Lecture 24

[[lecture-data]]

2024-10-28

5. Chapter 5

we are talking about norms.

Theorem

Suppose that $V$ is a finite dimensional vector space over a field $K$ . Then all norms are equivalent. ie, for any two norms, say $| | \cdot | |$ and $| | \cdot | |^{'}$ , on $V$ , there exist $m, M \in R$ such that for all $x \in V$ we have

m | | x | |^{'} \leq | | x | | \leq M | | x | |^{'}

note also that we also have then that

\frac{1}{M} | | x | | \leq | | x | |^{'} \leq \frac{1}{m} | | x | |

And norm equivalence is an equivalence relation!
(see norm equivalence)

Why is this important? for convergence purposes, this means that we can use ANY norm (that works and exists on the space) to prove convergence.

Example

${x_{k}}_{k = 1}^{\infty} \subseteq V, x \in V$ and we have that $lim_{x \to \infty} x_{k} = x$ then this means that

| | x_{k} - x | | \to 0

and since norms are equivalent then we can see that

| | x_{k} - x | | \geq m | | x_{k} - x | |^{'}

so we can show convergence for any norm and get any other norm

ALSO, the underlying topology generated by one norm will be the same as any other norm. (same open sets, closed sets, bounded sets, etc - so long as we are in a metric space. once we talk about convergence, each of these follow)

Example

Suppose we have $K^{n}$ on $K$ . Suppose $x_{k}$ is a sequence in $K^{n}$ and $x \in K_{n}$ . Then $x_{k} \to x$ if and only if for all $j = 1, 2, \dots, n$ we have $(x_{k})_{j} \to x_{j}$ ie convergence only happens if we have coordinate-wise convergence.

Proof

Let $| | \cdot | |$ be a norm on $V = K^{n}$ . Then $| | x_{k} - x | | \to 0 ⟺ | | x_{k} - x | |_{\infty} \to 0 ⟺$ for all $j$ we have that $| (x_{k})_{j} - x_{j} | \to 0$

(see convergence only happens if we have coordinate-wise convergence (for vector spaces))

Linear Transformation

Let $V, W$ be vector spaces over $K$ and let $T : V \to W$ . $T$ is a linear function means that for all $α, β \in K$ and $x, y, \in V$ we have $$T(\alpha x+\beta y) = \alpha T(x) + \beta T(y)$$

Example

$A \in M_{m, n} : C^{n} \to C^{m}$
$\frac{d}{d x} : c^{1} [a, b] \to c^{0} [a, b]$

(see linear function)

Isomorphic / Isomorphisc

Say $V$ is a vector space over $K$ with finite dimension and with basis $B = {b_{1}, b_{2}, \dots, b_{n}}$ . Then $V$ is isomorphic to $K^{n}$ over $K$ (denoted $V ≅ K^{n}$ ) if for all $x \in V$ there exists unique $α_{1}, α_{2}, \dots, α_{n} \in K$ such that $x = \sum_{i = 1}^{n} α_{i} b_{i}$ . We can call these coefficients " $x_{B}$ "

ie there is a one-to-one correspondence between $V$ and coefficients that form a vector in $K^{n}$ .

(see isomorphism)

Theorem

Say $V, W$ are finite vector spaces with respective bases $B_{V} = {b_{1}, \dots, b_{n}}$ and $B_{W} = {b_{1}^{'}, \dots, b_{m}^{'}}$ . Then $T : V \to W$ is linear function if and only if there exists an $A \in M_{m, n} (K)$ such that for all $x \in V$ and $y \in W$ , we have

T (x) = y ⟺ A x_{B_{V}} = y_{B_{W}}

Linear function

Suppose $V$ is a vector space over $K$ . A linear function $T : V \to K$ is called a linear function (ie the range is only 1 dimension)

Note

note linear functionals on $K^{n}$ over $K$ are precisely of the following form:
$T : K^{n} \to K$ where $T (x) = [a fixed 1 \times n matrix] x$ .

(see linear functional)

Vector's Linear Functional

Let $w \in K^{n}$ and define $\hat{w}$ the linear functional $\hat{w} : K^{n} \to K$ where for all $x \in K^{n}$ we have $\hat{w} (x) := w^{*} x$ .

(see vector linear functional)

Theorem

Let $V, W$ be NLS with respective norms $| | \cdot | |_{V}, | | \cdot | |_{W}$ . Then $T$ is continuous if and only if

sup_{x \in V \neq 0} \frac{| | T x | |_{W}}{| | x | |_{V}} < \infty

Note

(1) $T (0) = 0$ since $T (0 x) = 0 T (x) = 0$
(2) $$\begin{aligned}
\sup_{x \in V \neq 0} \frac{\lvert \lvert Tx \rvert \rvert_{W} }{\lvert \lvert x \rvert \rvert {V}} &= \sup{x \in V \neq 0} \left\lvert \left\lvert T \frac{1}{\lvert \lvert x \rvert \rvert_{V} } x \right\rvert \right\rvert_{W} \
&= \sup_{z \in V : \lvert \lvert z \rvert \rvert_{V}=1 }\lvert \lvert Tz \rvert \rvert {W} ;;;;;(*)
\end{aligned}$$
Where $(*)$ follows by letting $z = \frac{1}{| | x | |_{V}} x$ and observing that $$\lvert \lvert z \rvert \rvert = \left\lvert \left\lvert \frac{1}{\lvert \lvert x \rvert \rvert } x\right\rvert \right\rvert_{V} = \frac{1}{\lvert \lvert x \rvert \rvert _{V}} \lvert \lvert x \rvert \rvert _{V} = 1 $$

Proof

$(⟸)$ Suppose for all $x \in V \neq 0$ we have $\frac{| | T x | |_{W}}{| | x | |_{V}} \leq M, M < \infty$ . In particular, for any $y \neq z$ we have $$\frac{\lvert \lvert T(y-z) \rvert \rvert }{\lvert \lvert y-z \rvert \rvert } \leq M \implies \lvert \lvert Ty - Tz \rvert \rvert \leq M \lvert \lvert y-z \rvert \rvert $$

ie, $T$ is continuous. In fact, it is Lipschitz continuous!

$(⟹)$ Suppose $T$ is continuous. In particular, there exists $δ > 0$ such that for all $y \in V$ such that $| | y - 0 | | \leq δ ⟹ | | T y - T \cdot 0 | | \leq 1$ . (if continuous, then continuous at the origin. Set $ϵ > 0$ as 1). Thus, for all $x \in V \neq 0$ we have

\frac{| | T x | |}{| | x | |} = \frac{1}{δ} | | T (δ \frac{1}{| | x | |} x) | | \leq \frac{1}{d} \cdot 1 < \infty

(see continuity for linear functions)