How to prove that the Laplace transform goes to zero?

Question

Given a function $f:[0,\infty)\to\mathbb R$, let $F$ be the function defined by $$F(s)=\int_0^\infty e^{-st}f(t)\;dt,$$ whose domain is the set of points $s$ for which the integral converges.

Question: How to show that $F(s)\overset{s\to\infty}{\longrightarrow} 0$?

Remarks:

• The function $F$ is the Laplace transform of $f$.

• Simmons book says that the convergence $F(s)\overset{s\to\infty}{\longrightarrow} 0$ is true in general but proves it only if $f$ is piecewise continuous and of exponential order.

• A similar reasoning can be applied if $f\in L^p(0,\infty)$ for some $p>1$: from Hölder's inequality, $$|F(s)|\leq \|e^{-st}\|_{L^q}\|f\|_{L^p}=\frac{1}{(q s)^{1/q}}\|f\|_{L^p}\qquad(\tfrac{1}{p}+\tfrac{1}{q}=1)$$ which implies the desired result.

• I couldn't figure out how to treat the case where $f$ is not $p$-integrable as, for example, $f(t)=\frac{1}{\sqrt{t}}$ (for this particular example we can see that the condition is satisfied by calculating the integral, which gives $F(s)=\frac{\sqrt{\pi}}{\sqrt{s}}$).

Answer 1

If $x \mapsto G(x,K):=\exp(- K x) f(x)$ is Lebesgue integrable for some $K>0$, we know that also $x \mapsto G(x,k)$ is Lebesgue-integrable for all $k \ge K$. Note that for all $s=s'+K$ with $s'>0$. we have $$\exp(-st) f(t) = \exp(-ts') G(t,K) \le G(t,K).$$ Thus, we have found an integrable majorant and therefore we can apply the dominanted convergence theorem in order to conclude that $F(s) \rightarrow 0$ for $s \rightarrow \infty$.

In particular, this argument applies for $f(t) = t^{-1/2}$ and also $f(t) = (t+1)^{-1}$.

Answer 2

You need to assume something. As in the other answer, a reasonably weak assumption here is that there exists $K>0$ with $$\int_0^\infty e^{-Kt}|f(t)|\,dt<\infty.$$

Assuming that. the result is very simple, no need for dominated convergence. Let $\epsilon>0$. Choose $A>0$ so $$\int_0^Ae^{-Kt}|f(t)|\,dt<\epsilon.$$

If $s>K$ and $t>A$ then $$\frac{e^{-st}}{e^{-Kt}}=e^{-(s-K)t}<e^{-(s-K)A}.$$

So if $s>K$ then $$\int_0^\infty e^{-st}|f(t)|=\int_0^A+\int_A^\infty <\epsilon+e^{-(s-K)A}\int_A^\infty e^{-Kt}|f(t)|=\epsilon+ce^{-(s-K)A},$$which is less than $2\epsilon$ if $s$ is large enough.