2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.