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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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For 1, think about absolute values and how $e^{-sx}$ is bounded for fixed $s,x\geq 0$. For the second one, the idea is essentially the same. –  Alex R. Mar 27 '12 at 2:07
    
Your claim that "Since $f\in L^1$ ... then there is an $M>0$ such that $|f(x)|\leq M$" is false. Think about $1/\sqrt{x}$ on the interval $[0,1]$. You can however assert that $|f(x)|>M$ on a set of arbitrary small measure by Markov's Inequality. –  Alex R. Mar 27 '12 at 2:13
    
@Sam What I said was that $|f(x)|\leq M$ a.e. Do you mind elaborating on your first comment. Thanks. –  Kuku Mar 27 '12 at 2:27
    
What you're claiming is that there is some finite constant, $M$, such that $|f(x)|\leq M$ for every $x\in[0,\infty)$, except for possibly some $x$ in a set of measure zero. This is false, as @Sam has stated. For example, the function $f(x)=1/\sqrt x$ for $0\leq x\leq1$ and $f≡0$ for $x>1$ is in $L^1$, but it is not bounded as you say. However, you can say that for any fixed $\epsilon>0$, there exists $M_\epsilon<\infty$ such that $\left|\{x\in[0,\infty):|f(x)|>M_\epsilon\}\right|<\epsilon.$ –  Patch Mar 27 '12 at 5:20
    
What you do have, however, is that $e^{-sx}\leq1$, for all $x\geq0$, since $s\geq0$. Try using this to rework your argument for the first part. –  Patch Mar 27 '12 at 5:27

1 Answer 1

up vote 2 down vote accepted

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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Thanks: Can I do this for (2.)? $$|Lf_n(s)-f(s)|\le \int_0^\infty |f_n(x)-f(x)|e^{-sx}\leq \int_0^\infty |f_n(x)-f(x) \to 0, $$ since $f_n\to f$ in the $L^1$ norm –  Kuku Mar 27 '12 at 16:11
    
Despite the typos it looks like you are meaning the right thing... –  Dirk Mar 27 '12 at 21:25

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