# Cross correlation theorem for single-sided Laplace transform

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!

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

• Of course! Thanks. So obvious once you see it! Commented Feb 1, 2023 at 9:49

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}

• Thanks very much. Commented May 29, 2023 at 11:13