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I'm stuck on a problem I saw in a textbook:

Suppose $\phi$ is Lipschitz, non-negative and $c$ some positive constant. Consider the Cauchy problem on a ball $B$

\begin{align} -\Delta f = c \, \phi, \quad & \text{in } B \\ f = 0, \quad & \text{on } \partial B. \end{align}

Also for $\phi$ there holds $$-\Delta \phi - c \, \phi \leq 0.$$

(Hence $-\Delta(\phi - f) \leq 0$.)

Then there holds $$\sup_{B_{1/2}} \, (\phi - f) \leq \frac 1{|B|} \int_B \phi - f \, \mathrm d x.$$

How do I arrive at the final bound?

Thanks for your help!

cesare borgia
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    I do think I have a solution; one can approximate $\phi$ in $W^{1,p}$ for any $p<\infty$ by $C^\infty \supset \varphi_n = \phi * \rho_{1/n}$ with a standard mollifier $\rho_{1/n}(x) = \rho(x/n)$. Those are always subharmonic, too. Then one can use the property for subharmonic functions, that they're always less than their average, and finally pass to the limit. – cesare borgia Feb 26 '18 at 18:05

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