# 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...

• Reflexivity comes to mind. Commented Oct 26, 2013 at 13:17
• A norm is induced by an inner product iff the parallelogram identity holds. You have answered the question already, and that is all there is to it. Commented Oct 26, 2013 at 13:44
• @David - all $\ell^p$ spaces with $1 < p < \infty$ are reflexive, but their norms come from an inner product only for $p = 2$. And the vector space of all polynomials on $[0,1]$ equipped with the usual $L^2$ norm is not reflexive (since it is not complete) although its norm comes from an inner product. So reflexivity has little to do with it. Commented Oct 26, 2013 at 13:47
• @HansEngler $C[0,1]$ is complete and not reflexive. Complete inner product spaces are reflexive. Reflexivity and completeness are preserved by isomorphisms. Commented Oct 26, 2013 at 13:51

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).

• So, we are to provide a closed subspace $X_0\subset C[0,1]$ such that there is no bounded projection onto $X_0$. Could you name some 'famous' closed subspaces of $C[0,1]$? ($\{f: f(0)=0\}$ does not suit us... It is the first one I remembered)
– user74574
Commented Oct 26, 2013 at 18:23
• @Nikita, every finite (co)dimensional subspace is complemented. The space of linear functions on $[0,1]$ is one dimensional, so it is also complemented. In fact every separable Banach space can be isometrically embedded into $C([0,1])$. In particular $\ell_1$. But as Mazur proved every isometric copy of $\ell_1$ is not complemented in $C([0,1])$ Commented Oct 26, 2013 at 18:35
• Norbert, thank you for your help! So, could you provide any explicit embedding $\mathcal{l}_1\subset C[0,1]$?:)
– user74574
Commented Oct 26, 2013 at 19:54
• No, embeddings that I know how to construct are highly non-explicit Commented Oct 26, 2013 at 20:35
• But I think I have an explicit example. These are the space of continuous functions on $[0,1]$ that vanish at $0$ and $1$ and have negative Fourier coefficients. Commented Oct 26, 2013 at 20:50