Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 3 / d / Solution

At the origin, the Poisson kernel is one, so the mean value property for harmonic functions givesFor $0<r<1$, the denominator in the Poisson kernel satisfies $1-2r\cos \vartheta +r^2\ge (1-r{)}^2$. ThusThe nonnegativity of $f$ allows this upper bound to be inserted under the Poisson integral on the unit disk:This is the upper bound in the Harnack inequality on the unit disk. The same argument with the minimum of the Poisson kernel gives the corresponding lower bound $(1-r)u(0)/(1+r)$.