Laplace transform with (real) compact support

Given the standard unilateral Laplace transform defined on $L^1(\mathbb R ^+)$ $$\mathscr Lf(s) = \int_0^\infty e^{-st}f(t)~dt,$$

are there any functions in $L^1$ such that $\mathscr Lf$ is "compactly supported", where with compact support I mean that there exists an $M >0$ such that $$\mathscr Lf(s) = 0 \quad\text{if}\ \operatorname{Re} s > M.$$

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$f\equiv 0$. (Sorry, just being a dick.) – Nick Thompson Apr 18 '12 at 16:32

No, because $\tilde f(z) = \int e^{-zt}f(t)dt, Re(z) > 0$ is an analytic function of z and so is not going to vanish on any set with a limit.