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So I have the following question:

Let f be analytic in an open set which contains the closed unit disc $\overline{\mathbb{D}},$ and assume $M:=\max\{\textrm{Re}(f):|z|=1\}\geq0.$ Prove that for $z\in\mathbb{D},$ $$|f(z)|\leq \frac{1+|z|}{1-|z|}[M+|f(0)|].$$

So I am very inclined to believe this is an application of Schwarz's Lemma on some composition of conformal maps. However I don't feel like I have information regarding $f(\mathbb{D}).$ I wanted to shift $f$'s image into either the right or left half-plane, and Reimann's mapping Theorem guarantees such a conformal map, but I won't know what it looks like.

Also I know I draw conclusion about the range of conformal maps based on their behavior on the boundaries, but I only know that $f$ is analytic. So I can't be sure of what $f(\mathbb{D})$ looks like.

I tried looking at another post here, but they have stronger suppositions than I do, or at least so it seems to me.

As always, I appreciate any help or hints. Thanks.

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    $\begingroup$ Well, if Re$(f)\le M$ on $|z|=1$ then it holds for $z\in\mathbb{D}$ by Maximum modulus theorem for harmonic functions. Therefore, $g(z)=M-f(z)$ would be such a function you are looking for (with the image in the right half plane). $\endgroup$
    – A.Γ.
    Jul 12, 2015 at 2:33
  • $\begingroup$ Thanks so much that is exactly what I was missing. I knew there had to be some justification for that, but I kept looking in the wrong places. I really appreciate it. If you want to rewrite this as an answer I will give you credit. $\endgroup$
    – Scott
    Jul 12, 2015 at 4:32
  • $\begingroup$ Glad that it helped. It's enough for me :-) $\endgroup$
    – A.Γ.
    Jul 12, 2015 at 10:53

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