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Is it true that if a holomorphic function in the unit disk converges uniformly to the $0$ function some connected arc of the unit circle, this function is globally null?

If that is true, this would stand for a uniqueness fact and so how to recover some holomorphic function from its boundary value on an arc (if the uniform convergence still holds on the arc)

user1611830
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3 Answers3

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Let the (open) arc on which the boundary values of $f$ vanish be $A$. Since the boundary values of $f$ on $A$ are real, by the Schwarz reflection principle we know that the function

$$g(z) = \begin{cases} \;\,f(z) &, \lvert z\rvert < 1\\ \quad 0 &, z \in A\\ \overline{f(1/\overline{z})} &, \lvert z\rvert > 1\end{cases}$$

is holomorphic on the connected open set $\mathbb{D} \cup A \cup (\mathbb{C}\setminus \overline{\mathbb{D}})$. Since $g$ vanishes on a non-discrete set, the identity theorem yields $g \equiv 0$, in particular $f \equiv 0$ follows.

Daniel Fischer
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  • sorry I mislead, using Schwarz' reflection principle leads to some uniqueness fact as you're describing here. So any function that respects this uniform convergence should then be reconstructed from its boundary value, but how ? not by analytic continuation from the boundary, since there are holomorphic function that converges uniformly to the boundary but that are not extendable beyond the unit circle. – user1611830 Aug 10 '14 at 17:24
  • Unless the boundary values on the arc are nice, I don't know of a way to reconstruct the function from its values on the arc. If you have decently behaving boundary values on the whole circle (minus null sets), you get it as a Cauchy integral, but you probably know that. If the boundary values are analytic at some point on the arc, you get $f$ from analytic continuation (in principle at least). Other than those cases, I don't know a way. For continuous values on the arc, you have a chance if you approximate the boundary values by analytic functions, and hope for convergence, but that can fail. – Daniel Fischer Aug 10 '14 at 17:51
  • yes indeed, for the cases you say, or the case of real part values on some arc it is rather trivial. But for more general case, using the schwarz principle imposes uniqueness and so that some reconstruction formula exists... – user1611830 Aug 10 '14 at 17:55
  • @user1611830 Do we even have a characterisation of the functions on an arc of the unit circle that are boundary values of a holomorphic function on the disk? If we have, we still only have an existence theorem, from that to a reconstruction formula is a long long way (to Tipperary). We know we can well-order $\mathbb{R}$, but as far as I know, we're no closer to an explicit construction of a well-order on $\mathbb{R}$ than we were 130 years ago. Which means we have no idea how. – Daniel Fischer Aug 10 '14 at 17:59
  • sure in a general way, you can have infinity of holomorphic functions that are entirely 0 on the boundary. But here the problem seems to be so constrained that this kind of fact must be known – user1611830 Aug 10 '14 at 18:03
  • I don't understand what you mean. The problem is not very constrained, as far as I can tell. I'm out of touch with the research, but if there were a known way to reconstruct the function from the boundary values on an arc for general boundary values, unless it's very new ($\leqslant 15$ years, say), I'd expect I'd have heard of it. – Daniel Fischer Aug 10 '14 at 18:08
  • I say constrained because, from what you answered (with which I am perfectly ok), I deduce there exists at most one function (holomorphic converging uniformly up to the boundary) that has some specific boundary values. Btw sorry, perhaps I was not clear : sure I am not looking for one single global reconstruction formula, but even some local extension would fit (that must not in general involves analytic continuation) – user1611830 Aug 10 '14 at 18:13
  • Right, we have uniqueness. But if there was a known way to get a local extension, there'd be one [though not explicit] to get the global function, namely analytic continuation from the local function. Except for the easy special cases, I know of nothing in that direction. – Daniel Fischer Aug 10 '14 at 18:25
  • @user1611830 ".... imposes uniqueness and so that some reconstruction formula exists" -- that's a huge logical leap there. Knowing that some map is injective and having a formula for its inverse are two very different things. –  Aug 11 '14 at 19:46
  • @900sit-upsaday,first that some reconstruction exists is sure.The question is to know if it is explicit or not.Second,since the inverse engineered holomorphic function remains the same as long as you shrink the boundary arc any small you want, provided that its measure is not null. This suggests strongly that the reconstructed values depend only either of the germ of the boundary function at one point or of some PDE that constraints the behavior of the boundary function.As far as I am experienced in inverse geometry, this could lead to the reasonable hope that some reconstruction process holds – user1611830 Aug 12 '14 at 16:43
  • Hi Daniel, shouldn't it be $1/\overline{f(1/\overline{z})}$? – user135520 Jul 03 '18 at 21:07
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    @user135520 Not in this case. Here we have a function that takes real values (more specifically, the value $0$) on an arc of the unit circle. So we reflect the argument of $f$ in the unit circle - that's $1/\overline{z}$ - and we reflect the value of the function in the real axis, so overall we get $\overline{f(1/\overline{z})}$. The formula $1/\overline{f(1/\overline{z})}$ reflects both, argument and value, in the unit circle. That's for when $\lvert f(z)\rvert = 1$ on an arc of the unit circle. – Daniel Fischer Jul 03 '18 at 21:12
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I believe this is correct, but I may be wrong. Say we know the precise values of a holomorphic function everywhere in an arc and say the arc ranges from an angle of $\theta_0$ to $\theta_1$ as measured ( for example, but this is arbitrary) clock-wise from the origin. Then, we define the new function $f(\theta)$ as the values of the initial holomorphic function at an angle $\theta$ which is within the range from $\theta_0$ to $\theta_1$. Now, $f(\theta)$ can be interpreted as a real function defined within an interval. It must be analytic within this interval. From there, one can derive a power series of $f$ and extend it to all real values of $\theta$ ( values should cycle every $2\pi$ ). Also, from this power series, one can obtain the values of $f$ at complex numbers ( for now, do not over-think the idea of having complex angles). Let $g(z)$ be the initial holomorphic function. Then, by our initial definition, within the arc ( of radius, say, $r$) we have $$f(\theta)=g(re^{i\theta}) $$ and since we have $f$ for all complex $\theta$, we can choose a $\theta$ to make $re^{i\theta}$ whatever we want, and so we can obtain the function $g$. I hope I have explained it clearly. On a more general note, if one knows the behaviour of an analytic function in any connected subset of the complex plane, one can find the values of that function in the entire plane ( analytic continuations are, after all, unique). I hope that helped.

Asier Calbet
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    This would work if the function was assumed to be holomorphic on the closed unit disk. Which isn't assumed here; the boundary values are interpreted as limits. –  Aug 10 '14 at 17:13
  • Its given that the function is holomorphic! – Asier Calbet Aug 11 '14 at 19:42
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    Holomorphic on what set? On the open disk. Is the boundary a part of the disk? No. –  Aug 11 '14 at 19:43
  • It indeed is true the the limit function on the circle arising from a holomorphic function on the disk is analytic in the $z$-variable. Taking $z = e^{i\theta}$, you have a complex analytic function on the real interval $[0,2\pi)$. I believe the rest of the argument also follows through. – Teebro Prokash Oct 29 '20 at 19:14
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Suppose $f$ is analytic on the disc, whose limits at the boundary yields a set of zeros with non-empty interior, and $f$ is continuous on some neighborhood (in the closed disc) of one such interior accumulation point. By reflecting through the circular boundary we can extend $f$ to be analytic on a neighborhood (in the plane) of the accumulation point. As was indicated by a comment, this is essentially the Schwarz reflection principle, though in a slightly different form than usually stated. Thus the extension has an accumulation of zeros in its domain, and so is identically zero.

zibadawa timmy
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  • thank you but as I mentioned above, the problem is the following : using Schwarz' reflection principle leads to some uniqueness fact as you're describing here. So any function that respects this uniform convergence should then be reconstructed from its boundary value, but how ? not by analytic continuation from the boundary, since there are holomorphic function that converges uniformly to the boundary but that are not extendable beyond the unit circle. Right ? – user1611830 Aug 10 '14 at 17:37
  • @user Check the reflection principle. The specific behavior on the boundary is what guarantees the reflected portion and the original function agree at the boundary, and are analytic there. – zibadawa timmy Aug 10 '14 at 17:40
  • ok but how do you explain that in one hand there exists some holomorphic functions uniformly converging on the boundary that are not extendable beyond and on the other hand that any functions that has this behavior should be reconstructed by its value on the boundary in an unique way ? so again, how can I reconstruct some function (holomorphic and converging uniformly on an arc) from its boundary values ? I am ok that if the boundary values are real, than analytic continuation does the trick. But, if the boundary values are not real, some formula must exist, but how can I get it? – user1611830 Aug 10 '14 at 17:46
  • @user You could probably conformally map the actual boundary values to real numbers by an invertible holomorphic map. Which is basically the Dirichlet problem, I think. – zibadawa timmy Aug 10 '14 at 17:54
  • yes but this would impose that some analytic continuation is processed and this would violate the existence of holomorphic functions uniformly converging on the boundary that are not extendable beyond the unit circle, right ? – user1611830 Aug 10 '14 at 17:57
  • @user1611830 I believe your issue comes from attempting to globally extend along the whole boundary, whereas here we are only doing it locally. Local extensions beyond the boundary are permissible, but trying to globally patch them together may reveal discontinuities (branch cuts and poles pop up). – zibadawa timmy Aug 10 '14 at 18:05
  • sorry I "misexpressed" myself : sure I am not looking for one single globale formula, but even some local extension (that must not in general involves analytic continuation) – user1611830 Aug 10 '14 at 18:10
  • @user As long as the boundary values (on an arc) are 'nice enough' that you can convert to the real case, then you can extend. Problem is that 'nice enough' need not be more than a local thing, and so the extension may only be local. I'm fairly sure uniform convergence is always locally 'nice enough', but I won't guarantee it. – zibadawa timmy Aug 10 '14 at 18:16
  • ok but sure if you have some nice jordan curve, you can transfer it to the real axis. But that does not guarantee that you can map some neighborhood of the arc (within the closed disk) preserving this mapping – user1611830 Aug 10 '14 at 18:33
  • @user1611830 'nice enough' is an intentionally vague term. I do not know how to make it precise, and Dan seems to have similar misgivings. Point is, your original question is answered, and your new concern about reconstruction theory is a different and more complicated thing. The original question is about a uniqueness proof, not an existence proof. – zibadawa timmy Aug 10 '14 at 18:46
  • ok I just thought you thought nice enough meant some nice regularity assumption. I understand better now and thanks for all – user1611830 Aug 10 '14 at 18:57