# Functions in $A(\mathbb{C})$ that vanish at zero [closed]

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.

• You are reading the book wrongly . It says for 'For every $f$ in $A$....'. Doesn't say anything about a particular function with $f(0)=0$ and $f \in A$. May 26 at 4:49
• You are probably thinking of the converse of the corollary. Converse are not always true. May 26 at 5:14
• If such a $\mu$ exists then the converse would follow. The way he’s writing makes it seem that the converse is the obvious direction. May 26 at 5:20
• Yes that's correct but the $0$ here is the one of the unit disc not the $0$ of $(-\pi, \pi)$ in $L^{\infty}(-\pi, \pi)$; in other words while the integral of $f$ is on the unit circle, its value is the value of the analytic extension of $f$ to the unit disc taken at the zero of the unit disc May 26 at 15:04
• the integral of the unit circle of $e^{in \theta}$ is zero by periodicity unless $n=0$ since the function has the antiderivative $e^{in \theta}/(in)$ which is periodic so has same value at the ends unless $n=0$ when you integrate $1$ so get $2\pi$ since the antiderivative of $1$ is not periodic anymore; so integrating term by term $\sum_{n \ge 0}a_n e^{in \theta}$ gives precisely $2\pi a_0$ May 26 at 15:07

• 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:

"We denote by $$A$$ the collection of functions which are continuous on the closed unit disc and analytic at each interior point. [...] $$\|f\|_\infty=\sup_{|z|\le1}|f(z)|=\sup|f(e^{it})|.$$ [...] Thus, we may identify the functions in $$A$$ with their boundary values".

• @BrianMoehring Thank you but the OP mentionned "Lebesgue measure", and the excerpt takes $d\theta,$ not $ie^{i\theta}d\theta.$ May 26 at 8:13
• Thank you @BrianMoehring for insisting: now I read Hoffman's text and I am completely lost, and you made me eager to understand. If $\mathrm d\theta=ie^{ix}\mathrm dx,$ I understand why the first term of the sum is $0,$ but I don't understand why the second one is, i.e. why $\int\mathrm d\theta=2\pi.$ May 26 at 10:29
• @Brian $f(0)$ refers to the value at $0$ in the unit disc not at anything to do with the circle; the OP missed the important stuff that $A$ are actually analytic functions in the unit disc continuous on the boundary but the reference makes it clear; then indeed $f(0)=a_0$ the constant Fourier term May 26 at 13:45
• @Conrad Thank you! What a relief! I shall improve my answer thanks to your comment. May 26 at 14:03
• @AnneBauval what I was confused about was $f(0)=a_{0}$. Even though this was mentioned several times, all the comments about other spaces, measures, etc. obscured the simple fact that this is meant as a purely formal identification. May 26 at 15:32