Functional Analysis Lecture 2

[[lecture-data]]

← Lecture 1 | Lecture 3 → Lecture Notes: Rodriguez, page 7

1. Normed and Banach Spaces

Summable, Absolutely Summable

Let $\{v_n\}$ be a seq of points in vector space $V$. The the series ${\sum }_n{v}_n$ is summable if ${\left\{{\sum }_{n=1}^m{v}_n\right\}}_{m=1}^{\infty }$ converges.

The series is absolutely summable if ${\left\{{\sum }_{n=1}^m||v_n||\right\}}_{m=1}^{\infty }$ converges.

see summable series

Theorem

Let $V$ be a vector space. If $\{{\sum }_n{v}_n{\}}_n\subset V$ is absolutely summable, then the sequence of partial sums ${\left\{{\sum }_{n=1}^m{v}_n\right\}}_{m=1}^{\infty }$ is Cauchy.

see absolutely summable series have Cauchy partial sums

Exercise

Prove it! (same proof as when $V=\mathbb{R}$)

Theorem

A normed vector space $V$ is a Banach space if and only if every absolutely summable series is summable.

Proof

$(⟹)$ Suppose $V$ is a Banach space. If ${\sum }_n{v}_n$ is absolutely summable, then the sequence of partial sums is a Cauchy sequence in $V$ (see the previous theorem). Thus ${\left\{{\sum }_{n=1}^m{v}_n\right\}}_m$ converges in $V$. Thus by definition, the series is summable.

$(⟸)$ Suppose every absolutely summable series is summable. Let $\{v_n\}$ be a Cauchy sequence in $V$. We show that a subsequence of $\{v_n\}$ converges in $V$. Then $\{v_n{\}}_m$ converges by metric space theory.

$\{v_n{\}}_m$ is Cauchy $⟹$ for all $k\in \mathbb{N}$, there exists $N_k\in \mathbb{N}$ such that for all $n,m\ge N_k$, we have $||v_n-v_m||<\frac{1}{2^k}$ (we choose this expression because it is summable). Now, define $n_k=N_1+N_2+\cdots +N_k$ Then $n_1<n_2<\dots$ and for all $k$, we have $n_k\ge N_k$. Thus, for all $k\in \mathbb{N}$, we have $||v_{n_{k+1}}-v_{n_k}||<\frac{1}{2^k}$ Thus

&\sum_{k} (v_{n_{k+1}} - v_{n_{k}}) \text{ is absolutely summable} \\ \implies &\sum_{k}(v_{n_{k+1}} - v_{n_{k}}) \text{ is summable} \\ \implies &\left\{ \sum_{k=1}^m (v_{n_{k+1}} - v_{n_{k}}) \right\}_{m=1}^\infty = \{ v_{n_{m+1} - v_{n_{1}}} \}_{m=1}^\infty \text{ converges} \end{align}$$ Thus the (sub)sequence is $\{ v_{n_{m+1}} \}_{m=1}^\infty$ converges in $V$. Thus $V$ is [[Banach space|Banach]]. $\blacksquare$

see banach spaces have all absolutely summable series are summable

Operators and Functionals

These are the analog of matrices.

Example (important!)

Let $K:[0,1]\times [0,1]\to \mathbb{C}$ be a continuous function. Then for any function $f\in C([0,1])$, we can define $Tf(x)={\int }_0^{1}K(x,y)f(y)dy$

Then $Tf\in C([0,1])$ and for all ${\lambda }_1,{\lambda }_2\in \mathbb{C}$ and $f_1,f_2\in C([0,1])$ $T({\lambda }_1{f}_1+{\lambda }_2{f}_2)={\lambda }_{1}T{f}_1+{\lambda }_{2}T{f}_2$

Linear, linear operator

Let $V,W$ be vector spaces. We say a map $T:V\to W$ is linear if for all ${\lambda }_1,{\lambda }_2\in \mathbb{K}$ and $v_1,v_2\in V$, $T({\lambda }_1{v}_1+{\lambda }_2{v}_2)={\lambda }_{1}T{v}_1+{\lambda }_{2}T{v}_2$

We call the map $T$ a linear operator (in the past, may have been called “linear transformation”)

see linear function

Continuous

Recall that $T:V\to W$ (a map) is continuous on $V$ if for all $v\in V$ and for all sequences $\{v_n\}$ with $v_n\to v$, then $T_{v_n}\to T_v$.

Or, equivalently,

For all open $U\subseteq W$, the set $T^{-1}(U)=\{v\in V|T_v\in U\}\subseteq V$ is open in $V$.

see continuous map

Recall that linear functions with finite dimensional domain are continuous. However, this is not always the case in infinite dimensions!

Theorem

A linear operator $T:V\to W$ is continuous if there exists $C>0$ such that for all $v\in V$, $||Tv|{|}_W\le C||v|{|}_V$ If this holds, we call $T$ a bounded linear operator.

see continuity for linear functions (aka bounded linear operator)

NOTE

this does not means the image of $T$ is bounded. it means bounded subsets of $V$ are sent to bounded subsets of $W$.

Proof

$(⟹)$ Suppose $||Tv|{|}_W\le C||v|{|}_V$. Let $v\in V$ and suppose $v_n\to v$. Then $0\le ||T{v}_n-Tv|{|}_W=||T(v_n-v)|{|}_W\le C||v_n-v|{|}_V$

And since $||v_n-v|{|}_V\to 0$, by the squeeze theorem, we have $||T{v}_n-Tv|{|}_W\to 0$ as well.

$(⟸)$ Suppose $T$ is continuous. Then $T^{-1}(B_W(0,1))=\{v\in V:Tv\in B_W(0,1)\}$ Since $T0=0$, we must have $0\in T^{-1}(B_W(0,1))$. And since these are open sets, we can find $r>0$ such that $B_V(0,r)\subset T^{-1}(B_W(0,1))$.

Now, take $C=\frac{2}{r}$. Then $\frac{r}{2||v|{|}_V}v\in B_V(0,r)$ $⟹T\left(\frac{r}{2||v|{|}_V}v\right)\in B_W(0,1)$ Thus, by homogeneity of the norm, we have ${\left|\left|T\left(\frac{r}{2||v|{|}_V}v\right)\right|\right|}_W<1⟹||Tv|{|}_W\le \frac{2}{r}||v|{|}_V$

NOTE

From now on, we will drop the norm subscripts (see Matrix Analysis Lecture 25). The appropriate norm should be determined on context

Example

The linear operator $T:C([0,1])\to C([0,1])$ defined in the example above is a bounded linear operator (and therefore continuous)

illustration

We can then verify. Let $f\in C([0,1])$ then

\lvert Tf(x) \rvert &= \left\lvert \int _{0}^1 K(x,y)f(y) \, dy \right\rvert \\ &\leq \int _{0}^1 \lvert K(x,y) \rvert \cdot \lvert f(y) \rvert \, dy \\ &\leq \int _{0}^1 \lvert K(x,y) \rvert \cdot \lvert \lvert f \rvert \rvert_{\infty} \, dy \\ &\leq \int_{0}^1 \lvert \lvert K(x,y) \rvert \rvert \cdot \lvert \lvert f \rvert \rvert_\infty \, dy \\ &= \lvert \lvert K(x,y) \rvert \rvert \cdot \lvert \lvert f \rvert \rvert_{\infty} \end{align}$$ And since this bound holds for all $x$, it holds for the supremum. Thus $$\lvert \lvert Tf \rvert \rvert_{x} \leq \lvert \lvert K \rvert \rvert_\infty \lvert \lvert f \rvert \rvert_{\infty} $$ Thus we can use $C = \lvert \lvert K \rvert \rvert_{\infty}$ to show boundedness/continuity.

NOTE

we refer to $K$ as a kernel

Bounded Linear Operator space

Let $V$ and $W$ be normed vector spaces. The set of bounded linear operators from $V$ to $W$ is $\mathcal{B}(V,W)$

Operator Norm

The operator norm of an operator $T\in \mathcal{B}(V,W)$ is defined as $||T||={\sup }_{||v||=1,v\in V}||Tv||$

see operator norm

Let’s verify that this is indeed a norm

Theorem

The operator norm is an norm, so the space of bounded linear operators is a normed vector space.

Proof

definiteness Consider the zero operator $Tv=0$ for all $v\in V$. Then $||T||={\sup }_{||v||=1,v\in V}||Tv||={\sup }_{v\in V}\left|\left|T\frac{v}{||v||}\right|\right|={\sup }_{v\in V}\frac{1}{||v||}||Tv||=0$

homogeneity Let $c\in \mathbb{K}$. Then

\lvert \lvert cT \rvert \rvert &= \sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert cTv \rvert \rvert =\sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert c \rvert \cdot \lvert \lvert Tv \rvert \rvert \\ &=\lvert c \rvert \sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert Tv \rvert \rvert = \lvert c \rvert \cdot \lvert \lvert T \rvert \rvert \end{align} $$ *triangle inequality* Let $T, S \in {\cal B}(V,W)$. Then $$\begin{align} \lvert \lvert T+S \rvert \rvert &= \sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert (T+S)v \rvert \rvert =\sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert Tv+Sv \rvert \rvert \\ &\leq \sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert Tv \rvert \rvert +\lvert \lvert Sv \rvert \rvert \\ &\leq \sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert Tv \rvert \rvert + \sup_{\lvert \lvert v \rvert \rvert =1, v \in V} \lvert \lvert Sv \rvert \rvert \\ &= \lvert \lvert T \rvert \rvert +\lvert \lvert S \rvert \rvert \end{align}$$

note/aside

If $v\ne 0$, then $\left|\left|T\left(\frac{v}{||v||}\right)\right|\right|\le ||T||⟹||Tv||\le ||T||\cdot ||v||$ see observations for continuous linear function space

Theorem

Let $V$ and $W$ be normed spaces. If $W$ is a Banach space, then the bounded linear operator space $\mathcal{B}(V,W)$ is a Banach space.

NOTE

this proof is similar to the one for complete metric spaces have banach continuous bounded function spaces.

• And, as the note indicated in Lecture 1, this is the basic outline for showing a space is Banach.

To show that this space is Banach, we will use the characterization that banach spaces have all absolutely summable series are summable.

Proof

Suppose $\{T_n\}\subset \mathcal{B}(V,W)$ is a sequence of bounded linear operator space such that $C={\sum }_n||T_n||<\infty$ ie, the sequence is absolutely summable. We want to show that the ${\sum }_n{T}_n$ is summable.

${\sum }_n{T}_n$ is summable Let $v\in V$ and $m\in \mathbb{N}$. Then

\sum_{n=1}^m \lvert \lvert T_{n}v \rvert \rvert &\leq \sum_{n=1}^m \lvert \lvert T_{n} \rvert \rvert\cdot \lvert \lvert v \rvert \rvert \\ &\leq \lvert \lvert v \rvert \rvert \sum_{n} \lvert \lvert T_{n} \rvert \rvert \\ &= C \lvert \lvert v \rvert \rvert \end{align}$$ ie, the sequence of (nonnegative, real) partial sums $\left\{ \sum_{n=1}^m \lvert \lvert T_{n}v \rvert \rvert \right\}$ is bounded and therefore **converges**. Thus the series $\sum_{n}T_{n}v$ is [[summable series|absolutely summable]] in $W$. Since $W$ is [[Banach space|Banach]], (and [[banach spaces have all absolutely summable series are summable]]), we know that $\sum_{n}T_{n}v$ is [[summable series|summable]].

So define our candidate limit $T:V\to W$ as $Tv:={\lim }_{m\to \infty }{\sum }_{n=1}^m{T}_{n}v$ We now need to show that this is a bounded linear operator.

$T$ is linear For all ${\lambda }_1,{\lambda }_2\in \mathbb{K}$ and $v_1,v_2\in V$, we have

T(\lambda_{1}v_{1}+\lambda_{2}v_{2}) &= \lim_{ m \to \infty } \sum_{n=1}^m T_{n}(\lambda_{1}v_{1}+\lambda_{2}v_{2}) \\ (*)&= \lim_{ m \to \infty }\left[ \lambda_{1} \sum_{n=1}^m T_{n} v_{1} + \lambda_{2}\sum_{n=1}^m T_{n} v_{2} \right]\quad \\ &= \lambda_{1} Tv_{1} + \lambda_{2} Tv_{2} \end{align}$$ > [!NOTE] > For $(*)$, we did not prove that the sum of the limit is the limit of the sums, but it is the same proof as the case in $\mathbb{R}$ replacing absolute value with the appropriate norm instead. >

$T$ is bounded Let $v\in V$. Then

\lvert \lvert Tv \rvert \rvert &= \left\lvert \left\lvert \lim_{ m \to \infty } \sum_{n=1}^m T_{n}v \right\rvert \right\rvert \\ &= \lim_{ m \to \infty } \left\lvert \left\lvert \sum_{n=1}^m T_{n}v \right\rvert \right\rvert \\ (*)&\leq \lim_{ m \to \infty } \sum_{n=1}^m \lvert \lvert T_{n}v \rvert \rvert \\ &\leq \lim_{ m \to \infty } \sum_{n=1}^m \lvert \lvert T_{n} \rvert \rvert \cdot \lvert \lvert v \rvert \rvert \\ &= \lvert \lvert v \rvert \rvert \sum_{n} \lvert \lvert T_{n} \rvert \rvert \\ (**)&= C \lvert \lvert v \rvert \rvert \\ &< \infty \end{align}$$ Where $(*)$ is by the [[triangle inequality]] and $(* *)$ is because we assume this series is [[summable series|absolutely summable]].

Thus $T\in \mathcal{B}(V,W)$

$T$ is indeed the limit Now, all we need to show is that $T$ is indeed the limit in the operator norm. ie, that ${\sum }_{n=1}^m{T}_n\to T$ as $m\to \infty$.

Let $v\in V$ with $||v||=1$. Then

\left\lvert \left\lvert Tv - \sum_{n=1}^m T_{n}v \right\rvert \right\rvert &= \left\lvert \left\lvert \lim_{ m' \to \infty } \sum_{n=1}^m T_{n}v - \sum_{n=1}^m T_{n}v \right\rvert \right\rvert \\ &= \left\lvert \left\lvert \lim_{ m' \to \infty } \sum_{n=m+1}^m' T_{n}v \right\rvert \right\rvert \\ &\leq \lim_{ m' \to \infty } \sum_{n=m}^{m'} \lvert \lvert T_{n}v \rvert \rvert \\ (*) &\leq \lim_{ m' \to \infty } \sum_{n=m+1}^{m'} \lvert \lvert T_{n} \rvert \rvert \cdot \lvert \lvert v \rvert \rvert \\ (**)&= \lim_{ m' \to \infty } \sum_{n=m+1}^{m'} \lvert \lvert T_{n} \rvert \rvert\\ &= \sum_{n=m+1}^{\infty} \lvert \lvert T_{n} \rvert \rvert \\ \implies \left\lvert \left\lvert T - \sum_{n=1}^m T_{n} \right\rvert \right\rvert & \leq \sum_{n=m+1}^\infty \lvert \lvert T_{n} \rvert \rvert \to 0 \text{ as } m \to \infty \end{align}$$ Where $(*)$ is by the [[triangle inequality]] and $(* *)$ is because $\lvert \lvert v \rvert \rvert =1$. The last line we get because this is the tail of a convergent series that we defined at the beginning.

Thus the bounded linear operator space $\mathcal{B}(V,W)$ is Banach \tag*{$\blacksquare$}

see bounded linear operator space is banach

review

functional

What is a summable series? -?- A series where the sequence of partial sums converges

What is en equivalent definition for a Banach space? (not using completeness) -?- Every absolutely summable series is also summable

What norm do we use to define the space of bounded linear operators? -?- The operator norm

What condition must be met for $\mathcal{B}(V,W)$ to be a Banach space? -?- $W$ must be Banach

What four (4) steps do we need to show that if $W$ is Banach, then $\mathcal{B}(V,W)$ is Banach? (Recall that $\mathcal{B}(V,W)$ is the space of bounded linear functions $T:V\to W$) -?-

1. Show an absolutely summable sequence in $\mathcal{B}(V,W)$ is summable
2. Define a candidate limit $T$
3. Show that the limit is in $\mathcal{B}(V,W)$ by showing it is bounded and linear
4. Show that $T$ is indeed the limit

← Lecture 1 | Lecture 3 →

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