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Suppose $f:[0,\infty)\rightarrow \mathbb{R}$ is lebesgue integrable with compact support. Define the laplace transform by $$F(t) = \int_0^\infty f(x) e^{-tx}dx.$$

Show that $F$ is continuously differentiable on $[0,\infty)$.

The best I've been able to do with this one is to show that it's a decreasing function, and thus is differentiable almost everywhere. And that's about it.

I found another solution here, but I don't really understand it.

Any thoughts on how I could make a statement about differentiability (and the subsequent continuity) for this problem would be greatly appreciated.

Thanks in advance.

Bears
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  • Is the Leibniz rule not applicable ? – Maximilian Janisch Aug 07 '20 at 22:40
  • Maybe I'm over complicating something pretty straightforward, but I don't think it is since we don't know anything about the differentiability of $f$. – Bears Aug 07 '20 at 22:47
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    The function $t\mapsto f(x)\exp(-t x)$ is differentiable for fixed $x$ so I think you can use something like this: https://en.wikipedia.org/wiki/Leibniz_integral_rule#Measure_theory_statement – Maximilian Janisch Aug 07 '20 at 22:49

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Hint: You can simply "do it" without relying on any results on differentiation under the integral:

$$\frac{F(t+h)-F(t)}{h}=\int_0^\infty f(x)e^{-xt}\frac{e^{-xh}-1}{h}\,dx$$

and note that the fraction in the last integral converges uniformly on the (compact) support of $f$.

quid
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David C. Ullrich
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  • Ahh. The uniform convergence is what I was overlooking.

    Thanks a bunch.

     – Bears Aug 07 '20 at 23:30
  • Or note that the improper integral of the differentiated integrand, namely $-\int_0^\infty xf(x)e^{-xt},dx$, converges uniformly for $t>t_0>0$. – Mark Viola Aug 08 '20 at 00:11