I'm reviewing for an upcoming qualifying exam in partial differential equations and have the following problem:
Prove that when $n=1$, the following pair is a Hilbert space. In this context $f'$ is the weak derivative of $f$.
$$ V = \left\{ f:(0,1) \rightarrow \mathbb{R} \,\, | \, \, \int_0^1 (f')^2 \,dx < \infty \,\,\, \text{and} \,\,\, \int_0^1 f \,dx = 0 \right\} $$ and $$ \langle f,g \rangle = \int_0^1 f'(x)g'(x) \,dx. $$
There are some hints given:
We can utilize that $W^{1,p}((0,1))$ functions must be absolutely continuous, so that given a $g \in L^p((0,1))$ there is a one dimensional family of functions:$$ f_c(x) = c + \int_0^x g(t) \,dt $$such that $ f'_c = g$. For this particular space $V$, there is a unique choice of c such that $f'_c = g$ and $f_c \in V$. With this unique $c$ you can more easily check that $(V, \langle \cdot ,\cdot \rangle)$ is an inner product space, and also complete. For completeness, if you have a Cauchy sequence $\{f_m\}$, you should invoke that face that $L^p((0,1))$ is a Banach space and thus you can identify a limit for the sequence $\{f'_m\}$.
The symmetry and linearity requirements are straightforward, the positive definiteness and the completeness are more troublesome. I find that $f \in V$ should take the form:
$$ f(x) = \int_0^x f'(t) \,dt - \int_0^1\int_0^t f'(s) \,ds\,dt $$
where the term with two integrals is the $c$ in the hint and in order for $\langle f,f\rangle = 0$ we must have $ c=0 $ thus $f=0$.
For completeness I use the other portion of the hint. Let ${f_m}$ be Cauchy in $(V, \langle \cdot ,\cdot \rangle)$ then through substitution and application of triangle inequality I get to
$$ |f_m(x) - f_k(x)| \leq \int_0^1 \int_0^x |f'_m(t) - f'_k(t)|\,dt\,dx + \int_0^x |f'_m(t) - f'_k(t)| \,dt. $$
Here is where I'm stuck. An outline of a solution has the following inequality:
$$ |f_m(x) - f_k(x)| \leq 2x^{1/2} \left( \int_0^1 |f'_m - f'_k|^2 \, dt\right)^{1/2} $$
and I am having trouble getting there from what I have. I suspect Holder's inequality somewhere but I have been going around in circles for the past day or two. It is within the realm of possibility that this inequality from the solution outline is incorrect.
Any help/guidance would be appreciated it.
I will prove that $|f_m(x) - f_k(x)| \leq 2 (\int |f_m'-f_k'|^{2})^{1/2}$ since the stated inequality is false. [The factor $x^{1/2}$ plays no role in the proof.
It is clear from Holder's inequality that the second term is bounded by $(\int_0^{1}|f_m'-f_k'|^{2})^{1/2}$.
For the first term interchange the integrals and note that $\int_t^{1} dx=1-t$, apply Holder's inequality again and observe that $\int_0^{1} (1-t)^{2}dt \leq 1$.
Now completeness of $L^{2}$ shows that $f_n'$ converges to some function $h$ in $L^{2}$ norm. The inequality we just proved shows that $f_n$ converges uniformly to some function $g$. Now $f_n(x)=\int_0^{x} f_n'(t)dt$ and passing to the limit we get $g(x)=\int_0^{x} h(t)dt$. This proves that $g$ is absolutely continuous and $g'=h$ a.e.. Thus $f_n' \to g'$ in $L^{2}$.