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I have the following problem: Let $K$ be the equilateral triangle centered at 0. Assume that $f$ is continuous on $K$ and analytic inside of $K$, and assume that $|f(z)|\leq 8$ on one of the sides and $|f(z)|\leq 1$ on two other sides. Show that $|f(0)|\leq 2$.

For a solution, I have that when you define maps $f_1(z)=f(e^{\frac{2\pi i}{3}}z)$ and $f_2(z)=f(e^{\frac{4\pi i}{3}}z)$, that the center stays in place yet the sides rotate. Using the maximum modulus principle, you can get that $|f(0)f_1(0)f_2(0)|\leq 8$, which gives that $|f(0)|^3 \leq 8$ or just $|f(0)|\leq 2$. I'm just a little confused on how the maximum modulus principle can be implied and was wondering if someone could provide a detailed explanation.

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Let $g(z) = f(z)f_1(z)f_2(z)$. Then $g$ is continuous on $K$ and analytic inside. On $\partial K$, $|g(z)| \le 8 \cdot 1 \cdot 1$. (On each edge of the triangle, one of the factors is bounded by $8$ and the other two factors by $1$.)

The maximum principle shows that $|g(0)| \le 8$ and you finish off as you have already said yourself.

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But why don't the bounds move around? I don't get why $f(z)$ would only be less than 8 on one of triangles. –  Frank White Jan 10 '13 at 16:36
    
@FrankWhite: Draw a picture. The map $z \mapsto \exp(2\pi i/3)z$ is a rotation by $2\pi/3$, so it maps one side of the triangle to "the next". Similarly for the $4\pi/3$ one. –  mrf Jan 10 '13 at 16:38
    
Right, I understand that - but wouldn't it carry the bound and the point where the function has it's maximum with it? –  Frank White Jan 10 '13 at 16:40
    
@FrankWhite Yes. If $|f|$ is bounded by $8$ on the "first" side, then $|f_1|$ will be bounded by $8$ on the "second" side, and $|f_2|$ will be bounded by $8$ on the "third" side. In the same fashion, if $|f_1|$ is bounded by $1$ on the second side, then... –  mrf Jan 10 '13 at 16:42

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