Output of a stable LTI system
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 $$E_y \le M^2 E_g$$