9
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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