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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?

Alessandro
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    The dual of measures is not sets. https://math.stackexchange.com/questions/47544/double-dual-of-the-space-c0-1 – Anne Bauval Feb 01 '23 at 13:43
  • In my opinion, it'd be easier to say that the norm you chose doesn't satisfy the parallelogram identity, and as such cannot be associated to an inner product. However, I'll only leave it as a comment since it's a bit unrelated to your approach. – Bruno B Feb 01 '23 at 13:46
  • I was preparing an answer along the same lines https://en.wikipedia.org/wiki/Polarization_identity , using the Fréchet-Von Neumann-Jordan theorem you mention. But if you think it is unrelated to the OP's approach, let us consider my first comment as a proposition of duplicate. – Anne Bauval Feb 01 '23 at 13:48
  • A different and probably simpler argument is given here: A Banach space that is not a Hilbert space. It boils down to checking the parallelogram law. – Cm7F7Bb Feb 01 '23 at 13:54
  • Yes (and it is also @Salcio's answer) but this is what Bruno B and I agreed to consider as unrelated to the OP's approach. – Anne Bauval Feb 01 '23 at 13:55
  • It is simpler to check the parallelogram law, but the reason I didn't write an answer with this is that OP was going for a different appraoch and wanted to know if their proof was correct. Sure, it's nice to have one way of solving a question, but it'd be just as neat to have OP's approach work. Which I couldn't really evaluate because I don't have enough experience yet. (But that's just my opinion!) – Bruno B Feb 01 '23 at 13:59

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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 $y(t) = t^2$.

Bikhu
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Salcio
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