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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?

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    $\begingroup$ I don't think it's research level ... in the multiplication operator picture $e^{tA}$ is multiplication by $e^{-tx}$ on $[0, \infty)$ and $E_0$ is multiplication by $1_{\{0\}}$. Yes, the former converges strongly to the latter. $\endgroup$
    – Nik Weaver
    Commented Apr 1, 2019 at 11:31
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    $\begingroup$ It might be worthwhile to add that the following much more general result is true: If $(T(t))_{t\ge 0}$ is a bounded $C_0$-semigroup with generator $A$ on a reflexive Banach space (say, over $\mathbb{C}$) and if $\sigma(A) \cap i\mathbb{R} \subseteq \{0\}$, then $T(t)$ converges strongly as $t \to \infty$. This is (one version of) the so-called ABLV theorem. $\endgroup$
    – Jochen Glueck
    Commented Apr 1, 2019 at 17:25
  • $\begingroup$ To elaborate on Nik's comment, your claim follows from functional calculus (+ dominated convergence), since $\|e^{tA}x-E_0x\|^2=\int_{(-\infty, 0]} |e^{ts}-\chi_{\{ 0\} }(s)|^2\, d\rho(s)$ for some finite measure $\rho$. (And, indeed, this is certainly not research level.) $\endgroup$
    – Christian Remling
    Commented Apr 1, 2019 at 19:46

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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 $x\in H$.

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    $\begingroup$ This is not correct. $\endgroup$
    – Nik Weaver
    Commented Apr 1, 2019 at 14:44
  • $\begingroup$ @NikWeaver Maybe you can be a little bit more concrete. $\endgroup$
    – 0xbadf00d
    Commented Apr 1, 2019 at 18:37
  • $\begingroup$ For instance, $E_0$ is the orthogonal projection onto ${\rm ker}(A)$. So $\|E_0x\|^2 = \|x\|^2 \not\to 0$ for any nonzero $x \in {\rm ker}(A)$. $\endgroup$
    – Nik Weaver
    Commented Apr 1, 2019 at 21:36
  • $\begingroup$ @NikWeaver Sorry, that was just a typo. The "$+$" was supposed to be "$-$" on the right-hand side of $(6)$. $\endgroup$
    – 0xbadf00d
    Commented Apr 2, 2019 at 5:40
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    $\begingroup$ Okay, makes sense. It seems like the easiest way to infer convergence to $0$ in (6) is via spectral theory though. $\endgroup$
    – Nik Weaver
    Commented Apr 2, 2019 at 13:43

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