Let $H$ be the space of functions $\alpha: [0, T] \longrightarrow \mathbb{R}^n$ that are absolutely continuous and such that $\alpha(0)=0$. The statement that I have implicitly found in a paper is that this is a Hilbert space with the scalar product
$$ (\alpha, \beta) = \int_0^T \langle \dot{\alpha}(t), \dot{\beta}(t)\rangle \mathrm{d} t.$$
The first question is why this is even well-defined. I know that the product of absolutely continuous functions is absolutely continuous and that an absolutely continuous function has a derivative in $L^1$, but in the integral, this is something like a second derivative.
The second question is about the completeness. Are there good references?
Best Answer
The integral is certainly not finite in general, as there is no reason that the product of two $L^1$ functions is integrable. For a counterexample consider the case $n=1$, and choose $\alpha(t) = \beta(t) = 2\sqrt{t}$. Then the integral becomes $\int_0^T \frac{1}t \, dt = \infty$. This easily gives a counterexample for general $n$ as well.
As to completeness, if you have a sequence $\alpha_n$ which is Cauchy w.r.t. this inner product, then $\dot{\alpha}_n$ is a Cauchy sequence in $L^2$, so it converges to some $\gamma \in L^2$. Then by Cauchy-Schwarz, $\gamma \in L^1$, and $\alpha(t) = \int_0^s \gamma(s)\, ds$ is absolutely continuous with $\alpha(0)=0$ and $\dot{\alpha} = \gamma$, so $\alpha_n \to \alpha$ w.r.t. the inner product.
Summarizing, the subspace of absolutely continuous functions with $L^2$ derivatives is indeed a Hilbert space w.r.t. the inner product given.