Complex: If $|f|<\varepsilon$, then $\frac{1}{\pi}\int_E \frac{|f|}{|z-w|}<\varepsilon$?

Question

Is this true?

Let $f:E\xrightarrow{}\mathbb{C}$ be holomorphic on the interior of a compact set $E$, and let $\varepsilon>0.$

If $|f|<\varepsilon$ on $E$, then $$\frac{1}{\pi}\int_E \frac{|f(w)|}{|z-w|} dm(w)<\varepsilon$$ for $z\in E$, where $m$ is the two-dimensional Lebesgue measure.

Important edit: $E$ is COMPACT.

Answer 1

Less than $\epsilon$ looks very unreasonable. If $E$ is the closed unit disk, $w=0$ and $f(x)=\epsilon /2$ for all $x$ then the left side is $\pi \epsilon $ which is greater than $\epsilon$.