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?