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Let $f:[0,\infty)\to \mathbb{R}$ be supported on $[0,1]$, with $\int_0^1 f(x) dx = 1$. Let $\mathcal{L} f$ be its Laplace transform. How slowly may $$\int_{-\infty}^\infty |\mathcal{L} f(\sigma+i t)| dt$$ grow as $\sigma\to -\infty$? It is clear it could be $\ll e^{\epsilon |\sigma|}$ (just let $f$ be supported on $[0,\epsilon]$). Could it grow polynomially on $|\sigma|$? Linearly on $|\sigma|$?

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2 Answers 2

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It has exponential growth: if your integral is $O(e^{-\sigma\epsilon}), \sigma\to -\infty,$ then the support of your function $f$ is contained in $[0,\epsilon]$. This follows from the inversion formula for the Laplace transform: $$f(x)=\lim_{r\to\infty}\frac{1}{2\pi i}\int_{b-ir}^{b+ir}Lf(s)e^{sx}ds,$$ where the integration is on any vertical line $\Re s=b$. Under your assumptions, $$|f(x)\leq\frac{1}{2\pi} e^{-b\epsilon}e^{bx},$$ for every $b<0$, therefore, by letting $b\to-\infty$ we obtain $f(x)=0$ when $x>\epsilon$.

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The Fourier inversion formula says that $$f(x)e^{-\sigma x} = (2\pi)^{-1} \int \mathcal{L}f(\sigma+it) e^{itx} dt$$ so if your integral is $I(\sigma)$, one sees that $$|f(x)| \le (2\pi)^{-1} e^{\sigma x} I(\sigma)$$ It follows that if $I(\sigma)$ is $O(e^{\varepsilon\sigma})$, then f is supported on $[0,\varepsilon]$. So that can't hold for ALL $\varepsilon$. Thus the behavior that you mention is best possible.

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