The space $C([0,1])$ of continuous functions on $[0,1]$ is an inner product space under the $L^2$-norm, but not complete. Equipped instead with the $L^\infty$-norm, it becomes complete but the norm is no longer induced by an inner product (in particular it does not satisfy the parallelogram law). This got me wondering: can we equip $C([0,1])$ with a norm which makes it into a Hilbert space?

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    $\begingroup$ What would be the trivial examples that would make it a Hilbert space? $\endgroup$ – Theo Bendit May 15 '19 at 8:28
  • $\begingroup$ @TheoBendit My mistake, a non-trivial semi-norm would make sense. In my head I was thinking of a zero norm. Edited! $\endgroup$ – jl2 May 15 '19 at 8:32

As a vector space, $C\bigl([0,1]\bigr)$ is isomorphic to $L^2\bigl([0,1]\bigr)$, since they both have Hamel bases with the same dimension (the cardinal of $\mathbb R$). So, if $\psi\colon C\bigl([0,1]\bigr)\longrightarrow L^2\bigl([0,1]\bigr)$, define $\lVert f\rVert=\bigl\lVert\psi(f)\bigr\rVert_2$.

Of course, this uses the Axiom of Choice. I don't know how to solve the problem without it.


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