# Exponential series of bounded Hilbert space operator is convergent

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?

• Note that for $S_n-S_m$ there is no need to apply to a particular $v$, you can just estimate directly. May 11 '20 at 18:15

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 $$\left(e^{zA}\right)^*=\left(\sum_{k=0}^\infty \frac{z^kA^k}{k!}\right)^* =\sum_{k=0}^\infty\left( \frac{z^kA^k}{k!}\right)^*=\sum_{k=0}^\infty \frac{\overline{z}^k(A^*)^k}{k!}=e^{\overline z A^*}.$$
• 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? May 11 '20 at 18:13
• 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. May 11 '20 at 18:20
• 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? May 11 '20 at 19:14
• Well, yes and no. You are using the completeness of $B(H)$ to say that $e^{zA}$ exists. May 11 '20 at 19:25