I am looking at the following convolution integral (with $\theta > 0$):
$$ f(t) = \int_{-\infty}^t \exp(-(t-s)\theta) g(s) ds = (\textbf{1}_{\{.\geq 0 \}} \exp(-.\theta) * g)(t)$$
I want to find conditions on $g$ s.t.
$$ f \in C^{k, \alpha},$$ by which I denote the Hoelder space. That means, f should be $k$-times differentiable and the $k$-th derivative should satisfy (for some $C \in \mathbb{R}, 0 < \alpha \leq 1$:
$$\lvert f^{(k)}(x) – f^{(k)}(y) \rvert \leq C \lvert x-y \rvert^\alpha$$
I found that, if $g$ is bounded (which would be okay to assume in my case), then $f$ is uniformly continuous and bounded also. But what about the Hölder space?
Any ideas, hints, references welcome. Thanks!
Best Answer
Edit: In the first part, $C^{k,\alpha}$ is understood locally. For a global result, see the final part.
The equation reads $$f(t) = \int_{-\infty}^t e^{-(t - s)\theta} g(s) ds,$$ is equivalent to $$e^{t \theta} f(t) = \int_{-\infty}^t e^{s \theta} g(s) ds,$$ so if $g$ is $C^{k-1,\alpha}$, then $s \mapsto e^{s \theta} g(s)$ is $C^{k-1,\alpha}$, its integral $t \mapsto e^{t \theta} f(t)$ is thus $C^{k,\alpha}$, and finally $f$ is $C^{k,\alpha}$.
Note that this is optimal: if $f$ is $C^{k,\alpha}$, then, by the same argument, $g$ is $C^{k-1,\alpha}$.
Edit: a global result is given below.
The first part shows that if $g \in C^{k-1,\alpha}$ on $[0, 1]$, then $f \in C^{k,\alpha}$ on $[0, 1]$, with $$\|f'\|_{C^{k-1,\alpha}([0,1])} \leqslant C(k,\alpha) \|g\|_{C^{k-1,\alpha}([0,1])}.$$ But the result is clearly translation-invariant, and so $$\|f'\|_{C^{k-1,\alpha}([a,a+1])} \leqslant C(k,\alpha) \|g\|_{C^{k-1,\alpha}([0,1])} \leqslant C(k,\alpha) \|g\|_{C^{k-1,\alpha}(\mathbb R)}.$$ It follows that $$\|f'\|_{C^{k-1,\alpha}(\mathbb R)} \leqslant C(k,\alpha) \|g\|_{C^{k-1,\alpha}(\mathbb R)}.$$ It remains to observe that $\|f\|_\infty \leqslant C \|g\|_\infty$, and the desired global estimate follows.