Find the entire function $f$.

Question

Suppose that $f:\mathbb{C}\to\mathbb{C}$ is entire and that $f(x,y)=u(x,y)+iv(x,y)$. If $u^2-v^2\geq x^2-y^2$ for all $z=x+iy$, what information can we have about $f$?

It seems Liouville's theorem must apply somewhere here. Now the inequality above gives us that the real part of $f(z)^2$ is greater than the real part of $z^2$ for all $z\in\mathbb{C}$. But how to proceed from here? Hints would be great.

Answer 1

As you observed, it requires that $$\Re [f^2(z)]\ge\Re [z^2]\implies\Re[f^2(z)-z^2]\ge 0$$

Let $G(z)=f^2(z)-z^2$. Since its real part is lower bounded, we have $G(z)=C$ by the lemma below, where $\Re [C]\ge0$.

Thus, we obtain $f^2(z)=C+z^2$.

If $C\ne0$, $f=\sqrt{C+z^2}$ must be not entire. Thus $C=0$.

The only possibilities of $f(z)$ are: $f(z)=z$ and $f(z)=-z$.

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Lemma. If an entire function $f$ satisfies $\Re [f(z)]\ge M$ for some constant $M$ and all $z$, $f$ is a constant.

Proof. Consider $g=e^{-f(z)}$. Then $|g|=e^{-\Re [f(z)]}$.

Since $\Re [f(z)]\ge M$, $|g|=e^{-\Re [f(z)]}\le e^{-M}$. By Liouville's theorem, $g$ is a constant, and thus so is $f$. Q.E.D.