## 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 $t→∞$

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?