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Let $f \in L^1(\mathbb{R}_+, \mathbb{R})$. It's Laplace transform is well defined for all $\Re(z) \geq 0$. Assume that: $$ \forall \Re(z) \geq 0,~ \widehat{f}(z) := \int_0^\infty{e^{-zt}f(t) dt} = G(z),$$ where $G$ is an Holomorphic function on $\Re(z) > -1$.

May I deduce that for all real number $\lambda < 1$, $t \mapsto f(t)e^{\lambda t} \in L^1(\mathbb{R}_+, \mathbb{R})$?

If not, do you have a counterexample? Thanks!

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Not in general. For $\Re(s) > 0$ let $$f(t) = \cases{(-1)^{\textstyle\lfloor e^t \rfloor} \text{ if } t > 0\\ 0 \text{ otherwise}}$$ $$ F(s) = \int_0^\infty f(t)e^{-st}dt= \sum_{n=1}^\infty (-1)^n \int_n^{n+1} x^{-s-1}dx=\sum_{n=1}^\infty (-1)^n \frac{n^{-s}-(n+1)^{-s}}{s}$$ As $n^{-s}-(n+1)^{-s} = s n^{-s-1}+O(s(s+1)n^{-s-2})$ the latter series for $F(s)$ converges and is analytic for $\Re(s) > -1$.

But $f(t)e^{-\sigma t}$ is $L^1(\Bbb{R})$ only for $\sigma > 0$.

For $\Re(s) \in (-1,0)$ $$ F(s) = \int_{-\infty}^\infty (f(t)-1/2)e^{-st}dt$$ and the inverse Fourier transform of $F(\sigma+i\omega)$ in the sense of distributions gives $(f(t)-1/2)e^{-\sigma t}$

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  • $\begingroup$ That is a very neat example, thanks! Do you know what kind of assumptions should we add on $f$ such that the result holds? $\endgroup$
    – Quentin
    Commented Mar 19, 2019 at 20:52
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    $\begingroup$ If $f \ge 0$ then there is a singularity at the abscissa of convergence of its Laplace transform. Otherwise if $\int_{\sigma-i\infty}^{\sigma+i\infty} F(s)e^{st}ds$ converges say absolutely and $F(s) \to 0$ as $\Im(s) \to \pm \infty$ on a strip then you can move the contour to the right obtaining independence in $\sigma$ $\endgroup$
    – reuns
    Commented Mar 19, 2019 at 20:57
  • $\begingroup$ Makes sense for both points! Thanks! Also, how to prove that $\sum_{n \geq 1}{(-1)^n n^{-z}}$ is analytic for $\Re(z)> 0$? $\endgroup$
    – Quentin
    Commented Mar 19, 2019 at 21:06
  • $\begingroup$ @Adri I'm confused, I thought $f$ was required to be absolutely integrable? $\endgroup$
    – Maxim
    Commented Mar 20, 2019 at 5:38
  • $\begingroup$ @Maxim if you want $f$ to be $L^1$, it suffices to multiply this example by $e^{- \epsilon t }$, with $\epsilon > 0$ small. $\endgroup$
    – Quentin
    Commented Mar 20, 2019 at 7:10

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