I have this question from CVF, I'll add a screen capture since the whole problem is really long:
Where $\tilde{x}= x+ r\cos(\theta)$ and $\tilde{y}=y+r\sin(\theta)$
I don't know what this hint is implying, or what I'm supposed to do right now.
I have this question from CVF, I'll add a screen capture since the whole problem is really long:
Where $\tilde{x}= x+ r\cos(\theta)$ and $\tilde{y}=y+r\sin(\theta)$
I don't know what this hint is implying, or what I'm supposed to do right now.
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Assuming that $f$ is continuous on the closure of $D$ there is some point in the closure where $|f|$ has its maximum value. Suppose this point $x+iy$ is in the interior. Consider a closed disk, say of radius $R$, around this point contained in $D$. Use the last inequality in the hint to see that $|f|$ must be a constant in this disk: the integrand on the right does not exceed $M\equiv \max \{|f(z)|:z \in D\}$ so we get $M=|f(x+iy| \leq \frac 1 {\pi R^{2}} \int_{D_R(x+iy)} M dx dy$. This means equality must hold throughout.
We have proved that $|f|$ is a constant in the disk and this implies that $f$ is itself a constant. Connectedness of $D$ now implies that $f$ is a constant on the whole of $D$.