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I'm reading Courant's: Calculus, Vol 2.

When he speaks about continuity, he gives the following example:

$$ f(x,y) = \left\{ \begin{array}{ll} \cfrac{2xy}{x^2+y^2} & \quad (x,y)\neq (0,0)\\ \quad\,\,\, 0 & \quad (x,y)= (0,0) \end{array} \right. $$

And argues why this function is not continuous, however, I'd like to test it's continuous using the definition. I can understand Courant's argument: He takes $f(x,0)=0, f(0,y)=0$ and show that along the line $y=x$, we have: $f(x,x)=1$ and with this, we could choose any value $c$ for $f(0,0)=c$ and we would always have one path for which the function is discontinuous.

My question is different of this question because I don't want to find some paths and show it is discontinuous (I already know how to do this in the last paragraph), I'm actually using Courant's example to ask about something more subtle: Suppose I have a function $f(x,y)$ and want to test It's continuity at some point, I know I can test all paths along a line, I know I could use polar coordinates but given what I've seen in my lectures, the positivity of such tests is not a proof of continuity - It seems that even when these two tests are positive, there are still infinite paths in which the function could be discontinuous.

I've set up the expressions according to the text:

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enter image description here

What do I do now? I tried to write:

$$|f(h,k)-f(0,0)|\leq \epsilon$$

$$\left|\cfrac{2hk}{h^2+k^2}\right|\leq \epsilon$$

And it should hold for all the pairs given in the text but I'm not sure about what I should do now. I guess that it would be counter productive to search for a pair that acts as a counterexample because in more complicated functions, it may not be so easy to find one. I'm completely lost at what I should do now.

EDIT: I have tried to put some thought on it and also tried to read the definition of continuity in Zorich's book:

enter image description here

So this seems almost like a sort of script which seems a little easier to grasp. I take one neighborhood $N(f(a))$, then I need to find a neighborhood $N(a)$ such that $f(N(a))\subset N(f(a))$, then if it's impossible to find a neighborhood $N(a)$ for some neighborhood $N(f(a))$, the function is discontinuous.

Now I guess I can suppose $N(f(a))$ has length $\epsilon$ (in $\Bbb{R}^1$) , I guess I need to pick a reasonable radius for $N(a)$, that is: $x^2+y^2\leq \delta$. For operational simplicity, I have decided to use: $x^2+y^2 = \delta$ but I'm totally unaware of how catastrophic that could be (supposing it could be catastrophic). Now:

$$x=\pm\sqrt{\delta - y^2}\quad \quad \quad\quad \quad y=\pm\sqrt{\delta - x^2} $$

$$f(\sqrt{\delta - y^2}, \sqrt{\delta - x^2} )=f(-\sqrt{\delta - y^2},- \sqrt{\delta - x^2} )= \frac{2\sqrt{\delta - y^2}\sqrt{\delta - x^2}}{\delta}$$

$$f(-\sqrt{\delta - y^2}, \sqrt{\delta - x^2} )=f(\sqrt{\delta - y^2},- \sqrt{\delta - x^2} )=\frac{-2\sqrt{\delta - y^2}\sqrt{\delta - x^2}}{\delta}$$

This is the operational simplicity I mentioned: If I used $x^2+y^2\leq \delta$, I guess I couldn't make that substitution which - at least for first sight, seems to be a little helpful and harmless. Now I guess I need to check if:

$$-\epsilon \leq \frac{2\sqrt{\delta - y^2}\sqrt{\delta - x^2}}{\delta}\leq \epsilon \quad \quad \quad -\epsilon \leq \frac{-2\sqrt{\delta - y^2}\sqrt{\delta - x^2}}{\delta}\leq \epsilon $$

In which I guess that solving just one of the cases, would prove the other one. So I'll pick the first one:

$$-\epsilon \leq \frac{2\sqrt{\delta - y^2}\sqrt{\delta - x^2}}{\delta}\leq \epsilon $$

And here, I suppose I picked one small $\epsilon>0$, I guess I need to find a $\delta$ in terms of $\epsilon$ which I'm not completely sure but seems to ammount to the same as solving for $\delta$? So:

$$-\epsilon\delta \leq 2\sqrt{\delta - y^2}\sqrt{\delta - x^2}\leq \epsilon\delta $$

Now I need to square all the sides:

$$\epsilon^2\delta^2 \leq 2(\delta - y^2)(\delta - x^2)\leq \epsilon^2\delta^2 $$

If this is a valid move, then:

$$2 \delta ^2-2 \delta x^2+2 x^2 y^2-2 \delta y^2=\epsilon^2\delta^2 $$

Solving for $\delta$ gives me:

$$\delta=\frac{-x^2-y^2\pm\sqrt{x^4+2 x^2 y^2 \epsilon ^2-2 x^2 y^2+y^4}}{\epsilon ^2-2}$$

Now if $\epsilon \to 0$, then:

$$\delta=\frac{1}{2} \left(x^2+y^2\pm\sqrt{\left(x^2-y^2\right)^2}\right)$$

But I'm not sure if this would imply something useful or if everything I did was a complete nonsense.

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  • 2
    $\begingroup$ One useful thing is directional limits - If the limit exists, then approaching in different directions must always give the same value. ​ ​ $\endgroup$
    – user57159
    Sep 17, 2016 at 5:56
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    $\begingroup$ Dear @Voyska: Did you try along the paths $y=mx$? Recall that if limit exists,then along every path the limit would be same as pointed out in the above comment too. $\endgroup$
    – Arpit Kansal
    Sep 17, 2016 at 5:59
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    $\begingroup$ This may be a duplicate, but the OP wanted to use the definition directly, not specified or referenced in that question or its answers. $\endgroup$
    – David P
    Sep 17, 2016 at 6:12
  • $\begingroup$ It's actually a nice question, especially if it elicits a proof that is effectively simpler than the "checking every path" proof in all of our calculus books. $\endgroup$
    – Lee Mosher
    Sep 17, 2016 at 13:41
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    $\begingroup$ It is basically the same thing. $B_\epsilon(a)$ is the neighborhood $x\in X$ with $d(x,a)<\epsilon$. Consider $B_{0.9}(f(0)) = B_{0.9}(0)$. Then for any delta, $B_\delta(0,0)$ contains the point $x_0=(0.5\delta, 0.5\delta)$. However $f(x_0)\not\in B_{0.9}(0)$ so for all $\delta$, we have $f(B_\delta(0,0))\not\subset B_{0.9}(f(0))$. $\endgroup$
    – David P
    Sep 18, 2016 at 15:15

1 Answer 1

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Let $\epsilon = 0.9$ and suppose a $\delta$ with the desired property exists. Then, for instance, $(h,k) = \left(\dfrac{1}{2}\delta, \dfrac{1}{2}\delta\right)$ is in the ball of radius delta centered at the origin.

However

$$|f(h,k)| = \dfrac{2\cdot h^2}{h^2+h^2}=1 > \epsilon$$

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  • $\begingroup$ Yes, you answered in the spirit I was looking for. But how did you guess $0.9$? I could have a function with a much smaller number, how would I guess at which point the test would fail? Just trial and error? $\endgroup$
    – Red Banana
    Sep 17, 2016 at 17:45
  • $\begingroup$ I noticed that the function attains $1$ when $x=y$ regardless of the size of the ball, so I chose something less than $1$ $\endgroup$
    – David P
    Sep 17, 2016 at 17:58
  • $\begingroup$ But what do we do when the function is very complicated and It's not possible to see that easily? $\endgroup$
    – Red Banana
    Sep 17, 2016 at 18:06
  • $\begingroup$ You need to play with the function, investigate cases. For any technique which may apply to an infinite class of functions, there will always be another function for which a technique does not apply. It may have been that $f(x,y)$ was $0.25$ when $x,y$ had the relationship $1-\Gamma(\tan^{-1}(x)) \ge e^{\sin(y)}$ or something else weird, and I;d choose $\epsilon=0.2$. That said, for rational functions as $x,y)\to 0$ it helps to look at $x=y^k$ for different powers of $k$ $\endgroup$
    – David P
    Sep 17, 2016 at 18:17
  • $\begingroup$ I tried to do something and added to my question, could you take a look? $\endgroup$
    – Red Banana
    Sep 18, 2016 at 5:15

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