C$([0,1])$ does not admit an inner product.

Question

Let $X:=C([0,1])$ be a normed vector space, with norm $||f(x)||_X=$sup$|f(x)|$. Then $X$ does not admit an inner product $\langle \cdot \ , \cdot \rangle $

Since $[0,1]$ is compact, $X$ is Banach, so if it were to admit an inner product, it would be a Hilbert space. Hilbert spaces are reflexive. On the other hand $X$ is not reflexive:

Indeed, $X^*$ consist of measures on $C([0,1])$, and the dual of measures is sets, via $T_A: \mu \mapsto \mu(A)$. Therefore X is not reflexive, which is a contradiction.

My proof makes sense, but I'm not sure how correct it is. How can I make it more formal?

Answer 1

Your proof is Ok but a much simple argument is enough. Namely, for inner product we have identity: $\|x-y\|^2 + \|x+y\|^2 = 2\|x\|^2 + 2 \|y\|^2$. That is, if $\|\|_{\infty}$ norm came from an inner product then for any functions $x(t), y(t) \in C[0,1]$ we would have: $ (\sup| x(t) - y(t)|)^2 + (\sup|x(t) + y(t)|)^2 = 2\sup|x(t)|^2 + 2\sup|y(t)|^2$
But this does not hold for, say, $x(t) = t$ and $y(t) = t^2$.