If $A$ is a dissipative self-adjoint operator with spectral decomposition $(H_λ)$, then $e^{tA}x$ tends to the projection of $x$ onto $H_0$ as $t→∞$

Question

Let $(T(t))_{t\ge0}$ be a strongly continuous contraction semigroup on a $\mathbb R$-Hilbert space $H$ with dissipative self-adjoint generator $(\mathcal D(A),A)$. In particular, $T(t)$ is self-adjoint for all $t>0$. By the spectral theorem, $$T(t)=e^{tA}\;\;\;\text{for all }t\ge0.\tag1$$ Let $(H_\lambda)_{\lambda\ge0}$ be the spectral decomposition related to $(\mathcal D(A),-A)$ (see, for example, Definition 1.8.1 and Theorem 1.8.2 on page 23 here) and $E_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$. Using the spectral theorem, I was able to show that $$\lim_{t\to\infty}\left\|T(t)x\right\|_H=\left\|E_0x\right\|_H\;\;\;\text{for all }x\in H\tag2.$$

How can we conclude that we even got $$\left\|T(t)x-E_0x\right\|_H\xrightarrow{t\to\infty}0\tag3$$ for all $x\in H$?

I know that in a Hilbert space, convergence is equivalent to weak convergence together with convergence of the norms. So, we would be done if we could show that $$\langle T(t)x,y\rangle_H\xrightarrow{t\to\infty}\langle E_0x,y\rangle\tag4\;\;\;\text{for all }x,y\in H.$$ If this is the correct approach, how can we show that?

Answer 1

By definition, $$\langle T(t)x,E_0x\rangle_H=\left\|E_0x\right\|_H^2+\underbrace{\int_0^\infty e^{-t\lambda}\:{\rm d}\underbrace{\langle E_\lambda x,E_0x\rangle_H}_{=\:\left\|E_0x\right\|_H^2}}_{=\:0}\tag5$$ and hence $$\left\|T(t)x-E_0x\right\|_H^2=\left\|T(t)x\right\|_H^2-\left\|E_0x\right\|_H^2\xrightarrow{t\to\infty}0\tag6$$ for all $x\in H$.