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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? – Sangchul Lee Nov 15 '12 at 0:59
@sos440 let $g_n = f(s)\cdot e^{-ns}$? it sems not working. – newbie Nov 15 '12 at 1:10
Note that the family $\{ f(s)e^{-zs}:\Re(z)>0\}$ is dominated by an integrable function $|f(s)|$. – Sangchul Lee Nov 15 '12 at 1:28
@sos440 Yes. My question is how to show $g_n$'s pointwise limit is 0? since $f\in L^1$. – newbie Nov 15 '12 at 2:12

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