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 $$ c\leq \frac{m(B_1(0)\cap\{u(x)>0\} )}{m(B_1(0)\cap\{u(x)<0\} )}\leq c^{-1} $$
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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
Alexandre Eremenko
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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? – Sam and Jim Apr 27 '23 at 04:16
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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. – Alexandre Eremenko Apr 27 '23 at 10:56
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Then the positive part is not simply connected in this example, which does not satisfy the condition. – Sam and Jim Apr 27 '23 at 16:03
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I repeat that the set ${ z:u(z)>0}$ is ALWAYS simply connected. – Alexandre Eremenko Apr 27 '23 at 18:12
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Oh, you are right and I am afraid I forgot to assume ${z: u(z)>0}$ is also a region. – Sam and Jim Apr 28 '23 at 00:27
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@Sam and Jim: do you also want ${ z:u(z)<0}$ to be connected? – Alexandre Eremenko Apr 28 '23 at 12:05
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Yes! I have been wondering if there is any proof or counterexample of this case ( the positive part and negative part are both connected) – Sam and Jim May 01 '23 at 14:52
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The conjecture is that if both { z:u(z)>0} and { z: u(z)<0} are connected then your assertion about areas must be true. But if only one of them is connected, probably not. Both statements look plausible but difficult to prove. – Alexandre Eremenko May 01 '23 at 16:22