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Let $H$ be a complex Hilbert space, $(f_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ be nondecreasing with $\lambda_n\xrightarrow{n\to\infty}\infty$. It is easy to see that $$T(t)x:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle x,f_n\rangle_Hf_n\;\;\;\text{for }x\in H\text{ and }t\ge0$$ is a well-defined contractive semigroup on $H$.

Are we able to show that

  1. $(T(t))_{t\ge0}$ is uniformly continuous?
  2. If $U$ is a complex Hilbert space compactly and densely embedded into $H$ and $$e_n:=\frac1{\sqrt\lambda_n}f_n\;\;\;\text{for }n\in\mathbb N$$ is an orthonormal basis of $U$, does it follow that $$S(t)u:=T(t)u\;\;\;\text{for }u\in U\text{ and }t\ge0$$ is a strongly/uniformly semigroup on $U$ as well?

Noting that $$\sup_{t\ge0}\left\|\sum_{i=1}^ne^{-\lambda_it}\langle x,f_i\rangle_Hf_i-T(t)x\right\|_H^2\le\sum_{i>n}\left|\langle x,f_i\rangle_H\right|^2\xrightarrow{n\to\infty}0\tag1$$ for all $x\in H$, we should be able to conclude that $(T(t))_{t\ge0}$ is at least strongly continuous.

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It is not uniformly continuous: $$ \frac1t (T(t)f_i-f_i) = \frac1t (e^{-\lambda_i t}-1)f_i \to -\lambda_i f_i $$ for $t\to0$. The operator $f_i \mapsto \lambda_i f_i$ is unbounded, so $T$ cannot be uniformly continuous. Or put differently, the convergence $\frac1t (T(t)f_i-f_i) \to -\lambda_i f_i$ is not uniform with respect to $i$, and $\frac1t\|T(t)-I\|$ does not exist.

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  • $\begingroup$ Thank you for your answer. (a) Why can we conclude that $\lim_{t\to0}\left\|T(t)-I\right\|$ does not exist from knowing that $\lim_{t\to0}\frac1t\left\|T(t)-I\right\|$ does not exist? (b) But you agree that it is at least strongly continuous, right? What can you tell me about $(U(t))_{t\ge0}$? Is it strongly continuous as well? $\endgroup$
    – 0xbadf00d
    Feb 15 '21 at 13:41
  • $\begingroup$ $T(t)$ is strongly but not uniformly continuous $\endgroup$
    – daw
    Feb 15 '21 at 15:06
  • $\begingroup$ It seems that $S(t)$ is just the restriction of $T(t)$ to the smaller space $U$ with stronger norm, so the contiuity results can only be worse than those for $T$. $\endgroup$
    – daw
    Feb 15 '21 at 15:14
  • $\begingroup$ (b) Sure, after your answer I didn't expect $S(t)$ to be uniformly continuous. The question is whether it is at least strongly continuous, but I think the answer is yes. And this should immediately follow from the same argument as in $(1)$. Using that $U$ is continuously embedded into $H$, we see that the convergence is uniform in $t$ for $x\in U$ when the norm is replaced by $\left\|\;\cdot\;\right\|_U$. And the uniform convergence allows us to pull "$\lim_{t\to0+}$" inside the sum. Should be correct?! (a) Would be great if you could reply to (a) as well. $\endgroup$
    – 0xbadf00d
    Feb 15 '21 at 18:54
  • $\begingroup$ Still be interested in a response. $\endgroup$
    – 0xbadf00d
    Feb 20 '21 at 9:16

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