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Let $f$ and $g$ be complex-analytic functions on the unit disk $D_1$, and suppose that $$ \sup_{|z|\leq 1} \big||f(z)|-|g(z)|\big| \leq \epsilon. $$ I am curious whether there exists some $\theta\in \mathbb{R}$ such that on the half-disk $$ \sup_{|z|\leq 1/2} |f(z)- e^{i\theta} g(z)| \leq C \epsilon. $$

I can show that this holds when $g=1$ is the constant function. In this case I can use the fact that $|f|+\epsilon -1$ is subharmonic and positive in order to apply Cacciopoli's inequality to get a bound on $\int_{D_{3/4}} \big|\nabla |f|\big|^2$. Then I can use the fact that $\big|\nabla |f|\big| = |f'|$ to show that on $D_{1/2}$, $$ |f'|\leq C\delta. $$ This shows that $f$ is close to a constant, as desired.

This argument does not work when $g$ is not constant because $|f|-|g|+\epsilon$ is not subharmonic. I am curious if there is another way to proceed.

One barrier that makes the argument difficult to find is the example $f(z)=z$ and $g(z)=\overline{z}$. In this case $g$ is not analytic so the hypotheses do not hold, but it does show that the desired bound will not follow from only using bounds such as the maximum principle on $f$ and $g$ in isolation (perhaps one can use that $fg$ is also analytic, for example).

Daniele Tampieri
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felipeh
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  • Hi. I have got two questions: Does the constant $C$ depend on the functions $f$ and $g$? How can I prove the equality $|\nabla |f||=|f'|$? – C94 Feb 22 '22 at 10:10
  • (1) I am curious whether there exists a constant $C$ that is independent of $f$ and $g$. In the case I do know, when $g=1$, such a constant exists. (2) Writing $f=u+iv$, one can use the Cauchy-Riemann equations to write $\nabla f$ in the form $\nabla f = Q(\partial_x u, \partial_y u)$ where $Q = (u^2+v^2)^{-1/2} ((u, -v), (v, u))$ is an orthogonal matrix. – felipeh Feb 22 '22 at 15:07

1 Answers1

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Some partial results; wlog we can assume $f,g$ analytic on the closed disc (so they have analytic extension slightly beyond) as we can just use $f(rz), r \to 1, r<1$ etc;

We claim that if $M\epsilon \le A \le |f|, |g| \le B, M \ge 2$ the result holds for any radius $r$ st $1-r \ge C, 8\epsilon \le AC$ (so for $r=1/2, M=16$) with a constant $K$ that depends only on $A,B,C$ so for example it holds for $r=1/2, g=1, \epsilon \le 1/16, K=16$ as in the OP and we will give a short proof below.

On the other hand if $f,g$ can have small absolute values (eg zeroes) then one has counterexamples for the unit disc but not for the disc of radius $1/2$ as requested in the OP of the type $f=z^k, g=z^{k+1}$ where $\max_{|z|=1}|f-\alpha g| =2$ but $\max_{|z| \le 1}||f|-|g||=c_k/(k+1), c_k \to 1/e$ (and of course one can slightly perturb them with small analytic functions to distribute the zeroes), so that case definitely bears further investigation

Assume $|g| \ge A \ge 2\epsilon$ (and also $1-r \ge C, 8\epsilon \le AC$ as above) and choose $|\alpha|=1,|\beta|=1, \alpha f(0), \beta g(0) > 0$ and replacing $f,g$ by $\alpha f, \beta g$ we can assume $f(0), g(0) >0$ and prove $\sup_{z\in D_{r}} |f(z)- g(z)| \le K\epsilon $ (hence in the original problem $\sup_{z\in D_{r}} |f-\bar \alpha \beta g|\le K\epsilon$) with $K$ depending on $A,B,C,r$

We note that $1/2<|f/g| \le 1+\frac{\epsilon}{A} \le 1+1/2$. Then $\log (f/g)$ is analytic in the unit disc where we choose the branch for which $\log f/g(0)$ is the usual one for positive numbers and $\Re \log (f/g)=\log |f/g| \le 2\epsilon/A$, while $\arg f/g(0)=0$ so $|\log f/g(0)|=\log |f/g(0)| \le 2\epsilon/A $

By Borel Caratheodory one has $\sup_{z\in D_{r}} |\log (f/g)(z)| \le \frac{4\epsilon }{A(1-r)} \le 4\epsilon/ AC \le 1/2$. But if $|w| \le 1/2$ one has $|e^w-1| \le e^{|w|}-1\le 2|w|$ so on $D_r$ one has $|f/g-1| \le 8\epsilon/ AC$ or $|f-g| \le K\epsilon$ with $K=8B/AC$

Conrad
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  • Wonderful analysis in the case that $A$ is bounded from below. I had not tried looking at $\log(f/g)$. I am not sure that the example $f=z^k$ and $g=z^{k+1}$ is a counterexample since I only want $f-e^{i\theta} g$ to be small on the half-disk (which I could have made more explicit in text, as is the $1/2$ is in a tiny font in the equation). In this case $\sup_{|z|<1/2} |f-g| \leq 2^{1-k}$, which is much smaller than $c/k$. – felipeh Feb 22 '22 at 15:09
  • yes that's a very good point though I still think that it's unlikely to get a good bound for $|f-\alpha g|$ even restricted to the $1/2$ disc when $f,g$ have zeroes (or are very small) but will think more on it – Conrad Feb 22 '22 at 16:51