Continuity of partial derivatives of $f(x,y) =\dfrac{xy(x^2-y^2)}{x^2+y^2}$ if $(x,y)\neq 0$ and $f(x,y)= 0$ if $(x,y) = (0,0).$

Question

So I got that the partial derivatives of \begin{equation*} f(x,y) = \begin{cases} \dfrac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \not= (0,0),\\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*} are \begin{equation*} \frac{\partial f}{\partial x}(x,y)= \begin{cases} \dfrac{-y^5 +4x^2y^3+x^4y}{\left(x^2+y^2\right)^2} & \text{if } (x,y) \not= (0,0),\\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*} and \begin{equation*} \frac{\partial f}{\partial y}(x,y)= \begin{cases} \dfrac{x^5 -4x^3y^2-xy^4}{\left(x^2+y^2\right)^2} & \text{if } (x,y) \not= (0,0),\\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*}

Now I would like to prove that these are continuous everywhere on $\mathbb R^2$.

My first approach is to just check continuity at the origin, but I am not sure how to use the definition to prove that both are continuous.

Any help, hint or advice would be appreciated. Thanks in advance. :)

Answer 1

In case you have difficulty to get book mentioned by me in comment I bring here main inequality for question (it's hard to type tex in comment)

$$\left| \frac{\partial f}{\partial x}\right| = \frac{|-y^5 +4x^2y^3+x^4y|}{\left(x^2+y^2\right)^2}\leqslant \frac{6\left(x^2+y^2\right)^{5/2}}{\left(x^2+y^2\right)^2} = 6\left(x^2+y^2\right)^{1/2}$$

Answer 2

I believe you want to check the continuity of the partial derivatives at $0$. Note that the function in question has mixed partial derivatives at $(0,0)$ which are not equal to each other. In fact, the function in question is the prototypical example of such a situation.

In your case, you want to prove that $$ \lim_{(x,y) \to (0,0)} \left|\frac{\partial f}{\partial x}(x,y)\right| = \frac{\partial f}{\partial x}(0,0)=0. $$ This is identical, by your (correct) calculations to $$ \lim_{(x,y) \to (0,0)} \left|\frac{-y^5+4x^2y^3+x^4y}{(x^2+y^2)^2}\right| =0. $$ If we factorize $y$ out from the numerator, then the expression above equals $$ \left|\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\right|. $$ Now, the idea is as follows : the $y$ will go to $0$. We will make sure that the rest of the fraction is not too bad in comparison to $y$. To do this, once you expand $(x^2+y^2)^2$, the expression above equals $$ \left|\frac{y(x^4+4x^2y^2-y^4)}{x^4+2x^2y^2+y^4}\right|. $$ Multiply the top and bottom by $2$. Isolate the $2y$ part to get $$ \left|2y\frac{x^4+4x^2y^2-y^4}{2x^4+4x^2y^2+2y^4}\right|. $$ However, this fraction here is clearly smaller than $1$. Indeed, we can easily prove that $$ |x^4+4x^2y^2-y^4| \leq |2x^4+4x^2y^2+2y^4| = 2x^4+4x^2y^2+y^4 $$ since the RHS is always non-negative. To prove this, observe that $$ x^4\leq 2x^4,4x^2y^2 \leq 4x^2y^2,-y^4 \leq 2y^4 \implies x^4+4x^2y^2-y^4 \leq 2x^4+4x^2y^2+2y^4. $$ Similarly, $$ -x^4 \leq 2x^4, -4x^2y^2 \leq 4x^2y^2,y^4\leq 2y^4 \implies -(x^4+4x^2y^2-y^4) \leq 2x^4+4x^2y^2+2y^4. $$ Suppose that $a = x^4+4x^2y^2-y^4$ and $b = 2x^4+4x^2y^2+2y^4$. The above inequalities have proved that $a \leq b$ and $-a \leq b$. As a consequence, $|a| \leq b=|b|$. If $(x,y) \neq (0,0)$ then $b \neq 0$, hence $$ \left|\frac{a}{b}\right| \leq 1 \implies \left|\frac{x^4+4x^2y^2-y^4}{2x^4+4x^2y^2+2y^4}\right| \leq 1. $$ But now, $$ \left|\frac{x^4+4x^2y^2-y^4}{2x^4+4x^2y^2+2y^4}\right| \leq 1 \implies \left|2y\frac{x^4+4x^2y^2-y^4}{2x^4+4x^2y^2+2y^4}\right| \leq 2\left|y\right| \implies \left|\frac{\partial f}{\partial x}(x,y)\right| \leq 2\left|y\right| $$ for all $(x,y) \neq (0,0)$. As a consequence of the squeeze theorem, it follows that $$ \lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial x}(x,y) =0. $$

An extremely similar method yields that $\frac{\partial f}{\partial y}$ is also continuous at $(0,0)$. There, you factorize $x$ out of the numerator and control the fraction appropriately. I leave readers to this task.

Alternately, observe that $f(y,x) = -f(x,y)$ for all $x,y$. This symmetry can also be used to show that $\frac{\partial f}{\partial y}$ is continuous at $(0,0)$.