Give f is entire, I have to show if $$\lim_{z\to\infty}\frac{\text{Re }f(z)}{z}=0$$, then $f$ is bounded.
I've proved if $\lim_{z\to\infty}\frac{f(z)}z=0 $ then $f$ is constant by constructing new function $g(z)$. But I really don't know how to define in my case or how to prove it. I've googled it and asked help from many one, but like me they also failed. Please provide a proper explanation in answer section rather comment. It would be a great help. Thanks in advance.
Hint: Write $f(z)=\sum_{n=0}^\infty a_nz^n.$ Let $u$ be the real part of $f.$ Then
$$\tag 1 u(z) = \frac{f(z)+\overline {f(z)}}{2} = \frac{1}{2}\left (a_0 + \overline {a_0} + \sum_{n=1}^{\infty}a_nz^n+\sum_{n=1}^{\infty}\overline {a_n}\cdot\overline {z}^n\right ).$$
Setting $z=re^{it}$ in $(1)$ gives a nice Fourier series for $u.$ Apply Parseval to this, using the orthogonality of the exponentials and $|u(re^{it})|= o(r)$ as $r\to \infty.$