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 $$\int_{-\infty}^{\infty}|h(t)|dt\lt \infty \nRightarrow \int_{-\infty}^{\infty}|h(t)|^2dt\lt \infty$$