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Let $X$ be a Banach space, and $(P_t)_{t \ge 0}$ a strongly continuous semigroup of bounded operators on $X$. Using the uniform boundedness principle, it's simple to prove that there are constants $M, \omega$ such that $$\|P_t\| \le M e^{\omega t} \quad (*)$$ for all $t$. Moreover, if $M=1$ you can get a contraction semigroup by studying $e^{-\omega t} P_t$, and I'm having trouble thinking of an example where that's not the case.

What is a simple example of a semigroup $P_t$ for which (*) cannot hold with $M=1$?

If possible, I'd like to see an example where $X$ is a separable Hilbert space.

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1 Answer 1

up vote 6 down vote accepted

There are many examples constructed with weighted shifts. The following is a Hilbert space example.

Let us consider the Hilbert space $L^2((0,1),\mu)$, where $\mu$ denotes the measure defined by $$\mu(A):=2\lambda(A\cap(0\tfrac12))+\lambda(A\cap(\tfrac12,1)).$$ for all Lebesgue measurable sets A. Here $\lambda$ is the Lebesgue-measure. Furthermore, let $T(t)$ be the nilpotent left shift semigroup. Obviously, $T$ satises the semigroup property and, since the norm $\|\cdot\|_{\mu}$ is equivalent to the norm $\|\cdot\|_{\lambda}$, $T$ is strongly continuous.

Clearly, $\|T(t)\|\leq 2$.

In addition we see that $T(t) = 0$ for all $t>1$.

Finally, consider the function $$f_t = \frac{1}{\sqrt{t}}\chi_{(\frac12,\frac12+t)}$$ for $t\in(0,\tfrac12)$. Clearly, $\|f_t\|_{\mu} = 1$ and $$\|T(t)f_t\|_{\mu} = 2.$$


Theoretically, you can always introduce an equivalent norm in your space which makes your semigroup a contraction semigroup, see Lemma II.3.10 in K. Engel, R. Nagel One-Parameter Semigroups for Linear Evolution Equations. The construction is, however, in most cases highly non-constructive and can only used for theoretical purposes.

It is more important that there are examples in Hilbert spaces where it is imnpossible to find an equivalent Hilbert space norm (i.e., the semigroup is not similar to a contraction semigroup), see

Packel, E.W. A semigroup analogue of Foguel's counterexample. (English) Proc. Am. Math. Soc. 21, 240-244 (1969).

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Thanks very much! – Nate Eldredge Jun 28 '13 at 18:50

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