Let $u$ be a harmonic function defined on $B_1(0)\subset\mathbb{R}^2$, $u(0)=0$, and $\{x\in B_1(0):u(x)>0\}$ is simply connected. Is there a universal constant $c>0$ satisfying that $$ c\leq \frac{m(B_1(0)\cap\{u(x)>0\} )}{m(B_1(0)\cap\{u(x)<0\} )}\leq c^{-1} $$
1 Answer
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The answer is negative. To construct a counterexample, use an entire function $f$ which satisfies $|f(z)|<1$ in ${\mathbf{C}}\backslash D$, where $D$ is a half-strip $\{ x+iy:x>0,|y|<\pi \}$. Then $v={\mathrm{Re}}f-1$ is harmonic, and negative in ${\mathbf{C}}\backslash D$. Let $D_1=\{ z: u(z)>0\}\subset D$, and $z_1\in\partial D_1$. Then $v_1(z)=v(z-1)$ satisfies $v_1(0)=0$ and for sufficiently large $R$, $u(z)=v_1(Rz)$ will be positive on a subset of the unit disk of arbitrarily small area.
A function $f$ with required property can be constructed as a Cauchy integral $$f(z)=\int_{\partial D}\frac{e^{e^\zeta}d\zeta}{\zeta-z},$$ for details about this construction, see, for example W. Hayman, Meromorphic functions, Chap. 4, 4.1. See also Entire function bounded at every line
answered Apr 26, 2023 at 12:51
Alexandre Eremenko
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$\begingroup$ But I'm wondering if the positive part $\{x\in B_1(0): u(x)>0\}$ of such a harmonic function $u(z)$ is always simply connected? $\endgroup$Re_0– Re_02023-04-27 04:16:21 +00:00Commented Apr 27, 2023 at 4:16
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$\begingroup$ Each component of the set $\{ z:u(z)>0\}$ for any harmonic function is simply connected; this followd from the Maximum/Minimum Principle. But the number of components can be large. $\endgroup$Alexandre Eremenko– Alexandre Eremenko2023-04-27 10:56:36 +00:00Commented Apr 27, 2023 at 10:56
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$\begingroup$ Then the positive part is not simply connected in this example, which does not satisfy the condition. $\endgroup$Re_0– Re_02023-04-27 16:03:50 +00:00Commented Apr 27, 2023 at 16:03
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$\begingroup$ I repeat that the set $\{ z:u(z)>0\}$ is ALWAYS simply connected. $\endgroup$Alexandre Eremenko– Alexandre Eremenko2023-04-27 18:12:11 +00:00Commented Apr 27, 2023 at 18:12
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$\begingroup$ Oh, you are right and I am afraid I forgot to assume $\{z: u(z)>0\}$ is also a region. $\endgroup$Re_0– Re_02023-04-28 00:27:57 +00:00Commented Apr 28, 2023 at 0:27