# Linear Operators

• Jun 6th 2011, 08:34 AM
liquidpaper
Linear Operators
I need help with this:

Let E a Banach Space and B(E) the Banach space of linear and bounded operators from E to E with the usual norm. Let S $\displaystyle$\in$$B(E). If the sum \displaystyle \sum\limits_{n = 1}^\infty {\frac{{{S^k}}}{{k!}}}$$ convergs to the operator $\displaystyle$e^S$$, show that \displaystyle Se^S$$ = $\displaystyle$e^S S$$. And if T,S \displaystyle \in  B(E) conmute then \displaystyle e^{S+T}$$= $\displaystyle$e^S$$\displaystyle e^T$$. In particular prove that $\displaystyle$e^S$$is invertible. Please somebody give me a hand. • Jun 6th 2011, 08:47 AM girdav For the first question, put \displaystyle S_n :=\sum_{k=0}^n\frac{S^k}{k!}. Show that for all \displaystyle n\in\mathbb N we have \displaystyle SS_n=S_nS. Then use the fact that the maps \displaystyle T\mapsto ST and \displaystyle T\mapsto TS are continuous for the usual norm on \displaystyle \mathcal B(E). • Jun 6th 2011, 11:30 AM orbit Quote: Originally Posted by girdav For the first question, put \displaystyle S_n :=\sum_{k=0}^n\frac{S^k}{k!}. Show that for all \displaystyle n\in\mathbb N we have \displaystyle SS_n=S_nS. Then use the fact that the maps \displaystyle T\mapsto ST and \displaystyle T\mapsto TS are continuous for the usual norm on \displaystyle \mathcal B(E). Sorry to bother again, but I canīt show \displaystyle S{S_n} = {S_n}S$$...
• Jun 6th 2011, 12:02 PM
girdav
Prove it by induction.
• Jun 6th 2011, 01:11 PM
orbit
Quote:

Originally Posted by girdav
Prove it by induction.

Done!

Could you explain to me how to use the continuity of T and S?

Thanks"!
• Jun 6th 2011, 01:17 PM
girdav
Show that $\displaystyle SS_n$ converges to $\displaystyle Se^S$ in $\displaystyle \mathbb{B}(E)$ and similarly that $\displaystyle S_nS$ converges to $\displaystyle e^SS$ in $\displaystyle \mathbb{B}(E)$
• Jun 6th 2011, 01:46 PM
orbit
Quote:

Originally Posted by girdav
Show that $\displaystyle SS_n$ converges to $\displaystyle Se^S$ in $\displaystyle \mathbb{B}(E)$ and similarly that $\displaystyle S_nS$ converges to $\displaystyle e^SS$ in $\displaystyle \mathbb{B}(E)$

Done! Thank you again.

And, could you explain a bit how to do the other part?

Thanks!
• Jun 6th 2011, 01:47 PM
orbit
Thank you..
Sorry double post.
• Jun 7th 2011, 01:05 AM
girdav
Since $\displaystyle S$ and $\displaystyle T$ commute, you can try a proof which is similar to the case $\displaystyle S,T\in\mathbb{C}$. The main step consists of showing that
$\displaystyle e^Se^T = \sum_{n=0}^{+\infty}\sum_{k=0}^n\dfrac{S^k}{k!} \cdot\dfrac{T^{n-k}}{(n-k)!}$.