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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.

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    $\begingroup$ 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 $f \in A$. $\endgroup$
    – geetha290krm
    May 26 at 4:49
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    $\begingroup$ You are probably thinking of the converse of the corollary. Converse are not always true. $\endgroup$
    – geetha290krm
    May 26 at 5:14
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    $\begingroup$ 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. $\endgroup$
    – SihOASHoihd
    May 26 at 5:20
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    May 26 at 15:04
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    $\begingroup$ 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 $2\pi a_0$ $\endgroup$
    – Conrad
    May 26 at 15:07

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  • 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".

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  • $\begingroup$ @BrianMoehring Thank you but the OP mentionned "Lebesgue measure", and the excerpt takes $d\theta,$ not $ie^{i\theta}d\theta.$ $\endgroup$
    – Anne Bauval
    May 26 at 8:13
  • $\begingroup$ 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 $\int\mathrm d\theta=2\pi.$ $\endgroup$
    – Anne Bauval
    May 26 at 10:29
  • $\begingroup$ @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 $\endgroup$
    – Conrad
    May 26 at 13:45
  • $\begingroup$ @Conrad Thank you! What a relief! I shall improve my answer thanks to your comment. $\endgroup$
    – Anne Bauval
    May 26 at 14:03
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    $\begingroup$ @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. $\endgroup$
    – SihOASHoihd
    May 26 at 15:32

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