We define the Laplace transform of a non-negative function $f : \mathbb{R_+} \to \mathbb{R_+}$ by $$\mathcal{L}f(q) \triangleq \int_0^{+\infty}f(t)e^{-qt}dt,$$ where $q$ is in the domain of convergence of the integral.

Suppose that $\mathcal{L}f(q) \leq C(q)$ for some $C(q) \in \mathbb{R}_+$ that can depend on $q$. Question : is it possible to find an upper bound on $f(t)$ for any $t > 0$ that involves $C(q)$ ?

One possible answer is to assume that $f$ in non-decreasing and then we obtain, for each $t > 0$ and $q$ in the domain, \begin{equation} \begin{split} q\mathcal{L}f(q) & = \int_0^t q f(s)e^{-qs}ds + \int_t^{+\infty} q f(s)e^{-qs}ds \\ & \geq f(t) \int_t^{+\infty} q e^{-qs}ds = f(t)e^{-qt}. \end{split} \end{equation}

Therefore, $f(t) \leq e^{qt}q\mathcal{L}f(q) \leq e^{qt}qC(q)$.

Now, in my case I know that $f(t) \leq 1$ for all $t > 0$ (which also implies $q\mathcal{L}f(q) \leq 1$, for all $q$, so we can assume that $C(q) < 1$). So, I would like to find an inequality like the one above but where $e^{qt}$ is replaced by something which is smaller or equal to $1$. Does anybody have any ideas ? Thanks a lot !

  • $\begingroup$ You might consider adding a top-level tag in order to make more people see this question. $\endgroup$
    – Stefan Kohl
    Mar 29, 2017 at 12:04
  • $\begingroup$ Thank you for your comment ! I added CA, but maybe GM is more appropriate ? $\endgroup$
    – kantadou
    Mar 29, 2017 at 16:20
  • $\begingroup$ I think ca.analysis-and-odes is fine. $\endgroup$
    – Stefan Kohl
    Mar 29, 2017 at 16:23
  • $\begingroup$ @kantadou did you ever solve this problem? I'm facing similar issue: my $f(t)$ is monotonically decreasing and less than 1), $\mathcal{L}(f)$ is easy to bound but hard to invert, how to bound $f$ $\endgroup$ Mar 24 at 22:27

1 Answer 1


Without any assumptions (like monotonicity of your function) you cannot get any nice bound. That is, there exists the sequence of functions $f_n$ such that $0\leq f_n(t)\leq 1$ for all $t \in \mathbb R_{\geq 0}$ and $C(q)(n)$ tends to 0 uniformly in every finite interval $(\varepsilon,M)$, but $\sup\limits_{t \in \mathbb R_{\geq 0}} f_n(t)=1$ for any n.

Proof: Take $f_n(t)=\cos^n(t)$. By the Euler's formula, we have

$$f_n(t)=2^{-2n}\sum_{k=0}^{2n} {2n\choose k}e^{i(2k-2n)t}={2n\choose n}2^{-2n}+\sum_{k=0}^{n-1} 2^{1-2n}{2n \choose k}\cos((2n-2k)t).$$

Now, as Lapace transform of $\cos(\omega t)$ equals $\frac{q}{q^2+\omega^2}$, we get

$$\mathcal{L}f_n(q)=\frac{{2n\choose n}}{q2^{2n}}+\sum_{k=0}^{n-1} 2^{1-2n}{2n\choose k}\frac{q}{q^2+(2n-2k)^2}.$$

Next, by the Stirling's approximation,

$${2n\choose n} \sim \frac{2^{2n}}{\sqrt{\pi n}}.$$

Also, for any $k$ we have ${2n \choose k} \leq {2n \choose n}$. Lets decompose the sum into the sum over those $k$ which have $0\leq n-k \leq n^{1/4}$(say) and all the others. For the $k$ of first type we have trivially

$$\frac{q}{q^2+(2n-2k)^2}\leq \frac{1}{q}$$

and $O(n^{1/4})$ summands of this type. Thus, for the first type sum $S_1$ we get

$$S_1\leq \sum_{0 \leq n-k \leq n^{1/4}}\frac{2^{1-2n}{2n\choose n}}{q}.$$

There is $O(n^{1/4})$ summands and each of them is $O(\frac{1}{\sqrt{n}q})$. So,


In the second sum $S_2$ for every $k$ we have $n-k \geq n^{1/4}$, thus

$$\frac{q}{q^2+(2n-2k)^2}=O(\frac{q}{\sqrt n}).$$


$$S_2=O(\sum_{k=0}^{2n} 2^{-2n}{2n \choose k}\frac{q}{\sqrt n})=O(\frac{q}{\sqrt n}).$$

So, for $q$ in each interval $(\varepsilon,M)$ we have



$$\mathcal{L}f_n(q)\to 0$$

as $n \to \infty$, uniformly in $q$, as needed.

(exponent $1/4$ can be improved to $1/3$ but that is not very useful)

  • $\begingroup$ Thank you for your answer ! This does show that it is not possible in general and so answers the initial question. I will still try to look for cases where there is monoticity. $\endgroup$
    – kantadou
    May 22, 2017 at 10:19

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