I have a function $ u(x, y) = \text{Re}(f(z)) $ that is bounded, harmonic on the UHP and continuous on the real axis. How can I show f is bounded?

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As stated above, I have a "bounded C-harmonic function in the upper half-plane." (C-harmonic meaning harmonic and continuous on the boundary). I would like to show that if $ u = Re(f) $ then $ f $ is bounded as well. Any ideas?

Edit: I need $ f $ to be bounded in order to solve this contour integral problem. The full problem can be seen here.

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The statement is false: $f(z)=i\log (i+z)$ (principal branch of $\log$) is a holomorphic unbounded function in the upper half-plane, but its real part is $\arg (i+z)$ which is strictly between $0$ and $\pi$ (and is continuous on $\mathbb R$).

The idea of counterexample was to map the upper halfplane onto an unbounded domain that has bounded projection onto the $x$-axis.