Output of a stable LTI system

Question

Let $\mathcal{L}$ be a stable LTI system. Is it true that if input has finite energy then output also has finite energy? I'm not sure about that. We know that $$\int_{-\infty}^{+\infty}|h(t)|dt\lt\infty \tag{1}$$Where $h(t)$ is the impulse response. Also we have $$\int_{-\infty}^{+\infty}|y(t)|dt = \int_{-\infty}^{+\infty}|Y(s)|ds = \int_{-\infty}^{+\infty}|X(s)||H(s)|ds \tag{2}$$Since $y(t) = \mathcal{L}(x(t)) = x(t)\star h(t)$ which implies $Y(s) = X(s)H(s)$. Applying Cauchy–Schwarz inequality to $(2)$, $$\left(\int_{-\infty}^{+\infty}|X(s)||H(s)|ds\right)^2 \le \left(\int_{-\infty}^{+\infty}|X(s)|^2ds\right)\left(\int_{-\infty}^{+\infty}|H(s)|^2ds\right) \tag{3}$$We know that $$\int_{-\infty}^{+\infty}|X(s)|^2ds = \int_{-\infty}^{+\infty}|x(t)|^2dt<\infty$$Since input is an energy signal but $$\int_{-\infty}^{+\infty}|H(s)|^2ds$$doesn't necessarily exists. So is this indicates that we can find a counterexample to the statement or we can prove that by other methods?

Edit: Here is a counterexample which shows $$\int_{-\infty}^{\infty}|h(t)|dt\lt \infty \nRightarrow \int_{-\infty}^{\infty}|h(t)|^2dt\lt \infty$$

Answer 1

I think I've found the answer. Please correct me if I'm wrong. First of all, I've made a silly mistake $$\int_{-\infty}^{+\infty}|y(t)|dt = \int_{-\infty}^{+\infty}|Y(s)|ds$$which is clearly false. Let $y(t) = x(t)\star h(t)$. We have $$E_y = \int_{-\infty}^{+\infty}|y(t)|^2dt = \int_{-\infty}^{+\infty}|Y(s)|^2ds = \int_{-\infty}^{+\infty}|H(s)X(s)|^2ds = \int_{-\infty}^{+\infty}|H(s)|^2|X(s)|^2ds$$Also we have $$|H(s)| = \left|\int_{-\infty}^{+\infty}e^{-2\pi ist}h(t)dt \right | \le \int_{-\infty}^{+\infty}|e^{-2\pi ist}h(t)|dt = \int_{-\infty}^{+\infty}|h(t)|dt \lt\infty$$So $\exists M\in\mathbb{R}:\ \ |H(s)|\le M$ for all $s$. This means that $|H(s)|^2\le M^2$ and then $$\int_{-\infty}^{+\infty}|H(s)|^2|X(s)|^2ds\le M^2\int_{-\infty}^{+\infty}|X(s)|^2ds$$By assumption $$E_x = \int_{-\infty}^{+\infty}|x(t)|^2dt=\int_{-\infty}^{+\infty}|X(s)|^2ds$$The result is $$E_y \le M^2 E_g$$

Answer 2

Rough outline of proof:

1. A (real-valued) system that's IIR has an impulse response that's $h(t) \ne 0$ for infinitely long
2. Since $\lvert h(t)\rvert \ge h(t) > 0$ for infinity, it follows that this impulse response doesn't have finite energy
3. there are stable LTI IIR systems