Let $A\subset L^{\infty}(-\pi,\pi)$ be the closed subspace spanned by $\{e^{inx}\}_{n\geq0}$, i.e., all continuous functions on the unit circle that are uniform limits of trigonometric polynomials of the form $P(x)=\sum_{k=0}^{n} a_{k}e^{ikx}$. Is it true that if $f\in A$ and $f(0)=0$, then $\int f\;dm=0$ where $m$ is Lebesgue measure? This fact is used several times in “Banach Spaces of Analytic Functions” but I have no idea why it is true.
If you understand $f(0)$ as the value at $0$ of a function defined on $(-\pi,\pi)$ then the function $$f(x):=1-e^{ix}$$ is a counterexample: $$f\in A,\quad f(0)=0,\quad\int_{-\pi}^{\pi}f\left(x\right)dx=2\pi.$$
However, this interpretation is not compatible with Hoffman's proof, which uses that $$\forall f\in A\quad\int_{-\pi}^\pi\left[f(\theta)-f(0)\right]d\theta=0.\tag{*}$$ For this property to hold, $f(0)$ must be interpreted as the value, at the center $0$ of the unit disc $D$, of the analytic function $F$ on $D$ whose limit on the circle $\partial D$ is $f.$ If $f(x)=e^{ikx}$ then $F(z)=z^k,$ hence $F(0)=0$ if $k\ge1,$ whereas $F(1)=1$ if $k=0.$ Thus, the property $(*)$ above reduces to the obvious: $$\forall n\ge1\quad\int_{-\pi}^\pi\left[e^{in\theta}-0\right]d\theta=0.$$
This was understandable from Hoffman's previous page: