Let $T_t\alpha$ be the unique solution of the cauchy problem $$\frac{\partial A(x,t)}{\partial t}=-\Delta A(x,t)$$ $$A(x,0)=\alpha$$ We know that $T_t\alpha$ converges in $L^2$ to an $L^2$ form $H\alpha$, that $||T_t||^2$ is (nonstrictly) decreasing and that $T_0\alpha=\alpha$.
I wanna show that $G(\alpha):=\int_0^\infty(T_t\alpha-H\alpha)dt$ is well defined. For that I'm trying to show that I'm trying to show that $||T_t\alpha(x)-H\alpha(x)||$ decays fastly enough. I see that this decays to zero, but how does it guarantee that that integral converges? Using that $H$ is harmonic, I've also shown that this norm is (nonstrictly) decreasing.