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Let $f\colon\overline{\mathbb{D}}\to\mathbb{C}$ be a continuous function but that $f\colon\mathbb{D}\to\mathbb{C}$ is holomorphic. My question is

Can the restriction of $f$ to $\mathbb{S}$ assume its values in the unit interval $[0,1]$, that is $f(\mathbb{S})\subseteq[0,1]$?

Specific/explicit examples — if any — will be helpful.

($\mathbb{D}$ is the open unit disc with $\mathbb{S}$ as its boundary).

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    $\begingroup$ Perhaps this is clear to you - I'm very much not an analyst - but if things are regular enough then this is impossible. If $f(z) = \sum_{n \geq 0}a_n z^n$ on the disk, and the Fourier expansion of $f$ on the unit circle is $\sum_{n \geq 0}a_n z^n$, then being real-valued on the unit circle means $a_n = \overline a_{-n}$ and so all coefficients except $a_0$ vanish and $f$ is constant. $\endgroup$
    – Dan Petersen
    Commented Jan 3, 2021 at 9:25
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    $\begingroup$ A fourth argument follows from the uniqueness of solutions to the Dirichlet problem: $v=\textrm{Im}\; f$ is harmonic on $D$ and $v=0$ on $S$, hence $v\equiv 0$ and thus $f\equiv a\in\mathbb R$. $\endgroup$
    – Christian Remling
    Commented Jan 3, 2021 at 17:45

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As Dan Petersen anticipated, any function satisfying the requirements is constant. Indeed, the function defined as $g(z):=f(z)$ for $|z|\leq 1$ and $g(z):=\overline{f(1/\overline{z})}$ for $|z|\geq 1$ is entire and bounded, hence constant by Liouville's theorem.

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    $\begingroup$ Is it clear though that $g$ is holomorphic when $\vert z\vert =1$? When I wrote my comment I worried about regularity issues like whether the Taylor expansion would necessarily coincide with the Fourier expansion without assuming $f$ to extend holomorphically to any neighborhood of the disk; proving that $g$ is holomorphic seems to run into similar issues. $\endgroup$
    – Dan Petersen
    Commented Jan 3, 2021 at 10:28
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    $\begingroup$ @DanPetersen: $g$ is holomorphic because of Morera's theorem: it is continuous, and its line integral is zero over any triangular contour (as can be checked easily using Goursat's lemma). $\endgroup$
    – GH from MO
    Commented Jan 3, 2021 at 10:31
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    $\begingroup$ Thanks, sorry for the naive question! $\endgroup$
    – Dan Petersen
    Commented Jan 3, 2021 at 10:48
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    $\begingroup$ @DanPetersen: your question was not naive at all. BTW I just realize that instead of "Goursat's lemma" I should have said "Cauchy's theorem". The point is to decompose every line integral over a triangular contour into line integrals over closed curves lying entirely in $|z|\leq 1$ or in $|z|\geq 1$. $\endgroup$
    – GH from MO
    Commented Jan 3, 2021 at 10:53
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    $\begingroup$ Indeed, the construction in the answer and the justifications in the comments are all part of the proof of the Schwarz reflection principle. Here the function is being reflected in the unit circle, while the classical formulation is a reflection in the real axis ... but "we all know" that circles and lines are the same thing in the complex plane, thanks to Möbius transformations! $\endgroup$
    – Greg Martin
    Commented Jan 3, 2021 at 17:14
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A fundamental property of non-constant analytic functions is that they are open maps: the image of an open set is open. In your case, the image of the disk must be an open bounded region, and its boundary must be contained in the image of the circle. Which is of course impossible if the image of the circle is contained in a segment.

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    $\begingroup$ I prefer this explanation over mine. $\endgroup$
    – GH from MO
    Commented Jan 3, 2021 at 21:06
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    $\begingroup$ It has an advantage of covering more general cases than a straight line segment. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 4, 2021 at 0:44
  • $\begingroup$ Respect for the adjective “non-constant” (although in the full generality it should be stated more cautiously: not locally constant anywhere). $\endgroup$
    – Incnis Mrsi
    Commented Apr 13, 2021 at 11:01

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