How to show that $u$ and $v$ have continuous partial derivatives at $(x_0,y_0)?$

Question

How to show that $f:D(\subset\mathbb C)\to\mathbb C:(x,y)\mapsto u(x,y)+iv(x,y)$ is differentiable at $z_0=(x_0,y_0)\implies u$ and $v$ have continuous partial derivatives at $(x_0,y_0)?$

Added: I can show $u$ and $v$ have partial derivatives at $(x_0,y_0).$ Please help me to show they are continuous there.

@Kasper Stop this copy paste in every question and using free downvotes on questions. — ABC, Apr 26 '13 at 14:21
@exploringnet Is this an order ? How to treat questions that have nothing besides a problem statement are discussed in this topic: http://meta.math.stackexchange.com/questions/9201/proposal-discourage-questions-that-are-nothing-besides-a-problem-statement I don't think it is a bad thing, that the OP improved his question a bit — Kasper, Apr 26 '13 at 14:26
@Kasper Discourage does not means that you write a pet statement and surf around to paste it in questions. You must ask them for their problem , work or where they got stuck before giving them a big statement. — ABC, Apr 26 '13 at 14:28
@exploringnet Please read that topic and express your concerns about this policy there. I didn't invent this template, and I'm not sure if I would like questions like these to be closed. The general opinion on meta DOES think questions like this (before it was edited) should be closed, if you disagree, feel free to share your opinion. — Kasper, Apr 26 '13 at 14:30
On $\mathbb{R}$ already, the discontinuity set of a derivative can be highly nontrivial. See here. So this result is false as stated. Just take a real function $g$ with nontrivial discontinuity set and consider $f(x,y):=g(x)$. The partial derivative with respect to $x$ will have discontinuity points. You need stronger assumptions on $f$. — Julien, Apr 26 '13 at 14:32

score 1 · Accepted Answer · answered Apr 26 '13 at 14:36

1

The result is false in general. What is true is that $u$ and $v$ are differentiable, but the partial derivatives may be discontinuous.

Counterexample: $$ f(z)=z^2\sin\frac{1}{|z|}. $$

answered Apr 26 '13 at 14:36

Julián Aguirre

76,354

Is the result true if $f$ is analytic at $z_0?$ – Sriti Mallick Apr 26 '13 at 14:37
Yes. In my counterexample $f$ is differentiable only at $z=0$. – Julián Aguirre Apr 26 '13 at 15:08
But in your example $f$ is not defined at $z=0!$ – Sriti Mallick Apr 26 '13 at 15:34
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I think it is clear I meant $f(0)=0$. – Julián Aguirre Apr 26 '13 at 16:33

How to show that $u$ and $v$ have continuous partial derivatives at $(x_0,y_0)?$

1 Answers1