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Let $f,g$ be holomorphic on a bounded domain $D$ and continuous on $\overline{D}$. Assume that $\Re(f)=\Re(g)\forall z\in\partial D$. Prove that $\Re (f-g)=0$.

My attempt
Consider $h=f-g$. Clearly $h$ is holomorphic on $D$, continuous on $\overline{D}$.
If $h$ is constant, then we finish.
If $h$ is non-constant. According to Maximum modulus principle, for all $z\in\overline{D}$, we have $$|h(z)|\leq|h(z_0)|\quad (\text{ for some }z_0\in\partial D).$$ which also means that $$\sqrt{\Re^2(f(z)-g(z))+\Im^2(f(z)-g(z))}\leq\sqrt{\Im^2(f(z_0)-g(z_0))}$$ Hence we obtain the following inequality $$\Re^2(f(z)-g(z))+\Im^2(f(z)-g(z))\leq\Im^2(f(z_0)-g(z_0))^2\quad\forall z\in\overline{D}.$$ I'm stuck at here. Could someone help me to continue or another way to deal with the problem? Thanks in advance!

Alex Nguyen
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2 Answers2

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Since $h = f-g$ is holomorphic the real-part $\Re(h)$ is harmonic. So, it attains its maximum and minimum on the boundary. But $\Re(h) = 0$ on $\partial D$, so $\Re(f-g) = 0$ on the whole $D$.

psl2Z
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Regarding the real part of a holomorphic function $f$, it is usually helpful to consider $\exp(f)$. So let us consider a function $$h(z)=\exp(f(z)-g(z)) $$ on $\overline{D}$. Then $|h(z)|=1$ on $\partial D$ and hence the maximum modulus principle, applied to both $h$ and $1/h$, shows that $|h|=1$ everywhere on $D$. Thus $f$ and $g$ have the same real part.

MintChocolate
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