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?

ABLV theorem. $\endgroup$ – Jochen Glueck Apr 1 '19 at 17:25