Product of Laplace transforms with disjoint Region Of Convergence (ROC)

Question

Let the Laplace transform of $x(t)$ be $$X(s) = \int_{-\infty}^{+\infty}x(t)\exp(-st)dt, \ \ \ \ \ s\in\mathbb{C}$$This integral converges when $s \in \text{ROC}_1$. Also suppose that the Laplace transform of $h(t)$ be $H(s)$ which converges when $s\in\text{ROC}_2$. I'm interested in the case $\text{ROC}_1 \cap \text{ROC}_2 = \emptyset$. I think in this case the convolution of $x(t)$ and $h(t)$ is divergent, i.e. $$\forall t\in \mathbb{R}: \ \ y(t)=\int_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau \ \ \text{diverges}.$$ Or at least it diverges almost surely. I want to find a rigorous proof for this hypothesis. So maybe we should consider a special class of functions such as $L^1(\mathbb{R})$ or $L^2(\mathbb{R})$?

Example: Take $x(t) = \exp(-\alpha t)u(t)$ and $h(t) = \exp(-\alpha t)u(-t)$ where $\alpha \gt 0$ and $u(t)$ denotes the step function. Then we have $$X(s) = \frac{1}{s+\alpha} \ \ \ \text{ROC}_1: \ \Re\{s\}\gt -\alpha, \\ H(s) = \frac{1}{s+\alpha} \ \ \ \text{ROC}_2: \ \Re\{s\}\lt -\alpha.$$ In this case the convolution is divergent: $$y(t)=\int_{-\infty}^{+\infty}\exp(-\alpha \tau)u(\tau)\exp(-\alpha (t-\tau))u(-(t-\tau))d\tau = \int_{0}^{+\infty} \exp(-\alpha t)u(\tau - t)d\tau \\ = \exp(-\alpha t)\int_{t}^{+\infty}d\tau = \infty.$$

Also if we consider $h(t) = \exp(-\alpha't)u(-t)$ where $\alpha' \not = \alpha$ the same steps show that $y(t)$ converges when $\alpha' \lt \alpha$, so $\text{ROC}_1 \cap \text{ROC}_2: \ \Re\{s\} \in (-\alpha , -\alpha')$.

Related question: This question has been asked previously (for discrete-time signals and $\mathcal{Z}$-transform) but Matt L.'s answer doesn't contain a rigorous proof.

Answer 1

Willie Wong's great answer shows that it's possible to construct uncountably many counterexamples. Let $S = \{ (-2)^k : k \in \mathbb{N} \}, 0\lt \epsilon \lt \frac14$ and define the function$$\chi(t) = \begin{cases} 1 & \exists s\in S, \quad |t-s| < \epsilon \\ 0 & \text{otherwise} \end{cases}$$Also let $x(t) = h(t) = \chi(t)$. It can be shown that in this case $\text{ROC}_1 = \text{ROC}_2 = \text{ROC}_1 \cap \text{ROC}_2 = \emptyset$ and the function $\tau \mapsto g(\tau, t-\tau)$ defined by $g(\tau , t-\tau) = \chi(\tau)\chi(t-\tau)$ has compact support for each $t\in \mathbb{R}$. This shows that for each $t\in \mathbb{R}$ the convolution converges. If we want to have examples for the case $\text{ROC}_1 \not = \emptyset$ and $\text{ROC}_2 \not = \emptyset$, we can consider $x(t) = \exp(-\eta|t|)\chi(t)$ and $h(t) = \exp(-\eta|t| + 5\eta t)\chi(t)$. In this case we have $\text{ROC}_1: \Re(s') \in (-\eta , \eta)$ and $\text{ROC}_2: \Re(s') \in (4\eta , 6\eta)$, so still $\text{ROC}_1 \cap \text{ROC}_2 = \emptyset$ holds. Since the factors $\exp(-\eta|t|)$ and $\exp(-\eta|t| + 5\eta t)$ are always positive, they can't change the support of $g(\tau , t - \tau)$ and so the convolution is well-defined for every $t\in \mathbb{R}$.

Answer 2

I’m not sure if this is as thorough of an answer as you are looking for, but my thoughts are something like this.

Convolution needs BIBO stability for the integral to converge, ie

\begin{equation}\int_{-\infty}^{\infty}|h(t)|dt<\alpha\end{equation}

where $\alpha$ is some finite constant.

In the Laplace domain, for a function to be BIBO stable, the ROC must contain the imaginary axis. The pole location requirements differ for causal, anticausal, and non-causal, but all must contain the imaginary axis in the ROC to converge. If the ROCs are disjoint, then at most one of the functions’ ROCs contains the imaginary axis. This implies one of the functions violates the absolute integrability criterion, which would lead to a divergent convolution.

Additionally, in general the ROC of two Laplace domain functions multiplied together is the intersection of the individual functions’ ROCs. A disjoint intersection between the ROCs implies there is no value where both functions converge.