The sup norm on $C[0,1]$ is not equivalent to another one, induced by some inner product

Let $\mathrm{C}[0,1]$ be the space of continuous functions $[0,1]\rightarrow \mathbb{R}$ endowed with the norm $||x||_{\infty}=\mathrm{max}_{t\in [0,1]}|x(t)|$. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for $x(t)=t$ and $y(t)=1$). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties...

Solution 1:

As David Mitra, pointed out this particular norm is not equivalent to norm induced by inner product because $C([0,1])$ is not reflexive. But reflexivity is not enough for space to be Hilbertable.

One can suggest that being isomorphic to its dual is enough, but $X \oplus_2 X^*$ with reflexive $X$ gives a bunch of counterexamples.

Characterisation in terms of geometry of Banach spaces was given by Lindenstrauss and Tzafriri: Banach space $X$ is isomorphic to Hilbert space iff every closed subspace of $X$ is complemented (i. e.the range of some bounded projection).