\prime }}$ where $\frac{1}{p}+\frac{1}{p^{\prime }}=1$ for $1\le p<\infty$
• $(c_0{)}^{\prime }={\ell }^1$

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

1. for all $e\in E$, $e⪯e$
2. for all $e,f\in E$, if $e⪯f$ and $f⪰e$ then $e=f$
3. 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 $e⪯f⟹ef$

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 chain in a nonempty, 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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