Lecture 23

[[lecture-data]]

2024-10-23

The Poll

The best date for the next midterm is November 5th

• Chapter 4
• Chapter 7

Somewhat narrow scope - less material but more detail

5. Chapter 5

inner products and norms on vector spaces over a field (either $\mathbb{R}$ or $\mathbb{C}$):

Recall

• inner product space is defined by a vector space and an inner product
• normed linear space is defined by a vector space and a norm

Cauchy-Schwarz Theorem

Let $V,⟨,⟩$ be an inner product space over $K$. Then for all $x,y\in V$ we have $|⟨x,y⟩|\le ⟨x,x⟩⟨y,y⟩$

$K=\mathbb{C}$) For any $t,\theta \in \mathbb{R}$ it is true that and

(here we prove for

0 &\leq \langle te^{i\theta} + y, te^{i\theta}x,y\rangle \\ &= t^2 e^{i\theta}\overline{e^{i\theta}} \langle x,x \rangle + t e^{i\theta} \langle x,y \rangle + te^{i\theta} \langle y, x \rangle + \langle y, y \rangle \\ &= t^2 \langle x,x \rangle + 2 t Re[e^{i\theta} \langle x,y \rangle] + \langle y, y \rangle \\ \text{choose }\theta \text{ s.t. } Re[e^{i\theta} \langle x,y \rangle] &= \lvert \langle x, y \rangle \rvert \\ &= t^2 \langle x,x \rangle + 2 t \lvert \langle x, y \rangle \rvert + \langle y, y \rangle \\ & \geq 0 \;\;\;\forall t \in \mathbb{R} \end{aligned}$$ And this is a quadratic in $\mathbb{R}$! So the discriminant must be negative, that is $$[2\lvert \langle x, y \rangle \rvert]^2 - 4[\langle x,x \rangle \langle y,y \rangle] \leq 0$$ $$\implies \lvert \langle x, y \rangle \rvert^2 \leq \langle x,x \rangle \langle y,y \rangle$$

(see cauchy-schwarz theorem)

Corollary

If $V,⟨,⟩$ inner product space over $K$, define $||\cdot ||:V\to \mathbb{R}\ge 0$ as $||x||=\sqrt{⟨x,x⟩}\forall x\in V$ Then $V,||\cdot ||$ is a normed linear space 😼 and we say that it is “induced/generated by” the inner product.

triangle inequality since non negativity and homogeneity will hold by the properties of the inner product. By the definition of our operator, we have

Very basic: only need to show

\lvert \lvert x+y \rvert \rvert ^2 &= \langle x+y, x+y \rangle \\ & = \langle x,x \rangle + \langle x,y \rangle + \langle y, x\rangle+ \langle y,y\rangle \\ &= \lvert \lvert x \rvert \rvert ^2 + \lvert \lvert y \rvert \rvert ^2 + 2 Re[\langle x,y \rangle] \\ & \leq \lvert \lvert x \rvert \rvert ^2 + \lvert \lvert y \rvert \rvert ^2 + 2 \lvert \lvert x \rvert \rvert\times \lvert \lvert y \rvert \rvert \;\;\;\text{ by Cauchy-Schwarz} \\ &= (\lvert \lvert x \rvert \rvert + \lvert \lvert y \rvert \rvert )^2 \end{aligned}$$ thus the triangle inner quality holds!

(see inner products define norms)

Metric Space

A metric space is a set $S$ and a distance function $d:S\times S\to {\mathbb{R}}_{\ge 0}$ such that for all $x,y,z\in S$ we have

• $d(x,y)=0⟺x=y$
• $d(x,y)=d(y,x)$ (symmetry)
• $d(x,z)\le d(x,y)+d(y,z)$

(see metric space)

Proposition

If $V,||\cdot ||$ is a NLS, then we can make a metric space on $V$ with $d$ as follows: for all $x,y\in V$, $d(x,y):=||x-y||$

Proof

Let’s check the properties!

• $||x-y||=0⟺x-y=0⟹x=y$
• By homogeneity we can see $||x-y||=||(-1)(x-y)||=||y-x||$ so symmetry holds
• $||x-z||=||x-y+y-z||\le ||x-y||+||y-z||$ by the triangle inequality for norms, but then it also shows that this holds for our defined metric!

(see norm spaces induce metric spaces)

If the metric space induced by an inner product is complete, then we call this a Hilbert space If the metric space induced by a norm is complete, then we call it a Banach space And if a metric space is complete then it is…complete.

And if the vector space $V$ is finite dimensional, then any normed linear space on $V$ is complete.

Note (""Reverse"" Triangle Inequality)

If you have a normed linear space $V,||\cdot ||$, then for all $x,y\in V$ we have $|||x||-||y|||\le ||x-y||$

In particular, this shows that norms are

• continuous

• 1-Lipschitz continuous (and thus uniformly continuous!)

$||x||\le ||x-y||+||y||$

Easy to see by observing by the triangle inequality that

(see reverse triangle inequality)

Theorem

Let $V,||\cdot ||$ be a NLS. The following are equivalent:

1. $V$ is finite-dimensional
2. The unit ball $\{x\in V:||x||\le 1\}$ is compact
3. The unit sphere $\{x\in V:||x||=1\}$ is compact
4. For all $S\subseteq V$, $S$ is compact if and only if $S$ is closed and bounded.

(see compact space equivalencies for normed linear spaces)