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?

  • $\begingroup$ Do you have a link to the paper? $\endgroup$ – Davide Giraudo Dec 17 '12 at 13:07

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.

  • $\begingroup$ Is this different to $H^1$? $\endgroup$ – Kofi Dec 17 '12 at 19:13
  • $\begingroup$ It is slightly different due to the normalization $\alpha(0)=0$, but otherwise it the same. And for the $n$-dimensional case it is just the direct sum of $n$ such Hilbert spaces. $\endgroup$ – Lukas Geyer Dec 17 '12 at 19:20

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