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Our professor taught us this way to prove the continuity of a function at a point: we have to find a suitable "distance" function $h$. Then $f$ is continuous at $x_0$ if there exists $h$ such that $$|f(x) - f(x_0)| \leq h(||x-x_0||)$$

And this $h$ can be written as $h(d)$ and $h(d) \to 0$ for $d \to 0$.

I'm having problems in applying this for the function $$ f(x, y) = \begin{cases} \frac{2x^2 y}{x^4+y^2} & (x, y) \neq (0, 0)\\ 0 & (x, y) = (0, 0) \end{cases} $$

I already proved in many ways that the function is not continuous (example: using parabolas, straigh lines and sequences). I cannot manage to prove it with this method though.

I thought of

$$\bigg|\frac{2x^2y}{x^4+y^2}\bigg| \leq \frac{2x^2|y|}{x^4+y^2} \leq \frac{2x^2y^2}{x^4+y^2} \leq \frac{2x^2y^2}{x^2+y^2}$$

But then I always obtain some function that goes to $0$. For example if I say

$$x^2y^2 \leq (x^2+y^2)^2$$

or using AM/GM inequality...

Heidegger
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  • Your post is unclear: the method you were taught is used to prove the continuity, but your function is not continuous (as yourself claim having "proved in many ways"). – Anne Bauval Feb 24 '23 at 13:16
  • For $0<x,y<1,$ $|y|>y^2$ and $x^4<x^2$ hence your two inequalities $\frac{2x^2|y|}{x^4+y^2} \leq \frac{2x^2y^2}{x^4+y^2} \leq \frac{2x^2y^2}{x^2+y^2}$ are false. – Anne Bauval Feb 24 '23 at 13:20
  • @AnneBauval Yes, indeed! I was trying to prove, with the same method, that the function is not contnuous. Like maybe obtaining a distance that does not go to zero... – Heidegger Feb 24 '23 at 13:21
  • That would be of no help. There are always functions $h$ such that $\lim_0h\ne0$ and (for $x$ close to $x_0$) $|f(x)-f(x_0)|\le h(||x-x_0||):$ just take $h$ big enough. – Anne Bauval Feb 24 '23 at 13:24
  • @AnneBauval Oh, so I could indeed prove that my function is $\leq$ of certain $h(d)$ which goes to zero, even if $f$ is not continuous? I then have to rely on paths... Thank you! – Heidegger Feb 24 '23 at 13:25
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    Yes, the trick with $h$ is used to prove continuity, and the trick with curves to prove discontinuity. But I am trying to build a more precise answer to the present post. – Anne Bauval Feb 24 '23 at 13:26
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    @AnneBauval I really appreciate that! Thank you in advance (and for all!) – Heidegger Feb 24 '23 at 13:27

1 Answers1

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As mentionned in the comments, the trick with $h$ you have been taught is used to prove continuity, hence is not adapted here to prove discontinuity.

Whatever the function $f$ (continuous or not at $x_0$), you will always find some function $h$ such that $\lim_0h\ne0$ and $|f(x)-f(x_0)|\le h(\|x-x_0\|)$ for $x$ close to $x_0$ (just take $h$ big enough) and that will be of no help.

On the opposite, since your function $f$ here is discontinuous at $(0,0),$ your claim having bounded it and "always obtain some function that goes to $0$" cannot be true. In fact, as explained in the comments, your inequalities $\frac{2x^2|y|}{x^4+y^2} \leq \frac{2x^2y^2}{x^4+y^2} \leq \frac{2x^2y^2}{x^2+y^2}$ were wrong.

However, the "trick whith $h$" can be used to prove discontinuity if you calculate the least upper bound $h$ and prove that this bound does not tend to $0.$ I.e. if you prove that $$h(R):=\sup_{\|x-x_0\|\le R}|f(x)-f(x_0)|\not\to0\text{ as }R\to0.$$ (This is nothing more than a reformulation of the discontinuity of $f$ at $x_0$.)

For your function $f(r\cos t,r\sin t)=\frac{2r(1-s^2)s}{r^2(1-s^2)^2+s^2}$ where $s=\sin t,$ you will easily find that for $r$ fixed, the maximum of $|f|$ is attained when $|s|=r(1-s^2)$ and is therefore equal to $1.$ So in your example, the lub is the constant function $h(R)=1.$ But this way of proving discontinuity is much more expensive than the usual one with a well-chosen curve, which does not require to compute the lub explicitely.

Incidentally, the same ideas may be used when you want to apply the dominated convergence theorem but you don't know whether there is some integrable dominating function: simply calculate (if possible) the least dominating function, and look if it is integrable or not.

Anne Bauval
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  • That was awesome, thank you so much!! I just have a rapid question: suppose I have some $f(x, y)$ such that the limit while approaching the origin trhough different paths (like $y = mx$ or $y = \alpha x^2$ or $x = \sqrt{|y|}$ and so on) is the same (say I studied three different paths). How can I be sure from this, that $f(x, y)$ is continuous at the origin? I mean, I did not studied "all" the paths... – Heidegger Feb 24 '23 at 20:33
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    You can be sure if (useless theorem) and only if (unrealistic dream) you studied all the (continuous) paths, or equivalently: all the sequences in $\Bbb R^2$ converging to $(0,0).$ – Anne Bauval Feb 24 '23 at 21:55
  • Haha ok! Then how are people really sure about those limits? – Heidegger Feb 25 '23 at 13:47
  • Well: not by studying sublimits on curves, but by direct global $\epsilon-\delta$-proofs. – Anne Bauval Feb 25 '23 at 14:27
  • Then I wonder why our professor taught us the "distance method" instead of the epsilon-delta method, which for what I could understand, it's 101% safe in any case... – Heidegger Feb 25 '23 at 15:17
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    Oh, but the "distance method" is as good as the epsilon-delta for proving continuity! What is not good is the "curves method", which should be used only to prove discontinuity. This was my answer to your "rapid question". – Anne Bauval Feb 25 '23 at 15:26