# on the boundary of analytic functions

Suppose I have a function $f$ that is analytic on the unit disk $D = \{ z \in \mathbb{C} : |z| < 1 \}$ that is also continuous up to $\bar{D}$. If $f$ is identically zero on some segment of of the boundary (e.g. $\{ e^{it}, 0 \leq t \leq \pi/2 \}$ ), is it then true that $f$ is identically zero on the entire boundary?

I know that analytic functions that are zero at an accumulation point inside the domain of analyticity are identically zero throughout the entire domain, but I don't know what (if anything) can be said if something similar occurs on the boundary of the domain.

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 if $f$ is not identically zero, then $f$ cannot be identically zero on $\{z : |z|=1\}$ (the radius of convergence about zero goes up to the nearest singularity) – yoyo Mar 1 '11 at 15:57 If you have part of the unit circle equal to zero this means that you can analytically continue it in this direction (you can always continue up to the next pole). Expanding around a point on the unit circle for which $f=0$ there will be a finite convergence radius and an accumulation point inside the domain. The function therefore is equal to 0. – Fabian Mar 1 '11 at 17:46 Fabian: Does Schwartz reflection imply that the Analytic continuation exists? – jennifer Mar 1 '11 at 20:13 @Fabian: Can you explain how exactly you would perform the analytic continuation to that part of the unit circle? – user1736 Mar 1 '11 at 20:17 @Myke & user1736: I guess one can use a Moebius transformation of the part of the circle having the accumulation point on the real line and then use Schwartz reflection... – Fabian Mar 1 '11 at 20:43

The only way the function can be zero on an arc of the circle is for the function to be identically zero. To see this, consider ${\displaystyle g(z) = \prod_{j = 0}^{n-1} f(e^{2\pi i j \over n}z)}$ for a large enough $n$ so that $g(z)$ is identically zero on the boundary of the disk. Then by the Cauchy integral formula, if $r < 1$ then for any $w$ in the interior one has $$g(w) = {1 \over 2\pi i}\int_{|z| = r}{g(z) \over z - w}\,dz$$ Since $g(z)$ is continuous on the compact set $\bar{D}$, $g(z)$ is bounded, so one can apply the dominated convergence theorem and let $r$ go to $1$. One obtains $$g(w) = {1 \over 2\pi i}\int_{|z| = 1}{g(z) \over z - w}\,dz$$ $$= 0$$ Thus $g(z)$ is identically zero. Since the zeroes of a nonzero analytic function are isolated, we conclude that $f(z)$ is identically zero as well.

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Nice argument introducing $g$. Perhaps even more obvious to write $g(z)=f(z)\cdot f(iz)\cdot f(-z)\cdot f(-iz) = 0$ (looking at the segment in the example..). – AD. Mar 1 '11 at 18:11
Can you explain why g(z) is identically zero on the boundary of the disK? I understand everything after and why the function should be zero inside, but I don't really get the first part. – user1736 Mar 1 '11 at 18:23
if $f(z)$ is zero on an arc $C$ then $f(e^{2\pi i j/n}z)$ is zero if $e^{2\pi i j/n}z$ is on $C$, or equivalently if $z$ is on $C$ rotated counterclockwise by ${2\pi j\over n}$. If you make $n$ large enough, then for any $z$ with $|z| = 1$ there will be some $j$ for which $z$ is on the rotated arc. So the product will be zero there. The case $n=4$ works for your example and $g(z)$ is the function AD wrote there. – Zarrax Mar 1 '11 at 18:46
Ah, I think that makes sense. And you didn't even use analytic continuation right? That's pretty cool. – user1736 Mar 1 '11 at 20:56
A continuous function on the closed disk that is analytic on the open disk and zero on the boundary is identically zero by the maximum modulus theorem. – Jonas Meyer Mar 3 '11 at 7:03

Let $E$ be a closed subset of the boundary of the disk. Then there is a nonzero function $f$ continuous on the closed disk and analytic on the open disk such that $f\vert_E=0$ if and only if the Lebesgue measure of $E$ is zero.

The fact that the Lebesgue measure has to be zero (which answers your question) follows from a more general theorem that if $f$ is in any of the Hardy spaces of the disk, then the log of the absolute value of the boundary function of $f$ is in $L^1$ of the circle with Lebesgue measure. This is in Theorem 17.17 of Rudin's Real and complex analysis, 3rd Edition.

The fact that every closed subset of the boundary with Lebesgue measure zero can consist of zeros of a nonzero function that is continuous on the closed disk and analytic on the interior is a theorem of Fatou. This appears in the section "Theorems of Fatou and Rudin" in Chapter 6 of Hoffman's Banach spaces of analytic functions.

Consider any function $\phi:S^1\to\mathbb C$ which is smooth and which vanishes on a segment of the circle. Define $f:D\to\mathbb C$ so that $$f(z)=\int_{S^1}\frac{f(\zeta)}{\zeta-z}\,\mathrm d\zeta.$$ This will be holomorphic in the interior of $D$ and will coincide with $\phi$ on $S^1$.
this function will not be continuous on all of $\bar{D}$ – Zarrax Mar 1 '11 at 16:09
I don't think this answer is correct. As Zarrax pointed out this function when extended to the boundary of $D$ will not coincide with $\phi$ on the boundary. – Fabian Mar 1 '11 at 17:43