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Hello: I have to prove that $C^1([0,1])$ is not a Hilbert space. (with respect to $\def\norm#1{\left\|#1\right\|}\norm\cdot_*$; the norm is given by $$ \norm f_* := \bigl(\norm{f}_2^2 + \norm{f'}_2^2\bigr)^{\frac 12} $$

So, I need to find a counterexample. Can you help me?

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    Please see the FAQ for help formatting using MathJax. – 2017-01-03
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    Take the functions from [$C[0,1]$ is not Hilbert space](https://math.stackexchange.com/q/112168) and integrate them. – 2017-01-03
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    Take $f_n(x) = \int_0^{x-{1 \over 2}} \arctan (nt) dt$. – 2017-01-03

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The idea of a counterexample is that continuous functions may converge to a discontinuous one with respect to an integral norm such as $L^2$. Since the space involves derivative, the example should involve a limit function with discontinuous derivative (or perhaps not differentiable at all). The simplest of these is probably $f(x)=|x-1/2|$.

How to approximate the absolute value by differentiable functions? Increase the exponent slightly, so the graph becomes flat at the bottom: $f_n(x)=|x-1/2|^{1+1/n}$.

Here $f_n\to f$ uniformly, and also $f_n'\to f'$ in $L^2$ norm, as one can see from $$f_n'(x) = (1+1/n)|x-1/2|^{1/n}\operatorname{sign}(x-1/2)$$

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    Thank you very much!! It was a great help :) – 2017-01-07