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](https://wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Joe_Neeman/main.pdf)) and $E\_\lambda$ denote the orthogonal projection of $H$ onto $H\_\lambda$. Using the spectral theorem, I was able to show that $$\lim\_{t\to\infty}\left\|T(t)x\right\|\_H=\left\|E\_0x\right\|\_H\;\;\;\text{for all }x\in H\tag2.$$



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> How can we conclude that we even got $$\left\|T(t)x-E\_0x\right\|\_H\xrightarrow{t\to\infty}0\tag3$$ for all $x\in H$?
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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?



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 $t→∞$
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 $x\in H$.


