I have been considering the Laplace transform $$\mathcal{L}(f)(s)=\int_{0}^{\infty}{f(t)\, e^{-st}dt}$$ defined on $s\in\mathbb{R}^{+}$ as an linear operator from $\mathcal{L}^{2}(\mathbb{R}^{+},\lambda)$ to $\mathcal{L}^{2}(\mathbb{R}^{+},\lambda)$. I have managed to prove that it is well-defined, linear and bounded in the sense that $$||{\mathcal{L}(f)}| |_{\mathcal{L}^{2}(\mathbb{R}^{+},\lambda)}\leq \sqrt{\pi}\cdot ||f||_{\mathcal{L}^{2}(\mathbb{R}^{+},\lambda)}$$ Hence I have established that the operator-norm of $\mathcal{L}$, denoted $||\mathcal{L}||$ is bounded by $\sqrt{\pi}$, however my question is if the constant $\sqrt{\pi}$ is essentially sharp?

If it is, how does one prove such a thing?

I have tried pluggin in $f(t)=\frac{1}{\sqrt{t}}\cdot 1_{(a,b)}$ , where $0<a<b<\infty$, but without any success.

  • $\begingroup$ What are you using as the norm, e.g. what would be $\|f\|$ for $f(x) \equiv 1$? If it is $\int_0^\infty f(t) dt$ then $\|1\| = \infty$? $\endgroup$
    – gt6989b
    Nov 1, 2015 at 1:23
  • $\begingroup$ I am using the standard $\mathcal{L}^{2}$ -norm over $\mathbb{R}^{+}$. i.e the norm induced by the scalar product $\int_{0}^{\infty}{f(t)\cdot \bar{g(t)} dt}$ $\endgroup$ Nov 1, 2015 at 1:27
  • $\begingroup$ So in that standard norm you get $$\|1\| = \int_0^\infty 1^2 dt = \infty?$$ It just feels a little weird, you certainly expect $\|1\| < \infty$... $\endgroup$
    – gt6989b
    Nov 1, 2015 at 1:28
  • 4
    $\begingroup$ Well, the norm is only defined for square-integrable functions on $\mathbb{R}^{+}$, so I'm not sure we can say that, strictly speaking $\endgroup$ Nov 1, 2015 at 1:30
  • $\begingroup$ you are right... $\endgroup$
    – gt6989b
    Nov 1, 2015 at 1:30

2 Answers 2


I consulted my professor yesterday and he managed to provide me with a rather elementary but very clever solution.

Consider the sequence of functions $\left\{f_{n}\right\}_{n\geq 1}$ defined by $f_{n}(t)=\frac{1}{\sqrt{t}}\chi_{[\frac{1}{n},n]}(t)$ and notice that $f_{n}\in \mathscr{L}^{2}(\mathbb{R}^{+})$ for each fixed $n\geq 1$ with $||f_{n}||_{2}^{2}=2\log(n)$

Performing a change of variable in the inner integral we have that $$||\mathcal{L}(f_{n})||_{2}^{2} = \int_{0}^{\infty}\left(\int_{\frac{1}{n}}^{n}\frac{1}{\sqrt{t}}e^{-st}\, dt\right)^{2}\,ds=\int_{0}^{\infty}\frac{1}{s}\left(\int_{\frac{s}{n}}^{ns}\frac{1}{\sqrt{u}}e^{-u}\, du \right)^{2} \, ds$$ Now here is the trick: Our aim is to force the inner integral arbitrarly close to $\sqrt{\pi}$ by choosing $n$ sufficiently large, due to the simple fact that $\int_{0}^{\infty}\frac{1}{\sqrt{u}}e^{-u}\, du=\sqrt{\pi}$.

This can of course be done for each fixed $s\in [0,\infty)$, however the issue is that the boundaries depend heavily on $s$, which varies over the positive Real-line as we integrate. The idea is to restrict the domain of $s$ in a sophisticated way so that the desired estimate can be done over the restricted domain of $s$, while the outer integral does not lose to much mass.

Now let $\delta\in(0,1)$ arbitrary but fixed. Clearly for each $n\geq 1$ $$\int_{0}^{\infty}\frac{1}{s}\left(\int_{\frac{s}{n}}^{ns}\frac{1}{\sqrt{u}}e^{-u}\, du \right)^{2} \, ds\geq \int_{(\frac{1}{n})^{1-\delta}}^{n^{1-\delta}}\frac{1}{s}\left(\int_{\frac{s}{n}}^{ns}\frac{1}{\sqrt{u}}e^{-u}\, du \right)^{2} \, ds\geq \int_{(\frac{1}{n})^{1-\delta}}^{n^{1-\delta}}\frac{1}{s}\left(\int_{(\frac{1}{n})^{\delta}}^{n^{\delta}}\frac{1}{\sqrt{u}}e^{-u}\, du \right)^{2} \, ds$$ since $(\frac{1}{n})^{1-\delta}\leq s \leq n^{1-\delta}$.

Since the boundaries of the inner integral does no longer depend on $s$, we can for any given $0<\varepsilon<\sqrt{\pi}$ , choose $n$ sufficiently large such that $$\int_{\frac{1}{n^{\delta}}}^{n^{\delta}}\frac{1}{\sqrt{u}}e^{-u}\, du \geq \sqrt{\pi} -\varepsilon$$

Thus we obtain $$||\mathcal{L}(f_{n})||_{2}^{2} \geq \int_{(\frac{1}{n})^{1-\delta}}^{n^{1-\delta}}\frac{(\sqrt{\pi}-\varepsilon)^{2}}{s} \, ds =2\log(n)(1-\delta)(\sqrt{\pi}-\varepsilon)^{2}=||f_{n}||_{2}^{2}(1-\delta)(\sqrt{\pi}-\varepsilon)^{2}$$ Now since $\delta\in(0,1)$ and $\varepsilon\in (0 ,\sqrt{\pi})$ were arbitrary, we finally conclude that $$||\mathcal{L}|| \geq \sqrt{\pi}$$ and the proof is done.


The norm is exactly $\sqrt\pi$. You can find a prof in Setterqvist E., Unitary Equivalence: A New Approach to the Laplace transform and the Hardy operator. Master’s Thesis, Department of Mathematics Lule ̊a University of Technology, 2005:329 CIV. here


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