Let $A \in B(H)$ be a bounded Hilbert space operator. For $z \in \mathbb{C}$ exponential is defined as follows:
$$e^{zA} = \sum_{k=0}^{+\infty}\frac{z^kA^k}{k!}$$
Show that series defined above is convergent.
My attempt:
Let
$$S_n = \sum_{k=0}^{n}\frac{z^kA^k}{k!}$$
I will show that $(S_n)$ is Cauchy in $ B(H)$.
Without loss of generality assume that $m > n$.
For every $v \in H$ such that $\lVert v \rVert = 1$ we have:
$$\lVert (S_m – S_n)v \rVert = \lVert \sum_{k=n+1}^{m} \frac{z^kA^kv}{k!} \rVert \leq \sum_{k=n+1}^{m}\lVert \frac{z^kA^kv}{k!}\rVert \leq \sum_{k=n+1}^{m} \frac{(\lvert z \rvert \cdot \lVert A \rVert _{op})^k}{k!} < \varepsilon$$
for large $N$ since $e^x$ is uniformly convergent in $\mathbb{R}$, which is also a complete space.
Which leads to convergence of $(S_n)$.
Is it correct so far?
The next step is to show that
$$(e^{zA})^* = e^{\bar{z}A^*}$$
where $^*$ denotes Hermitian conjugate.
Again, my attempt:
Let $v, w \in H$ then
$$\langle v, (e^{zA})^*w \rangle =\langle e^{zA}v, w \rangle = \langle \sum_{k=0}^{+\infty}\frac{z^kA^kv}{k!}, w \rangle = \sum_{k=0}^{+\infty}\langle \frac{z^kA^kv}{k!}, w \rangle = \sum_{k=0}^{+\infty} \langle v, \frac{(\bar{z}A^*)^k}{k!}w \rangle = \langle v, e^{\bar{z}A^*}w \rangle$$
Which gives us $$(e^{zA})^* = e^{\bar{z}A^*}$$
Where I used a property $(zA)^* = \bar{z}A^*$ and continuity of inner product.
Again, is it correct?
Best Answer
Yes, it is correct. An essential part of the argument is that $B(H)$ is complete. Note that the argument works in any Banach algebra.
For the second part you could just use that taking adjoints is continuous (it is actually an isometry: $\|T^*\|=\|T\|$), so $$ \left(e^{zA}\right)^*=\left(\sum_{k=0}^\infty \frac{z^kA^k}{k!}\right)^* =\sum_{k=0}^\infty\left( \frac{z^kA^k}{k!}\right)^*=\sum_{k=0}^\infty \frac{\overline{z}^k(A^*)^k}{k!}=e^{\overline z A^*}. $$