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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

$$\left|f(z)\right|\leq\Im(z) \tag{1}$$

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.

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    $\begingroup$ It is that simple. $\endgroup$ – Daniel Fischer Jul 6 '15 at 19:03
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    $\begingroup$ Ok, that's too easy, now find all the real-valued harmonic functions in the upper half-plane satisfying the same inequality... $\endgroup$ – David C. Ullrich Jul 6 '15 at 19:51
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Your solution is fine. (Answered to keep the question off the unanswered list.)

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Any harmonic function $u$ in the upper half space $\mathbb{H}^{+}\simeq\mathbb{R}_{+}^{2}$ which extends to a continuous function on $\overline{H^{+}}$, vanishing on the real line, extends to a harmonic function on the whole plane by Schwarz reflection. If such $u$ grows at most polynomially in bounded, then it defines a tempered distribution in $\mathcal{S}'(\mathbb{R}^{2})$.

We make use of the following two results. A proof of the first one can be found in [L. Grafakos, Classical Fourier Analysis, Third Edition, Proposition 2.4.1], and the second is elementary.

Theorem. Let $u\in\mathcal{S}'(\mathbb{R}^{n})$ be a tempered distribution. If $u$ is supported in the set $\left\{x\right\}$, then \begin{align*} u=\sum_{\left|\alpha\right|\leq k}c_{\alpha}\partial^{\alpha}\delta_{x} \end{align*} for some nonnegative integer $k$, where $c_{\alpha}\in\mathbb{C}$.

and under the Fourier transform convention

$$\widehat{\phi}(\xi)=\int_{\mathbb{R}^{n}}\phi(x)e^{-2\pi i x\cdot \xi}\mathrm{d}x, \qquad\forall \xi\in\mathbb{R}^{n}$$

Lemma. If $u=(-2\pi i x)^{\alpha}$, then $\widehat{u}=(-1)^{\left|\alpha\right|}\partial^{\alpha}\delta$.

Taking inverse Fourier transforms, we conclude that the harmonic function $u$ must be of the form \begin{align*} u(x,y)=\sum_{\left|\alpha\right|\leq k}c_{\alpha}(2\pi i xy)^{\alpha} \end{align*} for some integer $k$. The condition that $\left|u(x,y)\right|\leq\left|y\right|$ forces $c_{\alpha}=0\Leftrightarrow \alpha=(0,1)$ and for such $\alpha$, $\left|c_{\alpha}(2\pi i )\right|=1$. We conclude that $u(x,y)=e^{i\theta}y$, for some fixed $\theta\in [0,2\pi)$. If we require $u$ to be real-valued, then $u(x,y)=y$.

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