Let $u(z)$ be a harmonic function on $\mathbb{C}$ such that $|u(z)| \leq \sqrt{|z|}$. Show $u$ is constant.
We can say $u(z) = \text{Re}f(z)$ for some entire $f$. Liouville would give $f$ is constant, and thus $u$ is constant. However, I have had a hard time obtaining a bound for $f$, or for related function.
Another route I have thought through is to use Picard's Little theorem, but again I have trouble finding points $f$ will not take.
Any tips appreciated, thank you.
I will make use of some standard inequalities related to Maximum Modulus Principle. [It is perhaps possible to simplify this proof]. Let $f$ be continuous on $\{z:\left\vert z\right\vert \leq R\}$ and holomorphic on $B(0,R).$ Let $M(r)=\sup \{\left\vert f(z)\right\vert :\left\vert z\right\vert =r\}$ and $\phi (r)=\sup \{\Re f(z):\left\vert z\right\vert =r\}$ for $0\leq r\leq R$. Then $$M(r)\leq \frac{R-r}{R+r}% \Re f(0)+\frac{2r}{R+r}\phi (r)$$ and $$M(r)\leq \frac{R-r}{R+r}% \left\vert f(0)\right\vert +\frac{2r}{R+r}\phi (r)$$ for $0\leq r\leq R.$
Corollary: if $f$ is an entire function such that $\Re f(z)\leq B\left\vert z\right\vert ^{n}$ for $\left\vert z\right\vert \geq R$ then $f$ is a polynomial of degree at most $n.$
We have $\phi (r)\leq Br^{n}$ for $r\geq R$. Hence $M(r)\leq \frac{2r-r}{2r+r}\left\vert f(0)\right\vert +\frac{2r}{2r+r}\phi (r)\leq \frac{1}{3}\left\vert f(0)\right\vert +\frac{2}{3}Br^{n}$ and $\left\vert f(z)\right\vert \leq \frac{1}{3}\left\vert f(0)\right\vert +\frac{2}{3}B\left\vert z\right\vert ^{n}$ if $\left\vert z\right\vert \geq R.$ This implies that $f$ is a polynomial of degree at most $n$.
Now to answer the present question let $f$ be an entire function with real part $u$. Consider $f^{2}$. This is an entire function and its real part does not exceed $u^{2}(z)\leq |z|$. Hence $f^{2}$ is polynomial of degree at most $1$. From this it trivial to complete the proof.