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Consider an equilateral triangle T. Suppose that f is analytic inside of T and satisfies that |f(z)| ≤ 8 on one side of the triangle, while |f(z)| ≤ 1 on the other two sides. Prove that |f(c)| ≤ 2, where c is the center of T.

I know that all of the angles are π/3, or 60 degrees. I know that I have to find the relation between 2 and 8, but I don't know how or where to start. Apparently this can be done in one line, but I am completely stumped.

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WLOG, we may consider $T$ to be the origin. Then consider $g(z) = f(z) f(\omega z) f(\omega^2 z)$, where $\omega$ is a complex cube root of $1$. Note than that $g$ when evaluated on any side of the triangle, it involves the product of values of $f$ from all three sides.

So $|g(z)| \le 8$ on the sides of the triangle, and analytic inside. Then use the maximum modulus principle, to argue $|f(0)|^3 = |g(0)| \le 8$.

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  • $\begingroup$ So then how does that relate back to |f(c)|≤2? $\endgroup$ – Becky Nov 4 '15 at 14:16
  • $\begingroup$ We have translated the centre to origin, so it's the same as $|f(0)|\le 2$. $\endgroup$ – Macavity Nov 4 '15 at 14:18

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