Let $A\in\mathcal{B}(\mathcal{H})$ be a self-adjoint operator and consider $\lambda\in\mathbb{C}$ outside the spectrum of $A$, i.e., $\lambda\in\mathbb{C}\setminus\sigma(A)$. In this case, $A - \lambda I$ has a bounded inverse. I'm trying to prove that $||(A-\lambda I)||^{-1}=\dfrac{1}{d(\lambda,\sigma(A))}$, where $\displaystyle d(\lambda,\sigma(A)) = \inf_{\gamma\in\sigma(A)}|\gamma - \lambda|$.
The hint I've been given is to apply the functional calculus to $(A - \lambda I)^{-1}$. Here, the term "functional calculus" refers to the process of elevating a typical complex-valued function on complex numbers to an operator-valued function. Such a "lifted" function will look like the following:
As before, denote by $\sigma(A)$ the spectrum of $A$ and take the Borel $\sigma$-algebra on $\sigma(A)$. Suppose $g:\sigma(A)\to\mathbb{C}$ is a bounded, measurable, complex-valued function. Then, $\displaystyle g(A) = \int_{\sigma(A)}g(\lambda)\mu^A(\lambda)$, where $\mu^A$ is a unique projection-valued measure (see definition from the Wikipedia article) satisfying $\displaystyle \int_{\sigma(A)}\lambda d\mu^A(\lambda) = A$.
Where I'm encountering difficulty is in linking the norm of any bounded linear operator, for instance, $(A - \lambda I)^{-1}$, with its representation in functional calculus. To be specific, if we define $f(\gamma) = \gamma - \lambda$, then we have $A - \lambda I = f(A)$ and $(A - \lambda I)^{-1} = g(A)$, where $g(\gamma) = \dfrac{1}{\gamma - \lambda}$. It's important to remember that $\sigma(A)$ is a closed and bounded subset of $\mathbb{R}$ because $A$ is self-adjoint.
So, what steps should I take to proceed from here?