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If $\Omega$ is a region which is dense in $\mathbb{C}$, $f\in H(\Omega)$ and is continuous on $\mathbb{C}$, moreover $f$ is bounded on $\mathbb{C}$, can we claim that $f$ is a constant?

Danqing He
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2 Answers2

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The answer is no. Indeed, let $K$ be a nowhere dense compact subset of the plane with Hausdorff dimension strictly greater than one (for example, one can take some Julia set appearing in the following list) and let $\Omega$ be the complement of $K$. The fact that every $f \in H(\Omega)$ continuous and bounded in $\mathbb{C}$ is constant is equivalent to saying that the continuous analytic capacity of $K$ is zero (see my answer to this question). However, as remarked by Emil Jeřábek in my answer, every compact set $K$ with zero continuous analytic capacity must have Hausdorff dimension at most $1$.

EDIT : As remarked in the very nice comment by user85506, since the Hausdorff dimension of $K$ is strictly greater than one, $K$ supports a nontrivial measure with growth $\mu(B(z,r)) \leq C r^{1+\epsilon}$, by Frostman's Lemma. The growth of $\mu$ implies that the Cauchy Transform $$f(z):=\int_{K} \frac{1}{z-\zeta}d\mu(\zeta)$$ is continuous on $\mathbb{C}$ and holomorphic on $\mathbb{C} \setminus K$. Furthermore, $f(z)\rightarrow 0$ as $z \rightarrow \infty$ and so in particular, $f$ is bounded. However, $f$ is non-constant because $zf(z) \rightarrow \mu(K) \neq 0$.

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Malik Younsi
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    To make the answer more self-contained (without introducing continuous analytic capacity), I'll add: by Frostman's lemma $K$ supports a nonzero measure with the growth bound $\mu(B_r)\le Cr^{1+\epsilon}$. Thanks to this bound, the integral $\int (z-\zeta)^{-1},d\mu(\zeta)$ defines a Hölder continuous holomorphic function on $\mathbb C\setminus K$ (as usual, split the integral into "nearby" and "far-away" parts to estimate the modulus of continuity). The function extends to $\mathbb C$ by virtue of uniform continuity. – 40 votes Jul 08 '13 at 21:17
  • @user85506 : Ah indeed, very nice! Thank you. I should have thought about the Cauchy Transform :-) – Malik Younsi Jul 08 '13 at 21:20
  • @user85506 : I will edit my answer according to your comment. – Malik Younsi Jul 08 '13 at 21:21
  • Thanks for your answer. It looks to be a nice way to answer my question but I am not familiar with this part, do you have some suggestions about books where I can learn tools you used here? – Danqing He Jul 10 '13 at 01:02
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    Just looking at the special case with $K$ of positive Lebesgue measure, you can take $\mu$ to be absolutely continuous with bounded density, and you can see that the Cauchy transform has the required properties. No need to know Frostman's Lemma for that case. – George Lowther Jul 10 '13 at 01:27
  • @DanqingHe : I would recommend the very nice book "Vitushkin's conjecture for removable sets" by James Dudziak. It contains a lot of information on analytic capacity. Moreover, if you want to know more about user85506's argument and Frostman's Lemma, you can take a look at section 2.3. Hope this helps! – Malik Younsi Jul 10 '13 at 01:42
  • @MalikYounsi Thank you. I will look at that book. – Danqing He Jul 10 '13 at 15:11
  • @MalikYounsi : Thanks for the great answer. – bryanj Jul 11 '13 at 11:22
  • @bryanj : You're welcome, thank you for the bounty. It is a very nice question. – Malik Younsi Jul 11 '13 at 12:20
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This answer is incorrect.

You can consider the function $g(x):=\sum_{k=0}^{\infty}\frac{x^{2^k}}{2^{k^2}}$. This function, and many examples like it, is going to be analytic on the unit disc, and continuous and even differentiable infinitely many times on the unit circle (but not analytic there!). We can have a biholomorphism between the unit disc and the plane minus the non-negative real line. Composing $g$ with this biholomorphism we get a function analytic in the whole plane minus the non-negative real line, and continuous on the non-negative real line.

OR.
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  • Well, the values it is taking are the same values $g$ was taking on the closed unit disc, and $g$ was continuous on the closed unit disc. Being the closed disc compact, the values $g$ takes on it is a compact set, and in particular bounded. I am being a bit lazy. To really get continuity on the boundary after composing one should pick the biholomorphism carefully. But try it. One can write down one explicitly. – OR. Jul 04 '13 at 17:03
  • Thank you for your example. But I am afraid the function $g\circ f$, where the $f$ is your biholomorphism, is not continuous on $\mathbb{C}$, since for a point $x>0$, we have two sequences ${x+i\tfrac1n}$ and ${x-i\tfrac1n}$ goes to $x$ with maybe two difference limits. – Danqing He Jul 04 '13 at 17:10
  • I said above to choose carefully. Should I do all the work? – OR. Jul 04 '13 at 17:30
  • No, thanks. I understand it now since $g$ is an even function. There is one more question. Under assumptions of my main question, can we show that $f$ is bounded by the largest value of $f$ on $\partial\Omega$? – Danqing He Jul 04 '13 at 17:59
  • @DanqingHe: Indeed you can. This is just an application of the Maximum Modulus Principle. – Cameron Buie Jul 04 '13 at 18:04
  • Perhaps it is easier to arrange continuity by mapping to $\mathbb{C}\setminus[-2,2]$, instead of the complement of the non-negative numbers, and considering the map $w=z^2+1/z^2$. – OR. Jul 04 '13 at 18:04
  • But when we are applying the Maximum Modulus Principle, we need the region to be bounded (or more essentially, $f$ attains its maximum at some point in $\Omega$), so is this a direct application or do we need more arguments to finish it? – Danqing He Jul 04 '13 at 18:20
  • Points interior to $\Omega$ correspond to points in the open unit disc. So just apply the principle to $g$ and in the unit disc. – OR. Jul 04 '13 at 18:37
  • I know that in your example it of course is bounded just by the definition of $g$. But with only assumptions of my question, $\Omega$ is arbitrary and we may not be able to map it onto a unit disk, so we cannot apply the principle in this situation directly. – Danqing He Jul 04 '13 at 18:45
  • Assume $a\in\Omega$ is a maximum of $|f|$, even a local maximum. Apply the MMP to the function in a small disc around $a$ and contained insider $\Omega$. MMP then says that $f$ is constant in that disc, and since $\Omega$ is connected, it will be constant in all of $\Omega$. Since $f$ is not constant this is a contradiction. $|f|$ doesn't attain maximum in $\Omega$. – OR. Jul 04 '13 at 18:53
  • @DanqingHe: There is no need for $\Omega$ to be bounded in this case, since $f$ is bounded. Since $f$ is continuous, then the image of $|f|$ will be a connected subset of $\Bbb R,$ and since $f$ is bounded, then in particular the image of $|f|$ will be a bounded interval of non-negative reals. MMP tells us that the image of $\Omega$ under $|f|$ will not include the supremum of this interval. Thus, if $|f|$ takes on a maximum value on $\delta\Omega,$ then said maximum value is in fact global. – Cameron Buie Jul 04 '13 at 18:56
  • @CameronBuie I agree that $|f|$ cannot take the supremum $M$ of the interval, but how can we show that $|f|{\partial\Omega}$ is $M$, where $M$ actually is also $|f|{\Omega}$. – Danqing He Jul 04 '13 at 19:14
  • @DanqingHe: It should be clear (if I'm interpreting your notation correctly) that putting $M=\lVert f\rVert_{\Bbb C},$ we have $\lVert f\rVert_{\partial\Omega}, \lVert f\rVert_{\Omega}\le M.$ Let $I$ be the image of $\Omega$ under $\lvert f\rvert,$ so that $\lVert f\rVert_{\Omega}=\sup I.$ By continuity of $\lvert f\rvert$ and the fact that $\Omega$ is dense in $\Bbb C,$ we have that the image of $\lvert f$ is contained in the closure of $I$, so $M\le\sup I,$ and so $\lVert f\rVert_{\Omega}=M.$ ... – Cameron Buie Jul 04 '13 at 19:54
  • However, I don't know if we can conclude in general that $\lVert f\rVert_{\partial\Omega}=M$ unless we know for some reason that $\lvert f\rvert$ attains some maximum value on $\partial\Omega$ (such as if $\partial\Omega$ is compact). I'll think on it. – Cameron Buie Jul 04 '13 at 19:58
  • Thank you. This actually is problem 11 of chapter 12 of Rudin's Real and complex analysis. – Danqing He Jul 04 '13 at 20:25
  • @DanqingHe : Let $B$ be the biholomorphism. Is it not true that if $g \circ B$ composed is continuous on the whole plane and analytic off of the non-negative real axis, then integrals of $g \circ B$ over rectangles are all zero? In this case does not Morera's theorem say that $g \circ B$ is analytic, and from this you see that it must be constant using Liouville's Theorem? – bryanj Jul 07 '13 at 15:47
  • That seems to be the case. Will think about it later. – OR. Jul 07 '13 at 16:13
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    @bryanj Yep. By Morera's theorem, a continuous function on all of $\mathbb{C}$ that is holomorphic on $\mathbb{C}\setminus {z \colon z \leqslant 0}$ is indeed entire. – Daniel Fischer Jul 08 '13 at 00:58
  • @DanielFischer Does that imply that the proposed biholomorphism in this answer does not exist? – bryanj Jul 08 '13 at 01:14
  • @bryanj No, the biholomorphism exists. But the function you get on $\mathbb{C}\setminus{z\colon z \leqslant 0}$ via that can't be continuously extended to the negative real line. – Daniel Fischer Jul 08 '13 at 01:19
  • @DanielFischer : Sorry - I was unclear - I meant what you just wrote. Since the resulting composite function on the slit plane is not continuous on the entire plane, does this not invalidate this answer's attempt to construct a counterexample? – bryanj Jul 08 '13 at 01:24
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    @bryanj Yes, a counterexample would need a more complicated $\partial \Omega$. A finite union of line segments or analytic arcs that intersect transversally falls down due to Morera. – Daniel Fischer Jul 08 '13 at 01:30