Power Series of an Operator I'm working through some functional analysis problems and  am having trouble with the following. Let $f(z)=\sum_{n=0}^\infty a_n z^n$ be a power series with radius of convergence $R>0$. Let $A\in\mathcal{B}(\mathcal{H})$ with $||A||<R.$ Show the following: 
(a) There is a $T\in\mathcal{B}(\mathcal{H})$ with $\langle Tx, y\rangle=\sum_{n=0}^\infty a_n\langle A^nx,y\rangle$ for $x,y\in\mathcal{H}.$ Here we denote $T$ by $f(A)$. 
(b) That the partial sums of $f(A)$ converge to $T$ in the operator norm. 
(c) That $f(A)B=bf(A)$ whenever $B\in\mathcal{B}(\mathcal{H})$ with $BA=AB.$ 
(d) For $f(z)=e^z$  for $A$ self-adjoint we have $f(iA)$ is unitary.
Solution attempts:
(a) I think $T = \sum_{n=0}^\infty a_n A^n$,  but am  having trouble showing this is within $\mathcal{B}(\mathcal{H})$. In particular, I have for $h\in\mathcal{H}$ with $||h||=1,$
$|| \sum_{n=0}^\infty a_nA^n h||\leq |a_0|||h||+|a_1|||Ah||+|a_2|||A^2h||+\cdots$
My thought was to continue this chain of inequalites until I got a series that converged, but since I am taking the absolute values of the coefficients I don't think that this will work.
(b) I think I could get this part if I saw how part (a) is done.
(c) $$
    f(A)B = (\sum a_n A^n ) B = (a_0+a_1A+a_2A^2+\cdots)B\\
    =a_0B+a_1AB+a_2AAB+\cdots = Ba_0+Ba_1A+Ba_2AA+\cdots \\
    =B(\sum a_n A^n) = Bf(A)
    $$
(d) $f(iA)=\sum \frac{1}{n!} (iA)^n\Rightarrow f(iA)^*= (\sum \frac{1}{n!}(iA)^n)^*=\sum \frac{1}{n!}(i^n)^*(A^n)^*=\sum \frac{1}{n!}(i^*)^n(A^*)^n=\sum \frac{1}{n!}(-i)^nA^n =f(-iA)$. Since $A$ commutes with itself, we have $f(iA)f(-iA)=f(-iA)f(iA)=e^{iA}e^{-iA}=e^{0}=I$
 A: In part a), your inequality is too crude. And it wouldn't prove much, because what you need to show is that the series exists. You have 
$$
\left\|\sum_{n=m}^ka_nA^n\right\|\leq\sum_{n=m}^k|a_n|\,\|A^n\|\leq\sum_{n=m}^k|a_n|\,\|A\|^n.
$$
Since $\|A\|<R$, the series $\sum_n |a_n|\,\|A\|^n$ converges, and so the last term above can be made arbitrarily small if $m$ and $k$ are big enough. This shows that the partial sums $\sum_{n=1}^ka_nA^n$ are Cauchy, and so the series converges since $B(H)$ is complete. 
Now, since the inner product is continuous (in the sense that $\langle T_nx,y\rangle\to\langle Tx,y\rangle$ if $T_n\to T$), we have 
\begin{align}
\langle Tx,y\rangle&=\left\langle \left(\lim_{k\to\infty}\sum_{n=1}^ka_nA^n\right)x,y\right\rangle=\lim_{k\to\infty}\left\langle \left(\sum_{n=1}^ka_nA^n\right)x,y\right\rangle\\ \ \\
&=\lim_{k\to\infty}\left\langle \sum_{n=1}^ka_nA^nx,y\right\rangle=\lim_{k\to\infty}\sum_{n=1}^ka_n\left\langle A^nx,y\right\rangle\\ \ \\
&=\sum_{n=1}^\infty a_n\left\langle A^nx,y\right\rangle,
\end{align}
where the last series can be shown to be convergent as in the first part. 
The rest of your arguments are ok. Just never forget that a series is a limit of sums and not a sum. Maybe I'm wrong, but the way you wrote part c) suggests to me that such idea is not very clear in your mind. 
A: Define $T = f(A)$ with $f(A) = \sum_{n=0}^\infty a_nA^n$. This series converges absolutely since
$$\sum_{n=0}^\infty |a_n|\|A^n\| \le \sum_{n=0}^\infty |a_n|\|A\|^n < +\infty$$
because $\|A\| < R$ and the power series $\sum_{n=0}^\infty a_nz^n$converges absolutely on $B(0,R) \subseteq \mathbb{C}$.
Since $\mathcal{H}$ is complete, this implies that the series $\sum_{n=0}^\infty a_nA^n$ converges so $f(A)$ is well-defined.
$(a)$ now follows from the continuity of the inner product.
$(c)$ and $(d)$ are correct provided you know that $A \mapsto A^*$ and $A \mapsto AB$, $A \mapsto BA$ for fixed $B$ are continuous with respect to the norm topology.
