About the continuity of the partial derivative

Question

I'm trying to study if the following function has continuous partial derivatives at the origin: $$f(x, y) = \begin{cases}\frac{x^4y^3}{x^8 + y^4} & (x, y) \neq (0, 0) \\\\ \quad 0 & (x, y) = (0, 0)\end{cases}$$

I proved $f$ is continuous at the origin, and I also proved its partial derivatives exist at the origin.

Now to show the continuity of $f'_x$ at $(0, 0)$ here is what I did:

$$\frac{\partial f}{\partial x} = \frac{4x^3y^3(y^4 - x^8)}{(x^8 + y^4)^2}$$

Having observed it goes to zero along various paths, I did:

$$\bigg|\frac{4x^3y^3(y^4 - x^8)}{(x^8 + y^4)^2}\bigg| \leq \frac{4x^2y^2|x| |y| (x^8 + y^4)}{(x^8+y^4)^2} \leq \frac{4x^2y^2|x| |y|}{y^4} = \frac{4x^2|x|}{|y|}$$

But now I am stuck.

If for example I say $y = x^4$, this reduces to $\frac{4}{|x|}$ which does not goes to zero.

But the notes say the partial derivatives are continuous (without any proof...)

Any help? Thank you!

Notice We cannot use polar coordinates. We are demanded to find a distance function to make an upper bound to the function we have, in order to conclude it goes to $0$ as $(x, y) \to (0, 0)$.

Answer 1

In problems like this I find it useful to use a homogenized form of polar coordinates: Let $x^4=r\cos\theta$ and $y^2=r\sin\theta$, with $0\le\theta\le\pi/2$. Then $$\left|\frac{x^3y^3(y^4-x^8)}{(y^4+x^8)^2}\right| = \frac{r^{3/4}r^{3/2}r^2|\cos 2\theta||\cos\theta|^{3/4}|\sin\theta|^{3/2}}{r^4} \le r^{3/4+3/2-2} = r^{1/4}.$$ You still have to do a little bit of algebraic work to finish. How do we bound $r^{1/4}$ by a power of $(x^2+y^2)^{1/2}$?

EDIT: $r^2=x^8+y^4\le x^4+y^4\le (x^2+y^2)^2$ when $x^2+y^2\le 1$.

Answer 2

If you really don't like trigonometric functions, we can consider the rectangles instead; however, this is not really any different than Ted Shifrin's answer.

Consider the family of rectangles $$ \max(x^2,|y|)=r\tag1 $$ On each rectangle, $$ r^4\le x^8+y^4\le2r^4\tag2 $$ and for $r\le1$, $\mathrm{d}((x,y),(0,0))\ge r$.

Thus, we have $$ \begin{align} |f_x(x,y)| &=\left|\frac{4x^3y^3\left(y^4-x^8\right)}{\left(x^8+y^4\right)^2}\right|\tag{3a}\\ &\le\frac{4r^{3/2}r^3r^4}{r^8}\tag{3b}\\[9pt] &=4r^{1/2}\tag{3c}\\[12pt] &\le4\mathrm{d}((x,y),(0,0))^{1/2}\tag{3d} \end{align} $$ So we can choose a rectangle small enough so that $f_x(x,y)$ is as small as we wish.

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However, on the path $(x,y)=\left(r^{1/2},r\right)$, $$ \begin{align} |f_y(x,y)| &=\left|\frac{x^4y^2\left(3x^8-y^4\right)}{\left(x^8+y^4\right)^2}\right|\tag{4a}\\ &=\frac{r^2r^2(2r^4)}{4r^8}\tag{4b}\\[9pt] &=\frac12\tag{4c} \end{align} $$ Whereas, along the path $(x,y)=(0,r)$, $f_y(x,y)=0$. That is, $f_y$ cannot be defined at $(0,0)$ so as to make $f_y$ continuous.

Answer 3

As Ted and Rob already pointed out, summing their hints let me see if I can convince you. Notice that: $x^3 = (x^8)^{3/8} \leq (x^8 + y^4)^{3/8}$ and $y^3 = (y^4)^{3/4} \leq (y^4 + x^8)^{3/4}$. We can then write:

$$\bigg|\frac{4x^3y^3(y^4-x^8)}{(x^8 + y^4)^2}\bigg| \leq \frac{4(x^8+y^4)^{3/8} (x^8 + y^4)^{3/4} (x^8+y^4)}{(x^8 + y^4)^2} = 4(x^8 + y^4)^{3/8 + 3/4 + 1 - 2} = 4(x^8+y^4)^{1/8}$$

Which goes to $0$ as $(x, y) \to (0, 0)$.