Is this space a Hilbert Space? 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.
 A: 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.
A: 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.
A: 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.
