Lecture 5

Info

A normed space is a Banach Space when the corresponding metric space is complete. Any normed space can be completed to a Banach Space, see the real numbers from $\mathbb{Q}$.

Example: ${\mathbb{C}}^n$ vector space, then

1. $\Vert v{\Vert }_1\equiv {\sum }_{k=1}^n|v_k|,v=(v_1,\dots ,v_n)$
2. $\Vert v{\Vert }_{\infty }\equiv {\text{sub}}_{k=1,\dots ,n}\{|v_k|\}$

Which are norms on ${\mathbb{C}}^n$

Example

Question

Consider $C_c(X)\subset {\Pi }_X\mathbb{C}=\{f:X\to C\}$ on functions $X->\mathbb{C}$ that are non-zero for finitely only many $x\in X$.

Then $C_c(X)$ is a vector space under op.

$f\ne \text{only on}Y\subset X$
$q\ne \text{only on}Z\subset X$
$⟹f+q\ne 0\text{only on}Y\cup Z$
$a\times f\ne \text{only on}Y$
with ${\delta }_x(y)=\{\begin{array}{ll}1,& x=y\\ 0,& x\ne y\end{array}$
Has linear basis $\{{\delta }_x\}$, $x\in X$

Proof

Given that $f\in C_c(X)$ then

$f={\sum }_{x\in X}f(x)\times {\delta }_x$ is a finite sum, and the only possibility.

QED

So $\text{dim}{C}_c(X)=|X|$.

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Definite norms on $C_c(X)$ by $\Vert f{\Vert }_1={\sum }_{x\in X}|f(x)|$ and $\Vert f{\Vert }_{\infty }={\sup }_{x\in X}|f(x)|$ when $|X|=n$ we recover ${\mathbb{C}}^n$.

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We write $V\simeq W$and say that $V$ and $W$ are Isomorphic

1. As sets if $\exists$ bijection $f:V\to W$
2. As vector spaces if in addition $f$ is linear; $f(a\times u+b\times v)=a\times f(u)+b\times f(v),\forall a,b\in \mathbb{C},u,v\in V$
3. As normed spaces if in addition $f$ is Isometric; $\Vert f(v)\Vert =\Vert v\Vert ,\forall v\in V$.
   $f$ should preserve all relevant structure.

Hilbert Spaces

Triangle Inequality

Follows from the Cauchy-Schwarz Inequality

Example on $C_c(X)$ then

$(f\mid g)={\sum }_{x\in X}f(x)\times \overline{g(x)}$
This defines an inner product, which can be completed to a Hilbert space.
$\Vert f{\Vert }_2={\left({\sum }_{x\in X}\mid f(x){\mid }^2\right)}^{\frac{1}{2}}$

The inner product here ${\mathbb{C}}^n(u\mid v)={\sum }_{k=1}^n{u}_k\overline{v_k}$ ${\mathbb{R}}^n\Vert u{\Vert }_2={\left(\sum \mid u_k{\mid }^2\right)}^{\frac{1}{2}}$ $\vec{u}\times \vec{v}=(\vec{u}\mid \vec{v})$