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 ${\sum_{n = 1}^{m} v_{n}}_{m = 1}^{\infty}$ converges.

The series is absolutely summable if ${\sum_{n = 1}^{m} | | v_{n} | |}_{m = 1}^{\infty}$ 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 ${\sum_{n = 1}^{m} v_{n}}_{m = 1}^{\infty}$ is Cauchy.

see absolutely summable series have Cauchy partial sums

Exercise

Prove it! (same proof as when $V = R$ )

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 ${\sum_{n = 1}^{m} v_{n}}_{m}$ 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 $k \in N$ , there exists $N_{k} \in N$
such that for all $n, m \geq N_{k}$ , 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} + \dots + N_{k}

Then $n_{1} < n_{2} < \dots$ and for all $k$ , we have $n_{k} \geq N_{k}$ . Thus, for all $k \in N$ , we have

| | v_{n_{k + 1}} - v_{n_{k}} | | < \frac{1}{2^{k}}

Thus

\begin{aligned} \sum_{k} (v_{n_{k + 1}} - v_{n_{k}}) is absolutely summable \\ ⟹ & \sum_{k} (v_{n_{k + 1}} - v_{n_{k}}) is summable \\ ⟹ & {\sum_{k = 1}^{m} (v_{n_{k + 1}} - v_{n_{k}})}_{m = 1}^{\infty} = {v_{n_{m + 1} - v_{n_{1}}}}_{m = 1}^{\infty} converges \end{aligned}

Thus the (sub)sequence is ${v_{n_{m + 1}}}_{m = 1}^{\infty}$ converges in $V$ . Thus $V$ is Banach. $◼$

see banach spaces have all absolutely summable series are summable

Operators and Functionals

These are the analog of matrices.

Example (important!)

Let $K : [0, 1] \times [0, 1] \to C$ be a continuous function. Then for any function $f \in C ([0, 1])$ , we can define

T f (x) = \int_{0}^{1} K (x, y) f (y) d y

Then $T f \in C ([0, 1])$ and for all $λ_{1}, λ_{2} \in C$ and $f_{1}, f_{2} \in C ([0, 1])$

T (λ_{1} f_{1} + λ_{2} f_{2}) = λ_{1} T f_{1} + λ_{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 $λ_{1}, λ_{2} \in K$ and $v_{1}, v_{2} \in V$ ,

T (λ_{1} v_{1} + λ_{2} v_{2}) = λ_{1} T v_{1} + λ_{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 $v \in V$ ,

| | T v | |_{W} \leq 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 $| | T v | |_{W} \leq C | | v | |_{V}$ . Let $v \in V$ and suppose $v_{n} \to v$ . Then

0 \leq | | T v_{n} - T v | |_{W} = | | T (v_{n} - v) | |_{W} \leq C | | v_{n} - v | |_{V}

And since $| | v_{n} - v | |_{V} \to 0$ , by the squeeze theorem, we have $| | T v_{n} - T v | |_{W} \to 0$ as well.

$(⟸)$ Suppose $T$ is continuous. Then

T^{- 1} (B_{W} (0, 1)) = {v \in V : T v \in B_{W} (0, 1)}

Since $T 0 = 0$ , 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 (\frac{r}{2 | | v | |_{V}} v) \in B_{W} (0, 1)

Thus, by homogeneity of the norm, we have

{| | T (\frac{r}{2 | | v | |_{V}} v) | |}_{W} < 1 ⟹ | | T v | |_{W} \leq \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

\begin{aligned} | T f (x) | & = | \int_{0}^{1} K (x, y) f (y) d y | \\ \leq \int_{0}^{1} | K (x, y) | \cdot | f (y) | d y \\ \leq \int_{0}^{1} | K (x, y) | \cdot | | f | |_{\infty} d y \\ \leq \int_{0}^{1} | | K (x, y) | | \cdot | | f | |_{\infty} d y \\ = | | K (x, y) | | \cdot | | f | |_{\infty} \end{aligned}

And since this bound holds for all $x$ , it holds for the supremum. Thus

| | T f | |_{x} \leq | | K | |_{\infty} | | f | |_{\infty}

Thus we can use $C = | | K | |_{\infty}$ 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 $B (V, W)$

Operator Norm

The operator norm of an operator $T \in B (V, W)$ is defined as

| | T | | = sup_{| | v | | = 1, v \in V} | | T v | |

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 $T v = 0$ for all $v \in V$ . Then

| | T | | = sup_{| | v | | = 1, v \in V} | | T v | | = sup_{v \in V} | | T \frac{v}{| | v | |} | | = sup_{v \in V} \frac{1}{| | v | |} | | T v | | = 0

homogeneity
Let $c \in K$ . Then

\begin{aligned} | | c T | | & = sup_{| | v | | = 1, v \in V} | | c T v | | = sup_{| | v | | = 1, v \in V} | c | \cdot | | T v | | \\ = | c | sup_{| | v | | = 1, v \in V} | | T v | | = | c | \cdot | | T | | \end{aligned}

triangle inequality
Let $T, S \in B (V, W)$ . Then

\begin{aligned} | | T + S | | & = sup_{| | v | | = 1, v \in V} | | (T + S) v | | = sup_{| | v | | = 1, v \in V} | | T v + S v | | \\ \leq sup_{| | v | | = 1, v \in V} | | T v | | + | | S v | | \\ \leq sup_{| | v | | = 1, v \in V} | | T v | | + sup_{| | v | | = 1, v \in V} | | S v | | \\ = | | T | | + | | S | | \end{aligned}

note/aside

If $v \neq 0$ , then

| | T (\frac{v}{| | v | |}) | | \leq | | T | | ⟹ | | T v | | \leq | | 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 $B (V, W)$ 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 ${T_{n}} \subset B (V, W)$ 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 $m \in N$ . Then

\begin{aligned} \sum_{n = 1}^{m} | | T_{n} v | | & \leq \sum_{n = 1}^{m} | | T_{n} | | \cdot | | v | | \\ \leq | | v | | \sum_{n} | | T_{n} | | \\ = C | | v | | \end{aligned}

ie, the sequence of (nonnegative, real) partial sums ${\sum_{n = 1}^{m} | | T_{n} v | |}$ is bounded and therefore converges.

Thus the series $\sum_{n} T_{n} v$ is absolutely summable in $W$ . Since $W$ is Banach, (and banach spaces have all absolutely summable series are summable), we know that $\sum_{n} T_{n} v$ is summable.

So define our candidate limit $T : V \to W$ as

T v := 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 $λ_{1}, λ_{2} \in K$ and $v_{1}, v_{2} \in V$ , we have

\begin{aligned} T (λ_{1} v_{1} + λ_{2} v_{2}) & = lim_{m \to \infty} \sum_{n = 1}^{m} T_{n} (λ_{1} v_{1} + λ_{2} v_{2}) \\ (*) & = lim_{m \to \infty} [λ_{1} \sum_{n = 1}^{m} T_{n} v_{1} + λ_{2} \sum_{n = 1}^{m} T_{n} v_{2}] \\ = λ_{1} T v_{1} + λ_{2} T v_{2} \end{aligned}

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 $R$ replacing absolute value with the appropriate norm instead.

T

is bounded

Let $v \in V$ . Then

\begin{aligned} | | T v | | & = | | lim_{m \to \infty} \sum_{n = 1}^{m} T_{n} v | | \\ = lim_{m \to \infty} | | \sum_{n = 1}^{m} T_{n} v | | \\ (*) & \leq lim_{m \to \infty} \sum_{n = 1}^{m} | | T_{n} v | | \\ \leq lim_{m \to \infty} \sum_{n = 1}^{m} | | T_{n} | | \cdot | | v | | \\ = | | v | | \sum_{n} | | T_{n} | | \\ (* *) & = C | | v | | \\ < \infty \end{aligned}

Where $(*)$ is by the triangle inequality and $(* *)$ is because we assume this series is absolutely summable.

Thus $T \in B (V, W)$

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

\begin{aligned} | | T v - \sum_{n = 1}^{m} T_{n} v | | & = | | lim_{m^{'} \to \infty} \sum_{n = 1}^{m} T_{n} v - \sum_{n = 1}^{m} T_{n} v | | \\ = | | lim_{m^{'} \to \infty} {\sum_{n = m + 1}^{m}}^{'} T_{n} v | | \\ \leq lim_{m^{'} \to \infty} \sum_{n = m}^{m^{'}} | | T_{n} v | | \\ (*) & \leq lim_{m^{'} \to \infty} \sum_{n = m + 1}^{m^{'}} | | T_{n} | | \cdot | | v | | \\ (* *) & = lim_{m^{'} \to \infty} \sum_{n = m + 1}^{m^{'}} | | T_{n} | | \\ = \sum_{n = m + 1}^{\infty} | | T_{n} | | \\ ⟹ | | T - \sum_{n = 1}^{m} T_{n} | | & \leq \sum_{n = m + 1}^{\infty} | | T_{n} | | \to 0 as m \to \infty \end{aligned}

Where $(*)$ is by the triangle inequality and $(* *)$ is because $| | v | | = 1$ . 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 $B (V, W)$ is Banach

\begin{matrix} ◼ \end{matrix}

see bounded linear operator space is banach

review

#flashcards/math/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 $B (V, W)$ to be a Banach space?
-?-
$W$ must be Banach

What four (4) steps do we need to show that if $W$ is Banach, then $B (V, W)$ is Banach?
-?-

Show an absolutely summable sequence in $B (V, W)$ is summable
Define a candidate limit $T$
Show that the limit is in $B (V, W)$ by showing it is bounded and linear
Show that $T$ is indeed the limit

← Lecture 1 | Lecture 3 →

Created 2025-05-29 Last Modified 2025-06-05