How to show that the Banach space $\left(C[a,b],\lVert.\rVert_{\scriptsize C[a,b]}\right)$ is not Hilbert space?

Question

I want to show that the Banach space $\left(C[a,b],\lVert.\rVert_{\scriptsize C[a,b]}\right)$ is not a Hilbert space. So I should show that it is not an inner product space. Most likely, The Parallelogram Equality does not work for some two elements of this space but I could not find these two elements. Thanks for your helps.

Answer 1

Take $$\begin{cases} f_1(x)=-1+2\frac{x-a}{b-a}\\ f_2(x)=\left\vert \frac{2x - (a+b)}{b-a} \right\vert \end{cases}$$

You have $\Vert f_1 \Vert = \Vert f_2 \Vert=1$ and $\Vert f_1+f_2 \Vert = \Vert f_1-f_2 \Vert = 2$. Hence the parallelogram law is not satisfied.

Answer 2

High level argument: The space $C ([a,b]) $ is separable, but its dual space (the space of finite measures) is not.

Indeed, we have $\|\delta_x -\delta_y\|\geq 1$ for $x\neq y $, where $\delta_x $ is the dirac measure at $x $. Note that we do not even need to know that the dual space is given by the space of finite measures for this argument to work. It suffices to know that each dirac measure is in the dual.

But if $C ([a,b]) $ was a Hilbert space, it would be self-dual, so that its dual space would be separable.

Answer 3

Almost any choice of $f,g$ works. Assume for example that $[a,b] = [0,1]$ and let $f = x^2$ and $g = x$. Then

$$ 2||f||^2 + 2||g||^2 = 2 + 2 = 4 $$

while

$$ ||f + g||^2 + ||f - g||^2 = 2^2 + \frac{1}{4^2} = 4 + \frac{1}{16}. $$