0
$\begingroup$

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 $u_x, u_y, v_x, v_y$

$\endgroup$
  • $\begingroup$ Calculate $\frac{\partial g}{\partial z}$. Use chain rule. $\endgroup$ – mlainz Oct 26 '15 at 2:44
  • $\begingroup$ What makes you think that the partial derivatives are both identically zero? $\endgroup$ – Giovanni Oct 26 '15 at 2:45
  • $\begingroup$ BTW, if it is identically zero, then it is holomorphic. $\endgroup$ – user99914 Oct 26 '15 at 2:46
  • $\begingroup$ Actually, it was a Multiple Choice question. The option identically zero is incorrect. $\endgroup$ – Rising Star Oct 26 '15 at 2:49
  • $\begingroup$ I mean derivative becomes zero if its a real valued function of a complex variable. $\endgroup$ – Rising Star Oct 26 '15 at 2:52
1
$\begingroup$

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".)

$\endgroup$
  • $\begingroup$ Ya, existence of $ \dfrac {\partial ^2 f} {\partial ^2 x} $ is guaranteed but what abt continuity ? $\endgroup$ – Rising Star Oct 27 '15 at 13:50
  • 1
    $\begingroup$ No, $C^2$ means that all second-order partial derivatives exist and are continuous. $\endgroup$ – mrf Oct 27 '15 at 14:27
  • $\begingroup$ @mrf Sir Harmonic function need not satify Cauchy Reimann equation , So how can we talk about holomorphicity ? $\endgroup$ – MathLover Sep 21 '18 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.