# Strongly continuous semigroups on weighted $\ell^1$ space

Let $$x=(x_i)$$ be a sequence in $$\ell^1$$ such that all $$x_i>0.$$

Let $$T(t):\ell^1 \rightarrow \ell^1$$ be a strongly continuous semigroup of, i.e. $$t \mapsto T(t)y$$ is continuous for every $$y \in \ell^1$$.

We can now study

$$F_x(t):=\left\Vert \frac{1}{\sqrt{x}} (T(t)x-x) \frac{1}{\sqrt{x}} \right\Vert_{\ell^{\infty}} = \left\Vert \frac{T(t)x}{x}-1 \right\Vert_{\ell^{\infty}},$$

here $$\frac{1}{\sqrt{x}}:=\Big(\frac{1}{\sqrt{x_i}} \Big)_i,$$ $$1/x$$ is defined accordingly and $$1$$ is the sequence that has all entries equal to one.

In general, this will be infinite for $$t \neq 0.$$

I ask: Under what natural conditions on $$x$$ and $$(T(t))$$, or rather the generator of the group, is this finite in a neighbourhood $$[0,\varepsilon)$$ of $$t=0$$, for fixed $$x$$, and we will have $$\lim_{t \downarrow 0} F_x(t)=0.$$

Observations:

The only danger that can happen is that an entry of $$\frac{1}{\sqrt{x_i}}$$ will be very large and $$T(t)x$$ will have a lot of mass in this entry, i.e. $$T(t)x_i$$ is then large and $$\frac{1}{\sqrt{x_i}}$$ is large and if this happens infinitely often, with increasing size, we are doomed.

So is there a way to ensure by looking at the generator that arbitrarily small entries do not get filled too quickly? Of course I still want my dynamics to mix the entries, just not in the way I described before.

• The premise of your question is rather restrictive: the only strongly continuous group of isometries on $\ell^1$ is the identity group. ;-) – Jochen Glueck Mar 8 at 0:12
• Invertible isometries on $\ell^1$ are very restricted: they must essentially all be given by permutation "matrices" composed with a diagonal operator where all entries have modulus 1. I suspect that this will force your SOT-continuous group to just consist of diagonal matrices – Yemon Choi Mar 8 at 0:12
• @JochenGlueck thanks for pointing that out, then I shall change my question quickly to semigroups :). Though in principle, I should also be able to multiply every entry by a phase $e^{it}$ no? – Sascha Mar 8 at 0:14
• @YemonChoi thanks to you too, $\ell^1$ is really a strange space. – Sascha Mar 8 at 0:14
• By the way, the isometry thing doesn't have too much to do with $\ell^1$ - the same argument also works on $\ell^p$ for $p \in [1, \infty) \setminus \{2\}$; see my answer here. But since we're back to semigroups now, I'll think about the question. – Jochen Glueck Mar 8 at 0:23