Analytic Functions on Upper Half Plane Satisfying Inequality 
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
$$\left|f(z)\right|\leq\Im(z) \tag{1}$$

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.

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).

An answer to this problem has been posted below.

Your solution is fine. (Answered to keep the question off the unanswered list.)
