= Harnack inequality on the unit disk
{c}
{title2=$\frac{1-|z|}{1+|z|}u(0)\le u(z)\le\frac{1+|z|}{1-|z|}u(0)$}
For a nonnegative <harmonic function> on the <unit disc>, the displayed inequalities are sharp. Apply the <Poisson integral on the unit disk> on radius $R>|z|$, bound its positive kernel between $(R-|z|)/(R+|z|)$ and its reciprocal, and use the <mean value property for harmonic functions>. Letting $R\uparrow1$ proves the assertion. Positive <Poisson kernels> with one boundary pole give equality in the appropriate radial directions.
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