Functional analysis question involving complex analysis Let $V$ be a Banach space and let $T \in L(V)$. Let $\Omega \supset \sigma(T)$ and let $f: \Omega \subset \mathbb{C} \rightarrow \mathbb{C}$ be analytic. Let $\gamma$ be a positively oriented simple closed $C^2$ path in $\Omega \cap \rho(T)$ whose interior contains $\sigma(T)$. Define $f(T)= \frac{1}{2 \pi i} \oint_{\gamma} \frac{f(z)}{(zI-T)} \ dz$.


*

*Let $P(z)= \sum_{j=1}^N a_jz^j$ be a polynomial. Prove that $P(T)= \sum_{j=1}^N a_jT^j$.


(Hint: $\frac{1}{2 \pi i} \oint \xi^k(\xi I -T)^{-1} \ d \xi  = \sum _{n=0}^\infty T^n \frac{1}{2 \pi i} \oint \xi ^{-n+k-1} \ d \xi = T^k $)


*Prove that the (holomorphic) functional calculus $f \rightarrow f(T)$ is a homomorphism between the algebra of functions analytic in $\Omega$ into the algebra $L(V)$
(Hint: express $(zI-T)^{-1}(wI-T)^-1$ as a difference. You're allowed to commute integrals without justification. Recall that if $f : \Omega \subset \mathbb{C} \rightarrow \mathbb{C}  $ is analytic then $\frac{1}{2 \pi i} \oint_{\gamma} \frac{f(z)}{z-w} \ dz$ equals $f(w)$, respectively 0, if $w$ is inside, respectively outside, of the contour $\gamma$).


*Show $\sigma(f(T))=f(\sigma(T))$
For question 1, I don't really see what am I supposed to do using the hint. So I'm not exactly sure how to start. 
For question 2, following the hint, I got $\frac{A}{(zI-T)} - \frac{B}{(wI-T)}  $, where $A=-B$ and $B=(z+w)^{-1}$. Then im not sure what to next, since our goal is to show that $f(T_1 + T_2)= f(T_1)+f(T_2)$. Also, I feel like we have to use part 1 somehow...
For question 3, I'm not too sure what to do... Moreover, what is $\sigma$?
Thank you for reading this! Any help is appreciated!
 A: I guess $L(V)$ is the space of bounded = continuous operators on the Banach space $V$.
(1) Since the formula to be show is linear in the polynomial $P$, it is enough to show it for the polynomials $P(T)=T^k$, $k$ running in natural numbers. For such a $k$ the formula follows from the hint.
(2) Let $f,g$ be holomorphic functions on the domain $\Omega$ containing the spectrum $\sigma(T)$ of an operator $T\in L(V)$. The formula defining $f(T)$ is linear in $f$, we obtain immediately $(f+g)(T)=f(T)+g(T)$. Let us show $(fg)(T)=f(T)\; g(T)$. We will use fractional calculus in usual notation, since the algebra of fractions / of fractional calculus (with invertible denominators) in $T$ is commutative. This is a $\Bbb C$-algebra, so we will use the notation $z-T$ instead of the explicit $zI-T$, by using the slightly abusive convention of no longer mentioning the structural morphism $z\to zI$ of this $\Bbb C$-algebra. 
The definition of $f(T)$ does not depend on the choice of the contour $\gamma$ around the spectrum $\sigma(T)$, which is a compact set. Let $\delta$ be an other contour containing in its interior (the image of) $\gamma$. The following argument is well known.
We compute the expression
$$
\begin{aligned}
f(T)\; g(T) 
&=
\left( \frac 1{2\pi i}\int_\delta\frac {f(z)}{z-T}\; dz\right)
\left( \frac 1{2\pi i}\int_\gamma\frac {g(w)}{w-T}\; dw\right)
\\
&=
\frac 1{(2\pi i)^2}
\int_\gamma
\int_\delta
f(z)\; g(w)\;\left( 
\frac {1}{z-T}\cdot
\frac {1}{w-T}
\right)
\; dz\; dw
\\
&=
\frac 1{(2\pi i)^2}
\int_\gamma
\int_\delta
f(z)\; g(w)\;\frac 1{z-w}\
\left( 
\frac {1}{z-T}-
\frac {1}{w-T}
\right)
\; dz\; dw
\\
&=
\frac 1{(2\pi i)^2}
\int_\gamma
\int_\delta
f(z)\; g(w)\;\frac 1{z-w}\cdot
\frac {1}{z-T}
\; dz\; dw
\\
&\qquad +
\frac 1{(2\pi i)^2}
\int_\gamma
\int_\delta
f(z)\; g(w)\;\frac 1{w-z}\cdot
\frac {1}{w-T}
\; dz\; dw
\\
&=
\frac 1{(2\pi i)^2}
\int_\gamma
\int_\delta
f(z)\; g(w)\;\frac 1{z-w}\cdot
\frac {1}{z-T}
\; dz\; dw
\\
&\qquad +
\frac 1{(2\pi i)^2}
\int_\delta
\int_\gamma
f(z)\; g(w)\;\frac 1{w-z}\cdot
\frac {1}{w-T}
\; dw\; dz
\\
&=
\frac 1{(2\pi i)^2}
\int_\gamma
\int_\delta
f(z)\; g(w)\;\frac 1{z-w}\cdot
\frac {1}{z-T}
\; dz\; dw
\\
&\qquad +0
\\
&\qquad\text{since each $z$ on $\delta$ is not in the interior of $\gamma$}
\\
&\qquad\text{and using the Residue Theorem}
\\
&=
\frac 1{2\pi i}
\int_\gamma
\left[
f(z)\; g(w)\;
\frac {1}{z-T}
\right]_{\text{computed in }z=w}
\; dw
\\
&=
\frac 1{2\pi i}
\int_\gamma
f(w)\; g(w)\;
\frac {1}{w-T}
\; dw
\\
&=
(fg)(T)\ .
\end{aligned}
$$
(3) Assume $b\not \in f(\sigma(T))$, which is a compact set.
Then the function $z\to g(z)=1/(f(z)-b)$ is well defined in a suitable domain containing $\sigma(T)$, we apply the analytic  functional calculus and obtain an operator $g(T)$ with $g(T)(f(T)-b)=(f(T)-b)g(T)=1$. So $f(T)-b$ is invertible, i.e. $b\not \in \sigma(f(T))$. 
Assume $b\in f(\sigma(T))$. Then $b=f(a)$ for some $a\in \sigma(T)$.
We write $f(z)-f(a)=(z-a)\;g(z)$ in the space of analytic functions in $z$. Then analytic functional calculus gives $f(T)-b=f(T)-f(a)=(T-a)\; g(T)$, which is not invertible because of the "factor" $(T-a)$. So $b\in \sigma(f(T))$.
