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Let $ Lf(s) = \int_0^\infty e^{-sx}f(x)~dx$ be the Laplace transform of a measurable function $f$ on $[0,\infty).$
I would like to be able to show the following:

  1. If $f\in L^1[0,\infty]$ , then $Lf(s)$ exists and is bounded for all $s\geq 0$.

  2. Suppose $\{f_n\}~n\geq 1$ and $f\in L^1[0,\infty)$. If $f_n \to f$ in $L^1$-norm, then $Lf_n\to Lf$ uniformly on $[0,\infty)$.

For (1), these are my thoughts. Since $f\in L^1[0,\infty)$ and $f\geq 0$, $f$ is finite a.e. So there is an $M \gt 0$ such that $|f(x)|\leq M$. Infact, I can choose $M$ large enought so that $|f(x)|\leq M e^{at}$ for some $a\geq 0$. Then $\int_0^\infty |e^{-sx}f(x)|~dx \leq M\int_0^\infty e^{(a-s)x}~dx =\frac{M}{s-a}.$

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    A part of the mistake is interesting. Exercise: Let $f$ be integrable on $(0,+\infty)$. Does this imply that there exists some finite $M$ and $a$ such that $|f(x)|\leqslant M\mathrm e^{ax}$ for almost every $x\geqslant0$? (The answer is no.)2012-03-27

1 Answers 1

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Hints:

For 1.: $|Lf(s)| \leq \int_0^\infty |\exp(-sx)f(x)|dx$ and use $s\geq 0$ (shouldn't it be that the real part of $s$ in non-negative?).

For 2.: Start with $|Lf_n(s) - Lf(s)|$ and bound this difference by something that goes to zero and is independent of $s$.

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    Despite the typos it looks like you are meaning the right thing...2012-03-27