Let $f : \mathbb{C} \to \mathbb{C}$ be given by

$$ f(z) = f(x + iy) = \frac{xy(x + iy)}{x^2 + y^2}, ~~~~ (x, y) \neq (0, 0) $$ and $f(0) = 0$. It is easy to show that $f$ is continuous at 0 and satisfies the Cauchy-Riemann equations at 0. But, I know that the function is not differentiable at $z = 0$. How do I prove this?

I know that $\mathbb{C}$ can be identified with $\mathbb{R}^2$ and that a function from $\mathbb{R}^2$ to $\mathbb{R}$ is not differentiable at a point if its partial derivatives are not continuous there. However, my complex function $f$ can only be "transformed" to a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ (by using $f(x, y) = u(x, y) + i v(x, y)$). Does this mean $f$ has 4 partial derivatives (2 for $u$, 2 for $v$) and that I have to prove that one of them is not continuous at $z = 0$? Frankly, I have never even heard of partial derivatives of functions whose codomain is not $\mathbb{R}$.

Any ideas on how to prove that $f$ is not continuous at $z = 0$ are welcome.

  • $\begingroup$ Yes, it has four partial derivatives. $\endgroup$ – user98602 Mar 13 '16 at 15:36
  • $\begingroup$ It isn't true that if a function $\;f\;$ isn't differentiable at a point if its partial derivatives aren't continuous at that point: the other way around is true, and in fact it is enough to assume one of the two partial der's is continous (the other one has only to exist) to deduce differentiability $\endgroup$ – DonAntonio Mar 13 '16 at 15:50
  • $\begingroup$ Why don't you check CR outside zero? $\endgroup$ – Friedrich Philipp Mar 13 '16 at 15:52
  • $\begingroup$ @Friedrich: It's not true everywhwre else. Globally having CR means you are globally smooth. The point of this exercise is no doubt that the same is not true pointwise. $\endgroup$ – user98602 Mar 13 '16 at 16:13
  • $\begingroup$ @Joanpemo: wasn't it that if $f$ is differentiable at a point, its partials exists and are continuous at that point, but if the partials exists and are continuous at a point, $f$ is total differentiable at that point but not necessarily differentiable at that point. Thus, the partials being continuous at a point is a necessary criterion, but not sufficient. $\endgroup$ – limitIntegral314 Mar 13 '16 at 20:38

Note that $\lim_{x+iy\rightarrow 0}\frac{f(x+iy)-f(0)}{x+iy}=\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2}$, which does not exist. So, $f$ is not differentiable at 0.

  • 1
    $\begingroup$ This works for this exercise, but what if the $x + iy$ had not canceled out? What if, in the resulting limit, $i$ would still be present? Can I replace "$x + iy \to 0$" by "$(x, y) \to (0, 0)$" anyway? $\endgroup$ – limitIntegral314 Mar 13 '16 at 20:41
  • $\begingroup$ Yup in general there is nothing like calculating left hand or right handed limits to decide. First you need to guess, whether the limit will exist. If so, you need to prove only using the definition. However there has been a method to prove 'non existence of limit of functions of several variables' .... Just Google the above text in quotes... Finally note that $x+iy\rightarrow 0$ if and only if $(x,y)\rightarrow (0,0)$ $\endgroup$ – Singh Mar 15 '16 at 5:26

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