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I have a space of continuously differentiable functions on [a, b] with the dot product defined in this way: $ x \cdot y = \int_a^b \! [x(t)y(t) + x'(t)y'(t)] \, \mathrm{d}t. $ Is this space a Hilbert Space? I think that completness of the space should be checked, but i don't know how to do it.

Comparing with the space of continuous functions on [a, b] (not mandatory differentiable) which has the dot product $ x \cdot y = \int_a^b \! x(t)y(t) \, \mathrm{d}t $ i see that my space and dot product (with derivatives) exclude some standart functional sequences that help to prove that the space of continuous functions is incomplete. I mean that, for example, this functional sequence $ f_n(t) = \begin{cases} -1, & \text{if }t\text{ in [-1, -1/n]} \\ nt, & \text{if }t\text{ in [-1/n, 1/n]} \\ 1, & \text{if }t\text{ in [1/n, 1]} \end{cases} $ shows that the space of continuous functions is incomplete, but it is not appliable to my problem, because it is not continuously differentiable.

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  • $\begingroup$ Its completion is Sobolev space $H^1([a,b])$ space $\endgroup$ – Norbert May 25 '14 at 19:51
  • $\begingroup$ I am not so sure about completeness. Suppose you have a Cauchy sequence of square integrable continuous x_n'(t). Then it may have a discontinuous limit. $\endgroup$ – Urgje May 25 '14 at 19:54
  • $\begingroup$ Consider $f_n(t)=\arctan(nt)$ $\endgroup$ – Norbert May 25 '14 at 21:30
  • $\begingroup$ @Norbert: I don't think $f_n(t)=\arctan(nt)$ is Cauchy. (Its pointwise limit is discontinuous, but elements of $H^1([a,b])$ are all continuous.) $\endgroup$ – Nate Eldredge May 25 '14 at 22:04
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You're getting close with your example functions $f_n$ on $[-1,1]$. Try letting $g_n(t) = \int_{-1}^t f_n(s)\,ds$. Show $g_n$ converges in your norm to a function which is not continuously differentiable.

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The completion of your linear space consists of all functions which are equal a.e. to absolutely continuous functions on $[a,b]$ whose derivatives are square integrable. This semi-classical characterization of the Sobolev space is available only for $R^{d}$ where $d=1$. So, no, the space of continuously differentiable functions is not complete when using your inner product.

It's not hard to show that the set of absolutely continuous functions on $[a,b]$ with square integrable derivatives is a complete space. And, if $f$ is any such function, there exists a sequence of continuous functions $\{ g_{n}\}_{n=1}^{\infty}$ which converges to $f'$ in $L^{2}[a,b]$. Then $f(a)+\int_{a}^{x}g_{n}(t)\,dt$ converges in your norm to $f$. So $f$ is in the completion.

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Ok, i got it. I constructed the sequence $ f_n(t) = \begin{cases} -t, & \text{if }t\text{ in [-1, -1/n]} \\ \frac{nt^2}{2} + \frac{1}{2n}, & \text{if }t\text{ in [-1/n, 1/n]} \\ t, & \text{if }t\text{ in [1/n, 1]} \end{cases} $

It has the function which i wrote above as a derivative and converges to |t|, which is not in the space of of continuously differentiable functions.

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