Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$\int_0^\infty |e^{-sx}f(x)|~dx \leq M\int_0^\infty e^{(a-s)x}~dx =\frac{M}{s-a}.$$

share|cite|improve this question
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
up vote 2 down vote accepted


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 $s$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.