For an exercise in my analysis course, I have to show that the function $$\newcommand{\sgn}{\operatorname{sgn}}f: (x,y) \mapsto \begin{cases} \frac{(x \sin y)^2}{|x|+|y|},&(x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$$ is $C^1$. I tried to proceed by showing that the partial derivatives exist and are continuous.

I have shown that $$\frac{\partial f(x,y)}{\partial x} = \begin{cases} \sin^2 y\frac{2x(|x|+|y|)-x^2\sgn(x)}{(|x|+|y|)^2}, & (x,y)\neq (0,0)\\ 0, & (x,y) = (0,0)\end{cases} $$

and $$\frac{\partial f(x,y)}{\partial y} = \begin{cases} x^2\frac{\sin^2y\cdot \cos y(|x|+|y|)-\sin^2(y)\sgn(y)}{(|x|+|y|)^2}, &(x,y) \neq (0,0) \\ 0, &(x,y) = (0,0)\end{cases}$$

Now, I tried to show that those derivatives are continuous. To show that they are continuous in $(x,y) \neq (0,0)$ is easy since we can apply limit laws since de denominator is not $0$ close to $(x,y)$.

However, To show that they are continuous in $(0,0)$ I do not know how to proceed. I tried the following:

$$ \begin{align*}\lim_{(x,y)\rightarrow (0,0)} \sin^2 y\frac{2x(|x|+|y|)-x^2\sgn(x)}{(|x|+|y|)^2} &= \lim_{(x,y)\rightarrow (0,0)} \sin^2 y \cdot \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)-x^2sign(x)}{(|x|+|y|)^2}\\& = 0 \cdot \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)-x^2sign(x)}{(|x|+|y|)^2} \end{align*}$$

So if $$\begin{align*} \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)-x^2\sgn(x)}{(|x|+|y|)^2} &= \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)}{(|x|+|y|)^2}-\lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sgn(x)}{(|x|+|y|)^2}\\ &= \lim_{(x,y)\rightarrow (0,0)}\frac{2x}{|x|+|y|}-\lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sgn(x)}{(|x|+|y|)^2}\end{align*}$$ converges I'm done. I'm not sure how to proceed now though. Could anyone give me a pointer?


You only need that $\frac{2x}{|x|+|y|}$ and $\frac {x^2\operatorname{sgn}(x)}{(|x|+|y|)^2}$ are bounded.

  • $\begingroup$ Why would that suffice? $\endgroup$ – sxd Nov 15 '12 at 18:06
  • $\begingroup$ Because then for $|(x,y)-(0,0)|<\epsilon$ you have $|f_x|\le \sin^2 y \cdot L< L \epsilon^2$ and $|f_y|\le x^2L<L\epsilon^2$. $\endgroup$ – Hagen von Eitzen Nov 15 '12 at 18:09
  • $\begingroup$ Oh I see what you mean, thanks ! $\endgroup$ – sxd Nov 15 '12 at 18:18

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.