Suppose $u:B(0,1)\rightarrow \mathbb{R}$ is a nonnegative harmonic function such that $u(0,0)=1$. What is the best estimation I can get for $u(\frac{1}{2},0)$?
Since \begin{align} u(0)&=\frac{1}{\pi}\int_{B(0,1)}u(x,y)dxdy \\ &\geq\frac{1}{\pi}\int_{B(\frac{1}{2},\frac{1}{2})}u(x,y)dxdy \\ &=\frac{1}{4}u(\frac{1}{2}), \end{align} I can get an estimation that $0\leq u(\frac{1}{2})\leq 4$. But it seems far from the best. To get the equality from above, we need that $u(x,y)=0$ for $(x,y)\in B(0,1)\setminus B(\frac{1}{2},\frac{1}{2})$, which seems impossible to be true.
Any advice will be helpful!
If $u$ is harmonic and non-negative in the unit disk then Harnack's inequality gives $$ \frac{1-r}{1+r} u(0) \le u(z) \le \frac{1+r}{1-r} u(0) $$ for $|z| = r$. If $u(0) = 1$ then $$\frac 13 \le u\left(\frac 12\right) \le 3 \, .$$
These bounds are sharp, as the functions $$ u_1(z) = \operatorname{Re} \left( \frac{1- z}{1+ z}\right) \, , \, u_2(z) = \operatorname{Re} \left( \frac{1+ z}{1- z}\right) $$ show.