Lecture 25

[[lecture-data]]

2024-10-30

5. Chapter 5

Note

Will use the same norm symbol and context determines which norm we are using

Recall that if $T : V \to W$ is a linear function on NLS and our theorem/definition for continuity for linear functions - ie, if

sup_{x \in V \neq 0} \frac{| | T x | |}{| | x | |} < M, M \in R $ $ t h e n $ T $ i s c o n t i n u o u s . A n d f u r t h e r, t h e l e a s t s u c h b o u n d $ M $ i s a l i p s c h i t z c o n s t a n t . i e, [[C o n c e p t W i k i / c o n t i n u i t y f o r l i n e a r f u n c t i o n s ∥ c o n t i n u o u s l i n e a r f u n c t i o n s a r e l i p s c h i t z c o n t i n u o u s]] . > [! e x a m p l e] > $ $ \frac{d}{d x} : c^{1} [a, b] \to c^{0} [a, b]

is a linear operator.

Let $| | f | | = max_{t \in [a, b]} | f (t) |$ . This satisfies the conditions of the norm:

positive unless $f$ is uniformly 0

any scalar comes out with an absolute value around it

satisfies the triangle inequality

Unfortunately, this is not continuous. Take $f (t) = t^{k}, k > 0$ . Then suppose we have some $t$ such that $| | t^{k} | | = 1$ . But then $| | \frac{d}{d t} t^{k} | | = | | k t^{k} | | = k | | t^{k} | | = k$ and so
$\frac{| | \frac{d}{d t} t^{k} | |}{| | t^{k} | |}$
is unbounded.

Operator Norm

Let $V, W$ be NLS. Let $T : V \to W$ be a linear function. If $T$ is continuous, then the operator norm of $T$ is defined as

| | T | | := sup_{x \in V \neq 0} \frac{| | T x | |}{| | x | |}

(see operator norm)

Note: The Operator Norm is indeed a Norm

Let $V, W$ be NLS and suppose that $T, S : V \to W$ both linear, and $α, β \in K$ . Define function $α T + β S : V \to W$ . Then for all $x \in V$ , we have that

[α T + β S] (x) = α T (x) + β S (x)

ie we calculate the output of these functions pointwise.

Exercise

$α T + β S$ is linear (just think about it).

And if $T, S$ are also continuous then

$| | T | | = 0 ⟺ T = 0$ function
$| | α T | | = | α | \cdot | | T | |$
$| | T + S | | \leq | | T | | + | | S | |$

ie, the operator norm can create a normed linear space on the space of continuous linear functions. 🤯

Proof

(1) If $T \neq 0$ , then there exists fomr $x \neq 0$ such that $T x \neq 0$ . Then $\frac{| | T x | |}{| | x | |} > 0$ . Thus the $sup$ must be greater than 0.

(2) $sup_{x \in V \neq 0} \frac{| | α T (x) | |}{| | x | |} = sup_{x \neq 0} \frac{| α | | | T (x) | |}{| | x | |} = | α | sup_{x \neq 0} \frac{| | T (x) | |}{| | x | |}$ where we can take $α$ out by homogeneity of $W$ 's norm

(3) note that

\begin{aligned} sup_{x \neq 0} \frac{| | [T + S] (x) | |}{| | x | |} \leq sup_{x \neq 0} \frac{| | T (x) | | + | | S (x) | |}{| | x | |} \\ \leq sup_{x \neq 0} \frac{| | T (x) | |}{| | x | |} + sup_{x \neq 0} \frac{| | S (x) | |}{| | x | |} \end{aligned}

Continuous Linear Function Space

Let $V, W$ be NLS. Then define $B (V, W)$ as the normed linear space whose elements are continuous linear functions $V \to W$ , and whose norm is the operator norm.

(see continuous linear function space)

Dual Space

Let $V$ be a NLS. Then $B (V, K)$ of continuous linear functionals is called the dual space $V^{*}$

(see dual space)

Observations

If $T \in B (V, W)$ , then for all $x \in V$ we have $| | T x | | \leq | | T | | \cdot | | x | |$ . This result is trivial by definition of the operator norm
If $T \in B (V, W)$ and $S \in B (W, U)$ then $S \circ T \in B (V, U)$ . And $| | S \circ T | | \leq | | S | | \cdot | | T | |$ (careful, each of these norms are different!).
Since for all $x \in V, | | [S \circ T] (x) | | = | | S (T (x)) | | \leq | | S | | \cdot | | T x | | \leq | | S | | \cdot | | T | | \cdot | | x | |$ by the above result (1). But then we can see that $\frac{| | [S \circ T] (x) | |}{| | x | |} \leq | | S | | \cdot | | T | |$ - and since $T, S$ are continuous this is bounded (since this holds for any $x$ ). And also, $| | S \circ T | | \leq | | S | | \cdot | | T | |$

Note

Suppose $A \in M_{m, n} (K)$ and $B \in M_{m, n} (K)$ thought of as functions " $A$ " and " $B$ " both from $K^{n} \to K^{m}$ . Suppose $C \in M_{p, m} (K)$ is associated with function " $C$ " from $K^{m} \to K^{p}$ . Then

"CA" = C \circ A = C A

and " $α A + β B$ " is $α$ " $A$ " + $β$ " $B$ " is $α A + β B$ .

Theorem

Suppose $V, W$ are NLS where $\dim (V) < \infty$ . Then $T : V \to W$ is continuous.

Proof

First,

Let $b_{1}, \dots, b_{n}$ be a basis for $V$ . Let $| | \cdot | |^{'}$ be defined $V \to R$ as such:
$\forall y \in V$ , there exist unique coefficients $a l p h a_{1}, \dots, α_{n} \in K$ such that $y = \sum_{i = 1}^{n} α_{i} b_{i}$ . Let $| | y | |^{'} = \sum_{i = 1}^{n} | α_{i} |$ (ie, the $L^{1}$ norm on the column vector space, which is the same as the vector space).
Let $E := max_{i \in {1, \dots, n}} | | T b_{i} | |_{W}$ .
And let $M > 0$ be such that for all $y \in V$ we have $| | y | |^{'} \leq M | | y | |_{V}$ (by norm equivalence).

Then for all $x \in V$ , say $β_{1}, \dots, β_{n} \in K$ such that $x = \sum_{i = 1}^{n} β_{i} b_{i}$ , then

\begin{aligned} | | T x | |_{w} & = {| | T \sum β_{i} b_{i} | |}_{W} \\ = {| | \sum β_{i} T b | |}_{W} \\ \leq \sum β_{i} | | T b_{i} | |_{W} \\ \leq \sum_{i = 1}^{n} | β_{i} | E \\ = | | x | |^{'} E \\ \leq M E | | x | |_{v} \end{aligned}

So we can see that $\frac{| | T x | |_{W}}{| | x | |_{W}} \leq M E$ ie $T$ is continuous

(see linear functions with finite dimensional domain are continuous)

Next time: thinking about norms and things for continuous linear spaces