Lerch's theorem

Question

In the context of a stats exercice, I need to show that, if for some $h$ measurable and for every real $\mu$, $\displaystyle {\int_{-\infty}^{+\infty} h\left(x \right)\cdot \exp\left(-x^2\right)\cdot \exp\left(-\mu x\right)\,\mathrm{d}x = 0}$, then $h=0\:a.e.$
The correction simply mentions that the Laplace transform is a bijection (between what spaces?), so h is $0\:a.e.$
But the only injectivity statement for the LT I could find was for well-behaved functions. Is it even true in my case ? Can anyone point me to such a proof ? Thanks
EDIT: I changed the title since the result holds for continuous, of exponential order functions (Lerch).

Answer 1

The proposition is true for any function $h$ the Fourier transform of which exists, such as a $L^2$ or $L^1$ function.

$\forall \mu\in\mathbf R$, $$0=\int_{-\infty}^{+\infty} h\left(x \right)\exp\left(-x^2\right) \exp\left(-\mu x\right)\,\mathrm{d}x = e^{\frac{\mu^2}4}\int_{-\infty}^{+\infty}h(x)e^{-\left(\frac\mu2-x\right)^2}\,\mathrm dx,$$ or $$\int_{-\infty}^{+\infty}h(x)e^{-\left(\frac\mu2-x\right)^2}\,\mathrm dx = 0.$$ Take Fourier transform, we have $$\hat h(\omega) \hat g(\omega)=0,$$ where $\hat f$ stands for the Fourier transform of function $f$, and $g(x)=e^{-x^2}$. $\hat g(k)\propto e^{-\frac{k^2}4}\ne 0,\,\forall k$. So $\hat h\equiv0$. Therefore $h\equiv0,\ a.e.$, by the uniqueness of the preimage of the Fourier transform.