# Thread: Linear Operators

1. ## 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$$is invertible. Please somebody give me a hand. 2. For the first question, put \displaystyle S_n :=\sum_{k=0}^n\frac{S^k}{k!}. Show that for all \displaystyle n\in\mathbb N we have \displaystyle SS_n=S_nS. Then use the fact that the maps \displaystyle T\mapsto ST and \displaystyle T\mapsto TS are continuous for the usual norm on \displaystyle \mathcal B(E). 3. Originally Posted by girdav For the first question, put \displaystyle S_n :=\sum_{k=0}^n\frac{S^k}{k!}. Show that for all \displaystyle n\in\mathbb N we have \displaystyle SS_n=S_nS. Then use the fact that the maps \displaystyle T\mapsto ST and \displaystyle T\mapsto TS are continuous for the usual norm on \displaystyle \mathcal B(E). Sorry to bother again, but I can´t show \displaystyle S{S_n} = {S_n}S$$...

4. Prove it by induction.

5. Originally Posted by girdav
Prove it by induction.
Done!

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

Thanks"!

6. 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 $\displaystyle \mathbb{B}(E)$

7. 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!

8. ## Thank you..

Sorry double post.

9. 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
$\displaystyle e^Se^T = \sum_{n=0}^{+\infty}\sum_{k=0}^n\dfrac{S^k}{k!} \cdot\dfrac{T^{n-k}}{(n-k)!}$.