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 $$\lim_{t \downarrow 0} F_x(t)=0.$$


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.

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    $\begingroup$ The premise of your question is rather restrictive: the only strongly continuous group of isometries on $\ell^1$ is the identity group. ;-) $\endgroup$ – Jochen Glueck Mar 8 at 0:12
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    $\begingroup$ 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 $\endgroup$ – Yemon Choi Mar 8 at 0:12
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    $\begingroup$ @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? $\endgroup$ – Sascha Mar 8 at 0:14
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    $\begingroup$ @YemonChoi thanks to you too, $\ell^1$ is really a strange space. $\endgroup$ – Sascha Mar 8 at 0:14
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    $\begingroup$ 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. $\endgroup$ – Jochen Glueck Mar 8 at 0:23

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