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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?

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  • $\begingroup$ Note that for $S_n-S_m$ there is no need to apply to a particular $v$, you can just estimate directly. $\endgroup$
    – copper.hat
    May 11 '20 at 18:15
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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 $$ \left(e^{zA}\right)^*=\left(\sum_{k=0}^\infty \frac{z^kA^k}{k!}\right)^* =\sum_{k=0}^\infty\left( \frac{z^kA^k}{k!}\right)^*=\sum_{k=0}^\infty \frac{\overline{z}^k(A^*)^k}{k!}=e^{\overline z A^*}. $$

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  • $\begingroup$ Thank you, sir. Can I have one more question? The next point of this exercise is to show that $e^{z_1A}\cdot e^{z_2A} = e^{(z_1+z_2)A}$. Should I use Cauchy product rule for finite partial sums and then just take the limit afterwards? $\endgroup$
    – janusz
    May 11 '20 at 18:13
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    $\begingroup$ You will have to involve the expansion of the binomial somehow. The proof is exactly the same that for the usual exponential. You can use it to show that $e^{A+B}=e^A \,e^B$ provided that $A$ and $B$ commute. $\endgroup$ May 11 '20 at 18:20
  • $\begingroup$ Actually one more question. If I choose they way of showing exactly that $S_n \rightarrow e^{zA}$ in operator norm I will not have to use completeness of B(H), right? $\endgroup$
    – janusz
    May 11 '20 at 19:14
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    $\begingroup$ Well, yes and no. You are using the completeness of $B(H)$ to say that $e^{zA}$ exists. $\endgroup$ May 11 '20 at 19:25

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