3
$\begingroup$

If a function of two variables is discontinuous at a particular point, say $(x,y)$, does this mean that the graph of that function has some hole around the point $(x,y,f(x,y))$? Is there any break in the graph at this point in certain direction?

This question arises because I have one function which is discontinuous at $(0,0)$ but all of its partial derivatives and directional derivatives exist at $(0,0)$. While calculating its partial or directional derivatives, we naturally look in a certain plane with that point and specified direction and calculate the slope of the tangent line (as you would with one variable). In my example I have all directional derivatives, which seems to imply that there is no break around $(0,0,f(0,0))$ in any direction. Then why is the function discontinuous at $(0,0)$?

$\endgroup$
  • $\begingroup$ Directional derivatives are one dimensional, while continuity in this case is two dimensional (as is differentiability). All the one dimensional vertical slices through the graph can be perfectly continuous and differentiable as one dimensional functions, while the two dimensional continuity condition fails because the it requires that the two-dimensional limit at (0,0) equals $f(0,0)$. The concept of a "break" isn't as simple in two dimensions as one, since you approach (0,0) along curves as well the straight lines-- a "break" in a curved path, but not in any straight line path, is possible. $\endgroup$ – Ned Aug 28 '18 at 21:01
  • $\begingroup$ See this example too: math.stackexchange.com/q/827382/67710 $\endgroup$ – Ned Aug 28 '18 at 21:26
  • $\begingroup$ What is your function? $\endgroup$ – TonyK Aug 28 '18 at 22:48
  • $\begingroup$ Function is almost similar as mentioned by @zhw. $\endgroup$ – Believer Aug 29 '18 at 6:32
2
$\begingroup$

Example 1 (one variable): Define $f(x) = \sin (1/x),x\ne 0,$ $f(0) =0.$ Then $f$ is (badly) discontinuous at $0.$ But there is no hole or break in the graph of $f.$ In fact the graph of $f$ is a connected subset of $\mathbb R^2.$

Example 2 (two variables): Define $f(x,x^2) = 1$ for real $x\ne 0.$ Define $f(x,y)=0$ everywhere else. Then all directional derivatives of $f$ at $(0,0)$ exist and are $0.$ But $f$ is discontinuous at $(0,0),$ as $\lim_{x\to 0} f(x,x) =0,$ while $\lim_{x\to 0} f(x,x^2) =1.$

$\endgroup$
  • $\begingroup$ So hard to imagine that if I take plane almost coinciding curve $y=x^{2}$,still there will be some point close to (0,0)in that direction which will take value 0 (that's why all directional derivatives are zero) and not 1 suddenly.Am I right? $\endgroup$ – Believer Aug 29 '18 at 6:44
  • $\begingroup$ Right, in every direction there will be an interval of positive length centered at $(0,0)$ where $f=0.$ $\endgroup$ – zhw. Aug 29 '18 at 14:46
1
$\begingroup$

Consider $f(x,y) = \frac{x^2+y^2}{x^2+y^2}$, using Cartesian coordinates. This function is undefined at $(0,0)$ so it cannot be continuous there. However, it is constantly $1$ everywhere else.

Now consider $g(x,y) = \frac{x^2-y^2}{x^2+y^2}$. What's happening at $(0,0)$?

$\endgroup$
  • $\begingroup$ I don't think it's very helpful to exhibit a function that is discontinuous at a point only because it's not defined there. This is not what the OP meant at all (otherwise $f$ wouldn't have directional derivatives at $0$). $\endgroup$ – TonyK Aug 28 '18 at 22:46
  • $\begingroup$ @TonyK : OP asks "Is there any break in the graph at this point in certain direction?", which most clearly does not happen in the first example. $\endgroup$ – Eric Towers Aug 29 '18 at 1:27
  • $\begingroup$ But in first example, clearly we can say function is discontinuous since partial derivatives are not existing at (0,0).What spinning my head is if all directional derivatives exists,and still it is discontinuous,then there should be some break in certain direction? $\endgroup$ – Believer Aug 29 '18 at 6:23
  • $\begingroup$ @omkarGirkar : What do you mean by "certain direction"? What certain direction does a point have? $\endgroup$ – Eric Towers Aug 29 '18 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.