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...

  • 1
    $\begingroup$ Reflexivity comes to mind. $\endgroup$ Commented Oct 26, 2013 at 13:17
  • $\begingroup$ 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. $\endgroup$ Commented Oct 26, 2013 at 13:44
  • $\begingroup$ @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. $\endgroup$ Commented Oct 26, 2013 at 13:47
  • $\begingroup$ @HansEngler $C[0,1]$ is complete and not reflexive. Complete inner product spaces are reflexive. Reflexivity and completeness are preserved by isomorphisms. $\endgroup$ Commented Oct 26, 2013 at 13:51

1 Answer 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).

  • $\begingroup$ 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) $\endgroup$
    – user74574
    Commented Oct 26, 2013 at 18:23
  • $\begingroup$ @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 $C([0,1])$ $\endgroup$
    – Norbert
    Commented Oct 26, 2013 at 18:35
  • $\begingroup$ Norbert, thank you for your help! So, could you provide any explicit embedding $\mathcal{l}_1\subset C[0,1]$?:) $\endgroup$
    – user74574
    Commented Oct 26, 2013 at 19:54
  • $\begingroup$ No, embeddings that I know how to construct are highly non-explicit $\endgroup$
    – Norbert
    Commented Oct 26, 2013 at 20:35
  • $\begingroup$ 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. $\endgroup$
    – Norbert
    Commented Oct 26, 2013 at 20:50

You must log in to answer this question.