Is it possible to a real-valued function of two variables defined on an open set to have partial derivatives of all order and to be discontinuous at some point or maybe at each point?
Example of discontinuous function having all partial derivatives
5
$\begingroup$
multivariable-calculus
partial-derivative
examples-counterexamples
-
0If the partials exist and are continuous, then the function is differentiable, hence continuous at any point in the open set. – 2012-04-24
3 Answers
3
Try $\frac {xy}{x^2 + y^2},(x,y) \ne (0,0), f(0,0) = 0$
3
$f(x,y)= 2xy/ x^2+y^2$ when $x^2+y^2>0$ and $0$ when $(x,y)=0$ f is differentiable at each point of other than the origin and $D_1f(0,0)=0=D_2f(0,0)$ since $f(x,0)=0=f(0,y) \forall x,y$ but But $f(t,t)=1 \forall t\neq 0$ so f is not continuous at $0$
-
0This post seems related to this specific example: [Sanity check on example 6.5 from “Counterexamples in probability and real analysis” by Wise and Hall](https://math.stackexchange.com/q/2391016). – 2017-08-12
2
$f(x,y)=\frac{x^3y}{x^3+y^3}$ when $(x,y)\neq 0$ and $f(0,0)=0$