Consider $f(x,y) = x^2+y^2$ when $x-y \neq 0$ and $0$ if $(x,y) = (0,0)$.

This function is easily shown to be continuous along all paths, but along $x=y$ it is not defined!

So will it said to be continuous at $0,0$ ?

  • $\begingroup$ How have you shown 'easily' that $f$ is continuous along the path, $t \to (t,t)$, say? $\endgroup$ – preferred_anon Sep 2 '18 at 9:23
  • $\begingroup$ For any other path, we simply have to use first definition, which goes to 0 $\endgroup$ – jeea Sep 2 '18 at 9:35
  • $\begingroup$ Yes of course, but you have said $f$ is continuous along all paths. The function $t \mapsto (t,t)$ is a path, but $f$ is not continuous along it. There are many other paths which intersect the ray $x=y$. Do you see the problem? $\endgroup$ – preferred_anon Sep 2 '18 at 9:53
  • $\begingroup$ @jeea Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE $\endgroup$ – user Oct 23 '18 at 20:49
  • $\begingroup$ @gimusi sorry I totally forgot about that! $\endgroup$ – jeea Oct 24 '18 at 13:48

For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=y\neq 0$, indeed we have that trivially

$$\lim_{\substack{(x,y)\to(0,0)}\\\quad \:x\neq y}x^2+y^2=0=f(0,0)$$

therefore by definition $f(x,y)$ is continuous at $(0,0)$.

  • 1
    $\begingroup$ How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$? $\endgroup$ – preferred_anon Sep 2 '18 at 9:20
  • $\begingroup$ @DanielLittlewood Just take the path $x=t$ and $y=-t$ for example. $\endgroup$ – user Sep 2 '18 at 9:21
  • $\begingroup$ I think you are using a non-standard definition of continuity in $\mathbb{R}^{2}$. c.f. (en.wikipedia.org/wiki/…) $\endgroup$ – preferred_anon Sep 2 '18 at 9:24
  • $\begingroup$ @DanielLittlewood Of course we are referring to different definition but I'm not sure that mine is the "non-standard" one. If f(x,y) is not defined for $x=y\neq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/… $\endgroup$ – user Sep 2 '18 at 9:39
  • $\begingroup$ The various definitions given in the answer you link are satisfactory, and establish that the limit equals $0$ (which is true), although I suspect that the OP may have been given a definition which assumes that the domain of $f$ should be a nbhd of $(0,0)$, which would explain the confusion. My serious objection is with your comment: continuity along a path does not imply continuity of the function! $\endgroup$ – preferred_anon Sep 2 '18 at 9:52

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