CAL1_L4

Recall what an inner product space is. Now, define $C[a,b]\equiv \{f:[a,b]\to \mathbb{C}:f\text{ is continuous}\}$. Define

$$⟨f,g⟩\equiv {\int }_a^{b}f\overline{g}.$$

Given an inner product on a vector space, we can always define a norm on the vector space by

$$\Vert f{\Vert }^2\equiv ⟨f,f⟩.$$

On $C[a,b]$,

$$\Vert f{\Vert }^2\equiv {\int }_a^b|f{|}^2.$$

On $C[a,b]$, we can also define a sup-norm

$$\Vert f{\Vert }_{\infty }\equiv \underset{x\in [a,b]}{\sup }|f(x)|$$

Theorem 1(Theorem).

$C[a,b]=(C[a,b],\Vert \cdot {\Vert }_{\infty })$ is a complete normed linear space.

Proof
We have to show that every Cauchy sequence in $C[a,b]$ converges (in $C[a,b]$, of course). Suppose $(f_n)$ is a Cauchy sequence in $C[a,b]$. Then, given $ϵ>0$, $\exists N\in \mathbb{N}$ such that $n_1,n_2\ge N⟹\Vert f_{n_1}-f_{n_2}{\Vert }_{\infty }<ϵ⟹{\sup }_{x\in [a,b]}|f_{n_1}(x)-f_{n_2}(x)|<ϵ$. So, $(f_n)$ satisfies the Cauchy criterion for uniform convergence. Thus, $(f_n)\to f$, where $f$ is continuous.

Note that $(C[a,b],\Vert \cdot {\Vert }_{\infty })$ and $(C[a,b],\Vert \cdot \Vert )$, where the latter uses the norm induced by the inner product, are different spaces. The first one is complete, and the second one is not. As a counter example, consider the sequence of functions $(f_n)=x,x^2,x^3,\dots$ in $(C[0,1],\Vert \cdot \Vert )$. They converge to a function $f$ which is $1$ at $x=1$ and $0$ everywhere else - certainly not in $C[0,1]$. However, the sequence of functions is a Cauchy sequence with respect to $\Vert \cdot \Vert$: For every $ϵ>0$, $\exists N\in \mathbb{N}$ such that $n_1,n_2>N⟹\sqrt{{\int }_0^1|x^{n_1}-x^{n_2}{|}^{2}dx}<ϵ$. While we are at it, also note that $(f_n)$ is not a Cauchy sequence in $(C[a,b],\Vert \cdot {\Vert }_{\infty })$.

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Refer LADW, chapter 5 for parallelogram identity, polarization identity, and the relation between norms and inner products.