Q:

Banach algebra, map $f \mapsto {1\over{2\pi i}} \int_S f(z) \cdot (z - a)^{-1}dz$ well-defined?

Let $A$ be a Banach algebra over $\mathbb{C}$ and $N: A \to \mathbb{R}_{\ge0}$ the corresponding multiplicative norm. For any $a \in A$, we define$$\text{spec}(a) = \{\lambda \in \mathbb{C} : \lambda - a \text{ is not invertible}\}.$$Fix $r > N(a)$ and let $S = \{z \in \mathbb{C} : |z| = r\}$ be the circle in $\mathbb{C}$ of radius $r$. Further, let $\mathcal{H}$ be the algebra of holomorphic functions $\mathbb{C} \setminus \text{spec}(a) \to \mathbb{C}$. Is the map$$\mathcal{H} \to A,\text{ }f \mapsto {1\over{2\pi i}} \int_S f(z) \cdot (z - a)^{-1}dz$$even well-defined?

A:

Since $r>||a||$ and $||(a/z)^n||\leq (||a||/r)^n$ for $z\in S$ ,the series $\sum_{n\geq 0}z^{-1}(a/z)^n$ converges uniformly in norm to  $(z-a)^{-1}$ on $S . $ So we can replace $(z-a)^{-1}$ in the integral with the power series, and because the convergence in norm is uniform,we can integrate term-by-term and the operator $$g(x)=\left(\frac {1}{2\pi i}\int_{|z|=r}\frac {f(z)}{z-a}dz\right)(x)$$exists for each $x\in A$......(Remark: Now that you have it,what do you do with it?)

