Banach Algebra exponentials

Question

Let $\mathcal{A}$ be a unital Banach Algebra. Suppose $a,b \in \mathcal{A}$.

1. show that if $a$ and $b$ commute, i.e. ab=ba, then $\exp(a) \exp(b)= \exp(a+b)$.

2. Give a counterexample to show that $\exp(a) \exp(b)= \exp(a+b)$ is not necessarily true when $ab \ne ba$.

For 1, I know we have that $\exp(a)=\sum_{n=0}^{\infty}\frac{a^n}{n!}.$

So $\exp(a) \exp(b)=\sum_{n=0}^{\infty}\frac{a^n}{n!}\sum_{n=0}^{\infty}\frac{b^n}{n!}$

And $\exp(a+b)=\sum_{n=0}^{\infty}\frac{(a+b)^n}{n!}.$

But this is almost like proving in real numbers using Taylor series. If this is correct, what is the difference between the proof in the real numbers and in the unital Banach Algebra?

For 2, I couldn't come up with anything...

Thanks!

Answer 1

For (1), as Cameron Williams says in the comment, mimic the proof of the real number case. This is a strategy that might appear a few times in the elementary theory, see for example the accepted answer in this post: Why is $f(t) = e^{ta}$ differentiable in a unital Banach algebra?

For (2), whenever you need a counter-example for the non-commuting setting, you always look in $M_2(\mathbb{C})$ first. Keep this in mind, as I suspect you are now starting Murphy's book.

Here is a counter example: try $a=e_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $b=e_{21}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. Note that these do not commute.

since $a+b=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, we have that $(a+b)^2=I_2$, so $$e^{a+b}=\sum_{n=0}^\infty\frac{1}{n!}(a+b)^n=\sum_{n=0}^\infty\frac{1}{(2n)!}\cdot I_2+\sum_{n=0}^\infty\frac{1}{(2n+1)!}\cdot(a+b)=\begin{pmatrix}s_0&s_1\\s_1&s_0\end{pmatrix}$$ where $s_0=\sum_{n=0}^\infty\frac{1}{(2n)!}$ and $s_1=\sum_{n=0}^\infty\frac{1}{(2n+1)!}$.

On the other hand, since $a^2=0$ and $b^2=0$ we have that $$e^a=I_2+a=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ and $$e^b=I_2+b=\begin{pmatrix}1&0\\1&1\end{pmatrix} $$ so $e^a\cdot e^b=\begin{pmatrix}2&1\\1&1\end{pmatrix}$, so $e^ae^b\neq e^{(a+b)}$ (unless $2=1$).