# Show that $C([0,1],\mathbb{R})$ with the $L_2$ inner product norm is not a Hilbert space.

I need to prove that all continuous functions on the closed set $[0,1]$ is not a Hilbert space. Given the $L_2$ norm.

I guess I need to show that every Cauchy sequence in the space, does not converge under the given norm. But I am a bit lost on how, not asking for full solutions here. Just some tips on how to get started. Maybe some general tips on how tackle such problems?

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Show that the indicator function of the interval $[\tfrac{1}{2}, 1]$ is an element of the $L^2$ closure. Try the first sequence of continuous functions that springs to mind. – WimC Nov 7 '12 at 12:16
Well, it suffices to find a Cauchy sequence that does not converge. Note that you're talking about continuous functions here, so a Cauchy sequence that has a non-continuous pointwise limit will do the trick, since although it converges to a limit in a bigger class of functions, that limit isn't in the space you're considering. – MightyJohnson Nov 7 '12 at 12:20
It's not entirely correct that you need to show that every Cauchy sequence does not converge in that space, but merely, some don't. There are Cauchy sequences that do converge, you just need to find one that doesn't. – T. Eskin Nov 10 '12 at 19:33

You need to find a $\|\cdot\|_2$-Cauchy sequence which doesn't converge in $C([0,1])$. You can think of approximating a jump function (with jump in the interior of $[0,1]$) by piecewise affine functions.
I am thinking something along the lines of a trigonometric sequence. What about $x \Rightarrow \sin(1/x)$? – N3buchadnezzar Nov 7 '12 at 17:41