The <holomorphic functional calculus> assigns to every function $f$ holomorphic on a neighbourhood of $\sigma_A(x)$ the element
$$
f(x)=\frac1{2\pi i}\int_\Gamma f(z)(z1-x)^{-1}\,dz,
$$
where the oriented contour $\Gamma$ surrounds the spectrum inside that neighbourhood. The value is independent of the admissible contour, and $f\mapsto f(x)$ is a continuous unital algebra homomorphism sending the coordinate function $z$ to $x$.

Every $\varphi\in\Phi_A$ commutes with the contour integral, so the <Cauchy integral formula> gives
$$
\varphi(f(x))
=\frac1{2\pi i}\int_\Gamma f(z)(z-\varphi(x))^{-1}\,dz
=f(\varphi(x)).
$$
Part a applied to $f(x)$ now proves the <spectral mapping theorem>:
$$
\sigma_A(f(x))
=\{\varphi(f(x)): \varphi\in\Phi_A\}
=f(\sigma_A(x)).
$$

Let $\Omega$ be the unbounded component of $\mathbb C\setminus\sigma_A(x)$. On the spectrum, spectral mapping gives
$$
|f(z)|\leq r(f(x))\leq\|f(x)\|.
$$
Every remaining point of $\mathbb C\setminus\Omega$ lies in a bounded complementary component $D$. Its boundary is contained in $\sigma_A(x)$, and $f$ is holomorphic near $\overline D$. The <maximum modulus principle> therefore extends the same estimate from $\partial D$ to $D$. Hence
$$
|f(z)|\leq\|f(x)\|
\qquad(z\in\mathbb C\setminus\Omega).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 1 / b / Solution

{wiki}

For $x$ in a unital Banach algebra and $f$ holomorphic near $\sigma(x)$, the holomorphic functional calculus defines
$$
f(x)=\frac1{2\pi i}\int_\Gamma f(z)(z1-x)^{-1}\,dz,
$$
where $\Gamma$ winds once around the spectrum. It is a continuous unital algebra homomorphism and satisfies the <spectral mapping theorem> $\sigma(f(x))=f(\sigma(x))$.

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