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consider $f\in L^1(R_+)$ and define Laplace transform $$\mathcal{L}f(z):=\int_0^{\infty} f(s)e^{-zs}\mathbb{d}s. $$

How can I prove $$\lim_{\mathbf{Re}z\rightarrow\infty}\mathcal{L}f(z) = 0?$$

Intuitively it is obvious, but without exponential property I can't figure out how to tackle it.

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    Have you tried the dominated convergence theorem?2012-11-15
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    @sos440 let $g_n = f(s)\cdot e^{-ns}$? it sems not working.2012-11-15
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    Note that the family $\{ f(s)e^{-zs}:\Re(z)>0\}$ is dominated by an integrable function $|f(s)|$.2012-11-15
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    @sos440 Yes. My question is how to show $g_n$'s pointwise limit is 0? since $f\in L^1$.2012-11-15

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