Show that $L^1([0,1])$ is not a Hilbert space.

The problem, as it is originally stated (here at http://orion.math.iastate.edu/dept/grad/analysis_f06.pdf), implies that we should show the parallelogram law fails for some potential inner product defined from the norm in the usual way. In particular, I'm wanting to find functions $f, g \in L^1([0,1])$ that force the equality $$\|f+g\|^2+\|f-g\|^2=2(\|f\|^2+\|g\|^2)$$

to fail, where $\|f\|=\int_{[0,1]}|f|d\mu$ with $\mu$ the Lebesgue measure on $[0,1]$. It is difficult to find examples that fail to satisfy this. My idea is, of course, to find $f$ and $g$ such that one side is finite but the other side is infinite, but none seem to come to mind - although most of my attempts have tried to force something out of $\frac{1}{\sqrt{x}}$. Thanks much!


Let $I_1=[0,1/2]$, $I_2=[1/2,1]$ and $f_1=\chi_{I_1}$, $f_2=\chi_{I_2}$.

  • $\begingroup$ I was chasing down a rabbit-hole. Beautiful answer from The Book. Thank you! $\endgroup$ – Darrin Dec 30 '13 at 3:36
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    $\begingroup$ @Darrin For the record, this is not originally my solution, it came up in the chat some time ago. $\endgroup$ – Pedro Tamaroff Dec 30 '13 at 3:38

Every Hilbert space is separable and reflexive. Therefore if $L^1$ were to be a Hilbert space it must also be separable and reflexive.

However, $L^1$'s topological dual is $L^\infty$ which is not separable (which can be verified using the indicator functions $I_{[0,1/4]}$ and $I_{[0,1/3]}$) where-from the reflexivity of $L^1$ implies it is not separable either, a contradiction.

Therefore $L^1$ cannot be as Hilbert space.


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