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> **Original Problem.** Let $\mathbb{H}=\left\{z\in\mathbb{C} : \Im(z)>0\right\}$ be the open upper half plane. Determine all analytic functions $f:\mathbb{H}\rightarrow\mathbb{C}$ that satisfy the inequality
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> $$\left|f(z)\right|\leq\Im(z) \tag{1}$$
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This is an old qual problem which I at first found deceptively simple and asked for a fresh set of eyes. The solution is straightforward. The inequality implies $f$ has a continuous extension $\tilde{f}:\overline{\mathbb{H}}\rightarrow\mathbb{C}$ to the closed upper half plane by setting $\tilde{f}(x):=0$ for all $x\in\mathbb{R}$. By Schwarz reflection, we obtain an entire function $\tilde{f}:\mathbb{C}\rightarrow\mathbb{C}$ which vanishes on the real axis. By the identity theorem, $\tilde{f}\equiv 0$, implying $f\equiv 0$.

I tried coming up with another solution, for example using maximum modulus principle, but didn't get anywhere. So if anyone has a different proof, I would be interested in seeing it.

At the suggestion of David C. Ullrich, we consider the problem of determining harmonic functions in the upper half plane satisfying the same inequality.

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> **Problem.** With $\mathbb{H}$ as above, determine all harmonic functions $f:\mathbb{H}\rightarrow\mathbb{C}$ (i.e. $\Delta f=0$) that satisfy the inequality in (1).
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An answer to this problem has been posted below.