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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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    @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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