Let $T$ be an element of $B(X)$, $f$ be analytic in a neighbourhood of the $T$'s spectrum.
Why do we have the following
$$f(T)x=\frac{1}{2\pi i}\oint f(z)(z-T)^{-1}x \ dz \text{ ?}$$
I would think we could only say that $f(T)x=\frac{1}{2\pi i}\oint f(z)(z-T)^{-1} \ dz\ x$.
Is it because we know that $(f\circ g) (T)=f(g(T))$?
$$f(T)(x)=f(T(x))=\frac{1}{2\pi i}\oint f(z)(z-T(x))^{-1} \ dz=\frac{1}{2\pi i}\oint f(z)(z-T)^{-1}(x) \ dz$$
I'm learning by myself, so it's not easy to understand what type of context is helpful for this question. I'm reading the book Functional Analysis by Joseph Muscat.