real part of analytic function

Question

Assume $f(x) \in D$ is analytic, where $x \in R$ and $D$ denotes the unit circle, which means $x$ is real and $f(x)$ is complex. Then what kind of function is the real part of $f(x)$? harmonic or analytic or...? Can Ref(x) be extended to a analytic function g(Z) on the strip |Im z|< c, for some constant c?

Answer 1

Clearly $Re(f)$ is not analytic if $f$ is nonzero. To show this you can just show that it cannot admit a derivative : for any $z \in D$, $h \in \mathbb{C}$, suppose f is analytic, then

$$Re(f)'(z) = \text{lim } \, \frac{Re(f(z + h) - f(z))}{h}$$

But then by setting $h1 = \epsilon$ and $h2 = i \epsilon$ and letting $\epsilon \rightarrow 0$, we get, since then numerator is real, that the limit must be both real and imaginary, thus zero.

However, it is indeed harmonic. Indeed, $f$ being holomorphic, it satisfies the Cauchy-Riemann equations, ie.

$$(\frac{\partial}{\partial y} - i\frac{\partial}{\partial x}) f = 0$$

Applying $(\frac{\partial}{\partial y} + i\frac{\partial}{\partial x})$ to the above, we get that $\Delta Re(f) + i \Delta Im(f) = 0$, so both the real imaginary parts are harmonic.