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How can I show:

Let $Ω$ be a domain and assume that $f ∈ C(\bar \Omega)$ restricts to a holomorphic function on $Ω$.

Prove: The real and imaginary part of $f$ assume their maxima and minima on the boundary of $Ω$.

Maybe I could show that the real and imaginary part are again holomorphic and then I apply the maximum principle on the real and imaginary part?! Is it actually true that those functions are holomorphic?

Any hints would be great!

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    Real-valued functions are never holomorphic unless they're constant. But the real and imaginary parts of a holomorphic function *are* harmonic. – 2017-01-17
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    Your second sentence makes no sense. What restriction? As stated, the result is false. Consider $x+iy$ in the upper half plane; the function $y$ is $0$ on the boundary. – 2017-01-17
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    I edited a bar over the $\Omega$. Now it should make sense. Anyway, I know that the real and imaginary part is harmonic and therefore they satisfy the maximum principle. Thank You! – 2017-01-17

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Decompose $f(z)$ into its real and imaginary parts:

$$f(z) = f(x,y) \equiv u(x,y) + iv(x,y)$$

Apply the Cauchy-Riemann conditions:

$$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

From these, it easily follows that:

$$\nabla^2u=0\\\nabla^2v=0$$

This implies that both $u$ and $v$ are solutions to Laplace's equation, which implies that they have no local maxima or minima (their maxima/minima must lie on the boundary of the region).