Let $V$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$ (call it $\mathbb{K}$). Then $V$ comes with two operations

• addition $+:V\times V\to V,(v_1,v_2)\mapsto v_1+v_2$
• scalar multiplication $\cdot :\mathbb{K}\times V\to V,(\alpha ,v)\mapsto \alpha \cdot v$

• $\mathbb{R},{\mathbb{R}}^2,{\mathbb{R}}^n,\mathbb{C},{\mathbb{C}}^n$
• $C([0,1])=\{f:[0,1]\to \mathbb{C}|f\text{ is continuous}\}$

$V$ is finite if every linearly independent set is finite.

ie, if for every set $E\subseteq V$ such that $\mathrm{\forall }{v}_1,\dots ,v_N\in E$

and $E$ has finite cardinality, then $V$ is finite-dimensional. Otherwise, $V$ is infinite-dimensional.

$C([0,1])$ is infinite dimensional because

is a linearly independent set with non-finite cardinality.

A norm on a vector space $V$ is a function $||\cdot ||:V\to [0,\mathrm{\infty })$ such that

1. $||v||=0⟺v=0$ definiteness
2. $||\lambda v||=|\lambda |\cdot ||v||\mathrm{\forall }\lambda \in \mathbb{K},v\in V$ homogeneity
3. $\mathrm{\forall }{v}_1,v_2\in V||v_1+v_2||\le ||v_1||+||v_2||$ triangle inequality

A vector space with a norm $||\cdot ||$ is called a normed linear space.

A semi-norm is a function $||\cdot ||V\to [0,\mathrm{\infty })$ satisfying homogeneity and the triangle inequality, but not necessarily definiteness.

If $X$ is a set then $d:X\times X\to [0,\mathrm{\infty })$ is a metric if

1. (identifiability) $d(x,y)=0⟺x=y$
2. (symmetry) $\mathrm{\forall }x,y\in X,d(x,y)=d(y,x)$
3. (triangle inequality) $\mathrm{\forall }x,y,z\in X,d(x,y)\le d(x,z)+d(z,y)$

If $||\cdot ||$ is a norm on a vector space $V$, then $d(v,w):=||v-w||$ defines a metric on $V$ (and this metric is induced by the norm $||\cdot ||$)

• Definiteness for the norm implies $d(x,y)=0⟺x=y$
• By homogeneity we get symmetry:

• triangle inequality follows immediately since $x-y=(x-z)+(z-y)$

If $X$ is a vector space, the vector space

$C_{\mathrm{\infty }}([0,1])=C([0,1])$ since we know the set $X$ is bounded

Then $C_{\mathrm{\infty }}(X)$ is a vector space and we can define the norm

NTS that this is indeed a norm by verifying each of the properties. identifiability and homogeneity are satisfied from the definition of the norm. It suffices then to show that the triangle inequality holds.

If $(u,v)\in C_{\mathrm{\infty }}(X)$ then for all $x\in X$

Convergence in this norm means

Thus convergence in this metric is uniform convergence when the functions $u$ are bounded and continuous.

$${\ell }^p=\{a=\{a_j{\}}_{j=1}^{\mathrm{\infty }}|||a|{|}_p<\mathrm{\infty }\}\$\$isthespaceof(infinite)sequences.Wedefinethe\${\ell }^p\$normas\$\$||a|{|}_p:=\{\begin{array}{l}(\sum _{i=1}^{\mathrm{\infty }}|a_i{|}^p{)}^{1/p}1\le p<\mathrm{\infty }\\ \underset{1\le j<\mathrm{\infty }}{sup}|a_j|p=\mathrm{\infty }\end{array}$$

${\left\{\frac{1}{j}\right\}}_{j=1}^{\mathrm{\infty }}\in {\ell }^p\mathrm{\forall }p>1$ but not for $p=1$

A normed space is a Banach space if it is complete with respect to the metric induced by the norm.

ie, Cauchy sequences in the space converge in the space.

$\mathbb{Q}$ is not complete, but $\mathbb{R}$ is. ${\mathbb{R}}^n,{\mathbb{C}}^n$ are complete wrt any of the $||\cdot |{|}_p$ norms.

If $X$ is a complete metric space, then the continuous bounded function space $C_{\mathrm{\infty }}(X)$ is a Banach space.

NTS every Cauchy sequence $\{u_n\}\in C_{\mathrm{\infty }}(X)$ has a limit in $C_{\mathrm{\infty }}(X)$.

Let $\{u_n\}$ be a Cauchy sequence in $C_{\mathrm{\infty }}(X)$. Then for all $ϵ>0$ there exists $N\in \mathbb{N}$ such that for all $n,m\ge N$, $||u_n-u_m|{|}_{\mathrm{\infty }}<ϵ$.

In particular, $\mathrm{\exists }{N}_0\in \mathbb{N}$ such that for all $m,n\ge N_0,||u_n-u_m|{|}_{\mathrm{\infty }}<1$. Then for all $n\ge N_0$, we have

And therefore for all $n\in \mathbb{N}$, we have

Since for all $x\in X$ we have $|u_n(x)-u_m(x)|\le ||u_n-u_m|{|}_{\mathrm{\infty }}$, we have that
for each $x\in X$, the sequence$\{u_n(x)\}$ is a Cauchy sequence. And since $\mathbb{C}$ is complete, its limit is in $\mathbb{C}$.

(define a candidate limiting function)
Define $u:X\to \mathbb{C},u(x)=\underset{n\to \mathrm{\infty }}{lim}{u}_n(x)$
Then for all $x\in X$,

thus $u$ is a bounded function.

Finally, we need to show continuity and convergence.

Let $ϵ>0$. Since $\{u_n\}$ is Cauchy in $C_{\mathrm{\infty }}(X)$, there exists some $N_1\in \mathbb{N}$ such that $\mathrm{\forall }n,m\ge N$ we have $||u_n-u_m|{|}_{\mathrm{\infty }}<\frac{ϵ}{2}$.

So for any fixed $x\in X$, we have

Thus as $m\to \mathrm{\infty }$, we have for all $n\ge N_1$ and each $x\in X$,

Thus we have $||u_n-u|{|}_{\mathrm{\infty }}\to 0$ and this implies that $u_n\to u$ uniformly on $X$. And because each $u_n$ is continuous, this implies that the limit $u$ is also continuous.

Thus, $u\in C_{\mathrm{\infty }}(X)$ and $u_n\to u\in C_{\mathrm{\infty }}(X)$, thus $C_{\mathrm{\infty }}(X)$ is complete and therefore a Banach space.

the approach for this proof is basically the same as any proof to show that something is a Banach space.

• choose a candidate for the limit
• show that the limit is in the space

show $c_0=\{a\in {\ell }^{\mathrm{\infty }}|\underset{j\to \mathrm{\infty }}{lim}{a}_j=0\}$ is also a Banach space with norm $||a|{|}_{\mathrm{\infty }}=\underset{1\le j\le \mathrm{\infty }}{sup}|a_j|$