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?
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1In order that $\partial f/\partial x$ and $\partial f/\partial y$ be defined, $f$ would have to be continuous as a function of $x$ with $y$ fixed and vice-versa, so the discontinuity would have to happen as one approaches a point from a diagonal direction. – Michael Hardy Apr 24 '12 at 17:29
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If the partials exist and are continuous, then the function is differentiable, hence continuous at any point in the open set. – copper.hat Apr 24 '12 at 19:30
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Yes. See https://math.stackexchange.com/questions/456581/is-there-a-discontinuous-function-on-the-plane-having-partial-derivatives-of-all – subrosar Aug 19 '23 at 19:43
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$f(x,y)= 2xy/ x^2+y^2$ 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 $0$
Myshkin
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This post seems related to this specific example: Sanity check on example 6.5 from “Counterexamples in probability and real analysis” by Wise and Hall. – Martin Sleziak Aug 12 '17 at 08:17
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$f(x,y)=\frac{x^3y}{x^3+y^3}$ when $(x,y)\neq 0$ and $f(0,0)=0$
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The domain of this function is ${(x,y)\in\Bbb R^2,:, x\ne -y\lor x=y=0}$, which is not an open set. – Sassatelli Giulio Jan 14 '24 at 03:12