Can someone give me an example of a Banach space that is not a Hilbert space? I can't think of any because I don't know how to show one space that can not have inner product structure.

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    $\begingroup$ A Banach space is a Hilbert space if and only if its norm satisfies the Parallelogram Law. Check with, say $\ell_\infty^2$. $\endgroup$ Feb 27, 2014 at 10:10
  • 1
    $\begingroup$ To add to that, if you do satisfy the parallelogram law, you can recover the inner product through a polarization identity. $\endgroup$
    – Batman
    Feb 27, 2014 at 10:51
  • $\begingroup$ @DavidMitra: what space is $\ell_\infty^2$? $\endgroup$ Nov 22, 2017 at 23:27

3 Answers 3


The space $C[0,1]$ of continuous functions $f:[0,1]\to\mathbb R$ with the supremum norm is an example of a Banach space which is not a Hilbert space.

We need to check that the parallelogram law is not satisfied. Take $f(x)=x$, $x\in[0,1]$, and $g(x)=1$, $x\in[0,1]$. Then $2(\|f\|_\infty^2+\|g\|_\infty^2)=4$, but $\|f+g\|_\infty^2+\|f-g\|_\infty^2=5$.


If a Banach space $(X, \|\cdot\|$ is a Hilbert space, then the norm satisfies the "parallelogram identity" $$ \|x+y\|^2+\|x-y\|^2=2\big(\|x\|^2+\|y\|^2\big). $$ But the norm of $C[0,1]$ does not satisfy such an identity: for $f=1$ and $g=x$, $$ \|f\|=1,\,\,\|g\|=1,\,\,\|f+g\|=2,\,\,\|f-g\|=1, $$ and hence $$ 5=\|f+g\|^2+\|f-g\|^2\ne2\big(\|f\|^2+\|g\|^2\big)=4. $$


Banach spaces that are not a Hilbert space are, among many others, $L^{p}(Rⁿ,dⁿx)$ for $p∈[1,∞),p≠2$. Linear functionals on such spaces can be written as an integral similar to the Hilbert space inner product but in general the functional cannot be associated with an element of the space itself. But there exists the notion of a semi-scalar product which was used by Lumer and Phillips in a study of contraction semi-groups (see Yosida, Functional Analysis).


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