Perturbation of uniformly continuous semigroup

Question

I think I can prove the following claim about perturbing semigroups, but this must be known somehow, and I am just reinventing the wheel. Do you have any suggestions for references that treat this, before I include the proof in a publication? Note that the operators $A, B$ in the claim below are bounded linear operators.

Claim: Let $X$ be a Banach space, and let $A, B$ be bounded linear operators on $X$ to itself. The uniformly continuous semigroups generated by $A$ and $B$ satisfy

$$ \Vert e^{tA}- e^{tB} \Vert \leq t \Vert A - B \Vert e^{t \Vert A-B \Vert} e^{t \Vert A \Vert}, \qquad t >0 . $$.

Answer 1

This seems to be a straight forward consequence of the solution formula for linear ODEs (see below). Elementary estimates of this type are often used but hard to find in exactly the form you need. I would just call it a lemma.

Set $h(t):=e^{tB}-e^{tA}$. Then $$ h'(t)=Be^{tB}-Ae^{tA}=(B-A)e^{tB} + A(e^{tB}-e^{tA})=(B-A)e^{tB} + Ah(t). $$ Since $h(0)=0$ we have $$ h(t)=\int_0^t e^{(t-s)A} (B-A)e^{sB} ds. $$ Thus $$ \|h(t)\| \le \int_0^t e^{(t-s)\|A\|} \|B-A\|e^{s\|B\|} ds = \int_0^t e^{t\|A\|} \|B-A\|e^{s(\|B\|-\|A\|)} ds $$ $$ \le \int_0^t e^{t\|A\|} \|B-A\|e^{s\|A-B\|} ds \le \int_0^t e^{t\|A\|} \|B-A\|e^{t\|A-B\|} ds = t \|A-B\|e^{t\|A-B\|}e^{t\|A\|} $$

Answer 2

A sligthly sharper estimate can be obtained by entirely elementary methods, namely $$\|e^A-e^B\|\le \|A-B\|e^{\max\{\|A\|,\|B\|\}}$$ WLOG we may assume that $\|A\|\ge \|B\|.$ First we will consider the case $\|A\|>\|B\|.$ We have $$e^A-e^B=\sum_{n=1}^\infty {1\over n!} (A^n-B^n)\quad (*)$$ Next $$A^n-B^n=\sum_{k=0}^{n-1} [A^{n-k}B^k-A^{n-k-1}B^{k+1}]=\sum_{k=0}^{n-1}A^{n-k-1}(A-B)B^k$$ Thus $$\|A^n-B^n\|\le \|A-B\|\,\sum_{k=0}^{n-1}\|A\|^{n-k-1}\|B\|^k\quad (**)\\ ={\|A-B\|\over \|A\|-\|B\|}(\|A\|^n-\|B\|^n)$$ Applying $(*)$ we get $$\|e^A-e^B\|\le {\|A-B\|\over \|A\|-\|B\|}\sum_{n=0}^\infty{1\over n!}[\|A\|^n-\|B\|^n]\\ =\|A-B\|\,{e^{\|A\|}-e^{\|B\|}\over \|A\|-\|B\|}$$ By MVT the ratio is equal $e^x$ for some $x,$ such that $\|B\|<x<\|A\|.$ Therefore $$\|e^A-e^B\|\le \|A-B\|\,e^{\|A\|}$$ If $\|A\|=\|B\|$ the estimate $(**)$ takes the form $$\|A^n-B^n\|\le \|A-B\|\,n\|A\|^{n-1}$$ Thus $$\|e^A-e^B\|\le \|A-B\|\sum_{n=1}^\infty {\|A\|^{n-1}\over (n-1)!}=\|A-B\|\,e^{\|A\|}$$ By substituting $A:=tA$ and $B:=tB$, $t>0,$ we get $$\|e^{tA}-e^{tB}\|\le t\|A-B\|e^{t\max\{\|A\|,\|B\|\}} \quad (***)$$ The estimate is stronger than the one provided in OP as $$\|A-B\|+\min\{\|A\|,\|B\|\}\ge \max\{\|A\|,\|B\|\}$$ Indeed if $\|A\|\ge \|B\|,$ then $\|A-B\|+\|B\|\ge \|A\|$ by the triangle inequality.

Remark The approach provided by @Gerd is much more popular and gives the answer faster. It can be easily modified at one place to give $(***).$

Answer 3

I have now found a reference that can help with the integral equality that Gerd uses above, but in the context of Banach spaces, namely Engel and Nagel, A short course on operator semigroups. I post below the current proof I give. Gerd and Ryszard figure both in the acknowledgment section of the manuscript I am preparing.

Below is a screenshot of the current proof