# 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 $\in$ B(E).

If the sum $\sum\limits_{n = 1}^\infty {\frac{{{S^k}}}{{k!}}}$ convergs to the operator $e^S$, show that $Se^S$ = $e^S S$. And if T,S $\in$ B(E) conmute then $e^{S+T}$= $e^S$ $e^T$. In particular prove that $e^S$ is invertible.

Please somebody give me a hand.
• Jun 6th 2011, 08:47 AM
girdav
For the first question, put $S_n :=\sum_{k=0}^n\frac{S^k}{k!}$. Show that for all $n\in\mathbb N$ we have $SS_n=S_nS$. Then use the fact that the maps $T\mapsto ST$ and $T\mapsto TS$ are continuous for the usual norm on $\mathcal B(E)$.
• Jun 6th 2011, 11:30 AM
orbit
Quote:

Originally Posted by girdav
For the first question, put $S_n :=\sum_{k=0}^n\frac{S^k}{k!}$. Show that for all $n\in\mathbb N$ we have $SS_n=S_nS$. Then use the fact that the maps $T\mapsto ST$ and $T\mapsto TS$ are continuous for the usual norm on $\mathcal B(E)$.

Sorry to bother again, but I canīt show $S{S_n} = {S_n}S$...
• 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 $SS_n$ converges to $Se^S$ in $\mathbb{B}(E)$ and similarly that $S_nS$ converges to $e^SS$ in $\mathbb{B}(E)$
• Jun 6th 2011, 01:46 PM
orbit
Quote:

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

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 $S$ and $T$ commute, you can try a proof which is similar to the case $S,T\in\mathbb{C}$. The main step consists of showing that
$e^Se^T = \sum_{n=0}^{+\infty}\sum_{k=0}^n\dfrac{S^k}{k!} \cdot\dfrac{T^{n-k}}{(n-k)!}$.