3
$\begingroup$

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 $(x, y) \to (0, 0)$.

$\endgroup$
2
  • 1
    $\begingroup$ you need a better upper bound... The fact that your upper bound goes to infinity says nothing about the behaviour of the functions you are trying to bound. $\endgroup$ Mar 4 at 21:21
  • $\begingroup$ @PierreCarre Any help with that, considering I'm not allowed to use polars? :/ $\endgroup$
    – Heidegger
    Mar 4 at 21:30

3 Answers 3

4
$\begingroup$

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 $x^2+y^2\le 1$.

$\endgroup$
13
  • $\begingroup$ I forgot to say we cannot use polars. Our professor didn't teach them to us because "it's too mechanical" and we have to use $\leq$... $\endgroup$
    – Heidegger
    Mar 4 at 18:31
  • $\begingroup$ Well, I agree that many people on this site abuse the technique, but the clever homogenization here is powerful and is not the "usual" mechanical approach at all. Your professor must love algebraic contortions with the AMGM inequality. At any rate, remove the trig and use $x^4=u$, $y^2=v$ and work harder. $\endgroup$ Mar 4 at 18:33
  • $\begingroup$ AMGM we use it! I also tried with AMGM from the very beginning (after passing from $y^4 - x^8$ to $x^4 +y^8$ but what I get is $$\frac{2|y|}{|x|}$$ and the problem remains... $\endgroup$
    – Heidegger
    Mar 4 at 18:36
  • $\begingroup$ And this is what I don't get, because if I take $x = y^6$ then as $y \to 0$ I don't get a finite result. $\endgroup$
    – Heidegger
    Mar 4 at 18:36
  • $\begingroup$ I'm not a fan of AMGM contortions, so I'm going to let you ignore my solution if you insist. My approach does show the sensitivity of the question to each and every exponent that appears. One of my favorite problems to assign to my students was to determine for precisely which values of $\alpha,\beta,\gamma,\delta$ the function $|x|^\alpha|y|^\beta/(|x|^\gamma+|y|^\delta)$ (and $0$ at $0$) is continuous/differentiable. Good luck. $\endgroup$ Mar 4 at 18:38
3
$\begingroup$

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.


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.

$\endgroup$
13
  • $\begingroup$ It's not that I don't like trigonometric. It's that out professor demand us to use "$\leq$ things" that is, we have to upper bound functions with a distance function (possibly the Euclidean one), so I thank you too but unfortunately I cannot use what you wrote $\endgroup$
    – Heidegger
    Mar 4 at 23:11
  • $\begingroup$ Anyway, does your method prove the derivative is NOT continuous? $\endgroup$
    – Heidegger
    Mar 4 at 23:14
  • $\begingroup$ I don't understand what you mean by the "$\le$ things". Both Ted and I give bounds for $|f_x|$ using $\le$. $\endgroup$
    – robjohn
    Mar 4 at 23:14
  • $\begingroup$ "that is, we have to upper bound function with a distance function". So it should end like $$|f(x, y) - f(0, 0)| \leq d(x, y)$$ Where $d$ is a distance function. $\endgroup$
    – Heidegger
    Mar 4 at 23:16
  • $\begingroup$ In both our answers, $f_x(0,0)=0$; that is shown by the estimates given. If you need to, you can replace $|f_x(x,y)|$ with $|f_x(x,y)-0|$. $\endgroup$
    – robjohn
    Mar 4 at 23:18
1
$\begingroup$

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 $(x, y) \to (0, 0)$.

$\endgroup$
1
  • $\begingroup$ That was indeed clear! Thanks. $\endgroup$
    – Heidegger
    Mar 5 at 23:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .