# Is there a norm making $C([0,1])$ into a Hilbert space?

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?

• What would be the trivial examples that would make it a Hilbert space? – Theo Bendit May 15 '19 at 8:28
• @TheoBendit My mistake, a non-trivial semi-norm would make sense. In my head I was thinking of a zero norm. Edited! – 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 $$\lVert f\rVert=\bigl\lVert\psi(f)\bigr\rVert_2$$.
• ,Hi the key point then is why this isometric isomorphism $\psi\colon C\bigl([0,1]\bigr)\longrightarrow L^2\bigl([0,1]\bigr)$ exist? – yi li Nov 2 '20 at 11:03
• @yi_li First you define $\psi$ as a linear isomorphism from $C\bigl([0,1]\bigr)$ onto $L^2\bigl([0,1]\bigr)$; it exists, since these spaces have the same dimenssion. Then you define, for each $f\in C\bigl([0,1]\bigr)$, $\|f\|$ as $\bigl\|\psi(f)\bigr\|_2$. Then $\psi$ becomes an isometry from $\left(C\bigl([0,1]\bigr),\|\cdot\|\right)$ onto $\left(L^2\bigl([0,1]\bigr),\|\cdot\|_2\right)$. – José Carlos Santos Nov 2 '20 at 13:15