On the function $f(x,y)=x^2$ if $|x|<|y|$, $f(x,y)=y^2$ if $|x|\geq|y|$

Question

We have this function defined in $\mathbb{R}^2$: $$f(x,y)=\begin{cases} x^2 & \text{if $|x|<|y|$}\\ y^2 & \text{if $|x|\geq|y|$} \end{cases} $$ How to study on $(a,a)$: the continuity, partial derivatives? Thank you.

I have an answer for the continuity:

1) Continuity: The problem is when $|x|=|y|$. $$\lim_{(x,y)\to(a,a),(x,y)\in \{(x,y),|x|<|y|}\}{x^2}=a^2$$ and $$\lim_{(x,y)\to(a,a),(x,y)\in \{(x,y),|x|\geq|y|}\} {x^2}=a^2$$ then we obtain the continuity.

score 1 · Answer 1 · answered May 06 '17 at 18:42

1

As for continuity you are correct. for differentiability you would like to show $f_x$ and $f_y$ are continuous. however they are not. take the line $a=-b +2$ and the dot $(1,1)$ you will get: $$\lim_{b\implies1^-}f_x(a,b)=2\neq 0 =\lim_{b\implies1^+}f_x(a,b)$$ You can generlize this idea to the whole line $(a,a)$. Hence $f$ is not differentiable on the line $x=y$

answered May 06 '17 at 18:42

Alon Yariv

1,363

Thank you for your answer. The differentiability exist only for $a=0$? – Theory Nombre May 06 '17 at 19:20
Yes, it does. and for all $a\neq b$ $(a,b)$ – Alon Yariv May 06 '17 at 20:50

On the function $f(x,y)=x^2$ if $|x|<|y|$, $f(x,y)=y^2$ if $|x|\geq|y|$

1 Answers1