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Let $A,A_n$ be closed operators on a Hilbert space $\mathcal H$ and assume that

  • $-(A-{\rm Id}_{\cal H})$ generates a strongly continuous contraction semigroup
  • $-A_n$ generates a strongly continuous contraction semigroup for every $n$.
  • $\|(z-A_n)^{-1}-(z-A)^{-1}\|\to 0$ as $n\to\infty$ for every $z$ with $\Re(z)<0$.

Can one show that there exists a $\mu\in (0,1)$ such that $-(A_n-\mu)$ generates a bounded semigroup for $n$ large enough?

If not, are there simple counterexamples?

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  • $\begingroup$ Do you want to show that the norms of the semigroups $T_n(t)$, generated by $-A_n + \mu$, are uniformly bounded in $n$ and $t$? Because, for every $n$, $-A_n+\mu$ generates a $C_0$-semigroup anyway, for every $\mu\in\mathbb C$. $\endgroup$
    – DominikS
    Aug 31 '17 at 13:06
  • $\begingroup$ I want to show that there exist a $n\in\mathbb N$ and $\mu\in(0,1)$ such that $\|e^{-t(A_n-\mu)}\|$ is uniformly bounded in $t$. $\endgroup$
    – Frank
    Aug 31 '17 at 14:08
  • $\begingroup$ I am probably going to say something totally known to you, but just in case I'll try anyhow. If $A, A_n$ are matrices then the claim is true; indeed, decay of $e^{-At}$ is equivalent to $\Re \sigma(A)>0$, and matrices with this property form an open set. This strongly suggests that the claim should be true in the general case as well, but I am not sufficiently familiar with Hille-Yosida's stuff to prove it. $\endgroup$ Sep 10 '17 at 12:55
  • $\begingroup$ Yes, I'm aware that the statement is true for matrices. This follows also from the Lumer-Phillips theorem, since norm-resolvent convergence implies convergence of matrix elements which in turn implies that $-A_n+\mu$ is dissipative for large $n$ and $0<\mu<1$. But I'm not sure whether this suggests anything for the general case, since for unbounded operators there are infinitely many matrix elements which don't necessarily converge uniformly. $\endgroup$
    – Frank
    Sep 10 '17 at 15:40
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I think, I've found a counterexample. Let $\mathcal H:=L^2(\mathbb R)$ and consider the multiplication operators $$(Au)(t) = \left(1+it\right)u(t)$$ $$(A_nu)(t) = \left(1+it-\tanh(\tfrac{t}{n})^2\right)u(t)$$ on the common domain $\mathcal D=\{u\in L^2(\mathbb R) : tu\in L^2(\mathbb R)\}$.

Then $-A+1$ generates a contraction semigroup by the Lumer-Phillips theorem and its spectrum is the essential range of $(1+it)$ which is a vertical line through the point 1.

Similarly, $A_n$ generates a contraction semgroup by Lumer-Phillips, however, here the tanh-term prevents us from subtracting an identity operator. For every $n$ the spectrum of $A_n$ is a curve passing through 1 and approaching the imaginary axis for $\text{Im}(z)\to\pm\infty$.

The resolvents of $A$, $A_n$ are simply given by the multiplication operators $$(z-A)^{-1}f = \frac{f}{z-1-it}$$ $$(z-A_n)^{-1}f = \frac{f}{z-1-it+\tanh(\frac{t}{n})^2}$$ for $\text{Re}(z)<0.$

It is easy to check that $\|(z-A)^{-1}-(z-A_n)^{-1}\|_{\mathcal L(L^2(\mathbb R))}\to 0$ as $n\to\infty$.

However, since $\tanh(t/n)^2\to 1$ as $t\to\pm\infty$ for every $n$, one has $\sigma(A_n-\mu)\cap\{\text{Re}(z)<0\}\neq\emptyset$ for every $\mu>0$.

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  • $\begingroup$ I liked this answer. I would like to make a connection with the bounded operator case mentioned in the comments to the main question. One could say that the key feature of the present example is that the distance between the spectrum of $A_n$ and $\{ \Re(z)<0\}$ is zero for all $n$. This cannot occur for bounded operators, because their spectrum is compact. Such a distance is positive if and only if one can subtract a small multiple of the identity and still generate a contraction semigroup, thus obtaining uniform exponential decay. $\endgroup$ Oct 3 '17 at 7:02

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