Cross correlation theorem for single-sided Laplace transform

Question

I'm trying to prove the cross correlation theorem for the single-sided Laplace transform. Define the cross correlation for two functions $f, g: [0, \infty) \to \mathbb{C}^n$ to be $$ (f\star g)(t) = \int_0^\infty f(\tau)^* g(t + \tau)d\tau. $$ The cross correlation theorem (from Wikipedia) then states that the single-sided Laplace transform of $f\star g$ is given by $$ F^*(-s^*)\cdot G(s). $$

Proof attempt

$$ \mathcal{L}(f\star g)(s) = \int_0^\infty e^{-st} \int_0^\infty f(\tau)^*g(t + \tau)d\tau dt\\ = \int_0^\infty f(\tau)^* \int_0^\infty e^{-st} g(t + \tau) dt d\tau. $$ Set $u = t + \tau$ and evaluate the inner integral:

$$ \int_0^\infty e^{-st} g(t + \tau) dt = \int_0^\infty e^{-s(u - \tau)} g(u)du\\ = e^{s\tau} \int_0^\infty e^{-su} g(u) du\\ = e^{s\tau} G(s). $$ We are then left with $$ \int_0^\infty f(\tau)^* e^{s\tau} d\tau G(s). $$ If we were using the two-sided Laplace transform, the result would then follow by substituting $\omega = - \tau$. In the single-sided setting, however, this doesn't work, as $f$ only takes positive arguments.

I suppose one way around this issue is to extend $f$ and $g$ to $(-\infty, \infty)$, setting them to $0$ on $(-\infty, 0]$. The result can then be derived using the two-sided Laplace transform, which is equal to the single-sided Laplace transform in this case. It is a bit unsatisfying though, a direct proof would be nice.

Thanks in advance for pointing out where I'm being stupid!

Answer 1

This theorem does not appear to be true in general. A counterexample:

Take $g(t)=\exp(-ta)$ with $a$ such that $F^*(a^*)$ exists. Then \begin{align} (f\star g)(t) &= \exp(-ta) \int_0^\infty f(\tau)^* \exp(-\tau a) d\tau \\&= \exp(-ta) \mathcal{L}\{f^*\}(a) \\&= g(t) F^*(a^*). \end{align} So the Laplace transform is \begin{align} \mathcal{L}\{f\star g\}(s) &= F^*(a^*) \mathcal{L}\{g(t)\}(s) \\&= F^*(a^*) G(s). \end{align} But the theorem states that \begin{align} \mathcal{L}\{f\star g\}(s) = F^*(-s^*) G(s). \end{align} Since $F^*(a^*) = F^*(-s^*)$ is not true in general, the theorem is also not true in general.

In your proof, you forgot to update the limits of the integral after setting $u=t+\tau$: \begin{align} \int_0^\infty \exp(-st) g(t+\tau) dt = \int_\tau^\infty \exp(-s(u-\tau)) g(u) du \neq \int_0^\infty \exp(-s(u-\tau)) g(u) du. \end{align}

Answer 2

You can't make $\tau$ negative but what if we play around with $s$? $$\begin{align}\int_{0}^{\infty}f^*(\tau)e^{s\tau}d\tau&=\int_{0}^{\infty}f^*(\tau)e^{-(-s)\tau}d\tau\\\\&=\mathcal{ULT}\{f^*(t)\}_{s=-s}\\\\&=\mathcal{F}^*(s^*)_{s=-s}\\\\&=\mathcal{F^*}(-s^*)\end{align}$$

(As, $(-s)^*=-s^*$)