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I'm having difficulties in showing that, given the function $$f(x, y) = \begin{cases} 0 & y \leq 0 \qquad y \geq x^2 \\ 1 & 0 < y < x^2 \end{cases} $$

goes to zero when $(x, y) \to (0, 0)$ along any line for the origin, and then there does exist a curve passing though the origin such that $f(x, y) = 1$, except in the origin.

attempts

Since I have to study the straight lines for the origin, I though of restricting $y = mx$, then the function becomes defined by

$$ f(x, mx) = \begin{cases} 0 & mx \leq 0 \qquad mx \geq x^2 \\ 1 & 0 < mx < x^2 \end{cases} $$

The condition can be simplified, according if $m > 0$ or $m<0$ like for example if $m > 0$ I read in the first one: $0 \leq x \leq m$.

In any case, as $x\to 0$ I obtain $0 =0$ in the conditions.

What confuses me is how $f$ is defined. I also tried to "plot" it.

About the second point I thought for a logarithmic curve, since it's surely true that $\ln(x) < x^2$, but again it was just an intuition.

Perhaps this function is not even continuous at the origin...

Heidegger
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    Take limit along $(\frac 1n,\frac 1 {2n^{2}})$ to see that $f$ is not continuous at $(0,0)$ – geetha290krm Feb 24 '23 at 11:12
  • Since you already proved the first part, you could have shortened your post by clearly asking: find some continuous curve passing through the origin and such that every other point $(x,y)$ satisfies $0<y<x^2.$ Or even: find some continuous function $g$ s.t. $g(0)=0$ and $\forall x\ne0\quad0<g(x)<x^2.$ – Anne Bauval Feb 24 '23 at 11:23
  • This $y\leq 0, y\geq x^2$ is impossible, is there a typo? – user376343 Feb 24 '23 at 11:24
  • @geetha290krm Oh, so I can show by using sequence! Cool! – Heidegger Feb 24 '23 at 11:25
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    @user376343 I checked again, and no, it's written like that. It's a double condition, like when $y$ is negative OR $y \geq x^2$ – Heidegger Feb 24 '23 at 11:26
  • @AnneBauval Oh, so my first part was right? I thought I did some mess! – Heidegger Feb 24 '23 at 11:27
  • No, sorry, I was wrong, your first part is messy indeed. As for the 2nd part, a sequence is sufficient if you only want to prove the discontinuity, but of course not if you are really asked for a curve. – Anne Bauval Feb 24 '23 at 11:40
  • @AnneBauval Sigh! How can I then show it? I'm confused – Heidegger Feb 24 '23 at 11:46
  • For the 1st part, you want to prove that for $x$ small enough, $mx\le0$ or $x^2\le mx.$ A sufficient "smallness" is $|x|\le|m|.$ – Anne Bauval Feb 24 '23 at 11:47
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    @AnneBauval Sorry, my bad. I did not want the chat! Wanted to say that the proposed duplicate made me to understand well! Thank you! – Heidegger Feb 24 '23 at 12:42
  • @AnneBauval Could you please help me with this one? https://math.stackexchange.com/questions/4645590/continuity-of-this-function-though-the-distance – Heidegger Feb 24 '23 at 12:59

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