# Why real valued harmonic functions are holomorphic.

Let $f$ be a real valued harmonic function on $C,$ then Claim $g= \frac {\partial f}{\partial x} - i \frac {\partial f}{\partial y}$ as holomorphic and $h= \frac {\partial f}{\partial x} + i \frac {\partial f}{\partial y}$ as need not be holomorphic.

My Attempt

To prove g as holomorphic, we need to satisfy two conditions

1. C.R Eqns : $u_x = \dfrac {\partial ^2 f} {\partial ^2 x} = v_y = -\dfrac {\partial ^2 f} {\partial ^2 y}$ ($\because$ f is homomorphic )

2. $u_x, u_y, v_x, v_y$ exists and continous.

Here existence is guaranteed BUT HOW TO VERIFY THE CONTINUITY OF $u_x, u_y, v_x, v_y$

• Calculate $\frac{\partial g}{\partial z}$. Use chain rule. – mlainz Oct 26 '15 at 2:44
• What makes you think that the partial derivatives are both identically zero? – Giovanni Oct 26 '15 at 2:45
• BTW, if it is identically zero, then it is holomorphic. – user99914 Oct 26 '15 at 2:46
• Actually, it was a Multiple Choice question. The option identically zero is incorrect. – Rising Star Oct 26 '15 at 2:49
• I mean derivative becomes zero if its a real valued function of a complex variable. – Rising Star Oct 26 '15 at 2:52

Since $f$ is harmonic, $f \in C^2$ by definition, so your function is automatically smooth enough to Cauchy-Riemann's equations (in the "other direction".)
• Ya, existence of $\dfrac {\partial ^2 f} {\partial ^2 x}$ is guaranteed but what abt continuity ? – Rising Star Oct 27 '15 at 13:50
• No, $C^2$ means that all second-order partial derivatives exist and are continuous. – mrf Oct 27 '15 at 14:27