I'm struggling with the following question:
Suppose that $C$ is a positive constant, $u$ is harmonic in the upper half plane $\mathrm{Im}z>0$, and that $0 \le u(z) \le C\mathrm{Im}z$ for $\mathrm{Im}z>0$. Prove that there is a constant $c \in [0, C]$ such that $u(z)=c\mathrm{Im}z$ for $\mathrm{Im}z>0$. Hint: Consider the function $\tilde u$ defined by
$\tilde u(z) = \begin{cases} u(z), & \mathrm{Im}z>0 \\ 0, & \mathrm{Im}z=0 \\ -u(\bar{z}),& \mathrm{Im}z<0 \end{cases}$
Show first that $\tilde{u}$ has the mean value property in $\mathbb{C}$.
I already proved that $\tilde{u}$ has the mean value property, but I'm not sure what to do next. Since $\tilde{u}$ is harmonic in $\mathbb{C}$, I set an entire function $f$ such that $\mathrm{Re}f=\tilde{u}$, and I'm trying to deduce that $f$ is injective, so $f(z)=az+b$ for some constant $a, b$. I'm stuck in there and completely lost my way.
Does my approach correct? Or should I try another method?
Assume $f$ is entire, $\text { Re }f = u,$ and $|u(z)| \le A + B|z|.$ Then $f$ is linear polynomial.
Proof: Write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Letting $z= re^{it}$ and using $u = (f+ \overline f )/2,$ we get
$$u(re^{it}) = (1/2)[\sum_{n=0}^{\infty}a_nr^ne^{int}+\sum_{n=0}^{\infty}\overline {a_n} r^ne^{-int}].$$
Using the orthogonality of the exponentials, we then get
$$\int_0^{2\pi} |u(re^{it})|^2\, dt = 2\pi[(\text { Re }a_0 )^2 + \sum_{n=1}^{\infty}|a_n|^2r^{2n}].$$
Think now of $r\to \infty.$ The integral above is no more that $2\pi(A+Br)^2$ by our hypothesis on $u.$ That forces $a_n = 0, n>1,$ and implies $f$ is linear as desired.