Differentiable function $f$ such that one of its partials is not continuous

Asked Jun 24 '17 at 03:27

Active Jun 24 '17 at 05:01

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Is it possible to have a function $f:G\subseteq \mathbb R^2\to\mathbb R$, differentiable in a point $(x,y)\in G$, such that one of its partials is not continuous at $(x,y)$?

I think that it is not possible to find such a function $f$. Because if $f$ is differentiable at some point $(x,y)$, then it is continuous at that point.

Also $f'$ must be equal to ($\partial f_1,\partial f_2$).

Is my argumentation correct?

edited Jun 24 '17 at 05:01

Chris Culter

26,806

asked Jun 24 '17 at 03:27

user441848

1,658

5

You don't even need to several variables for that. – OR. Jun 24 '17 at 03:31
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So it is possible and the example would be f(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \ 0 & \mbox{if } x=0. \end{cases} ? @MlazhinkaShungGronzalezLeWy – user441848 Jun 24 '17 at 03:50
Well at least you have to put $g(x,y)=f(x)$ since required is a function on $\Bbb{R}^2$ :P – Li Chun Min Jun 24 '17 at 05:25
Ok so I just change $f(x)$ by $g(x,y)$ and in both cases should I write $y=0$ ? @LiChunMin :) – user441848 Jun 24 '17 at 05:40
@AnneliseToft I think the function is already well defined after you define for the cases $x=0$ and $x\neq 0$? – Li Chun Min Jun 24 '17 at 05:42
Oh ok, so the only change would be g instead of f ? @LiChunMin – user441848 Jun 24 '17 at 05:43
Put it this way then: you shall put $(x,y)$ as the argument of your function. – Li Chun Min Jun 24 '17 at 05:48
like this $f(x,y)=\begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \ 0 & \mbox{if } x=0. \end{cases}$ ? @LiChunMin – user441848 Jun 24 '17 at 05:51
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Yeah. Exactly. $$ – Li Chun Min Jun 24 '17 at 05:52
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ah finally.. thanks @LiChunMin – user441848 Jun 24 '17 at 05:53

Differentiable function $f$ such that one of its partials is not continuous

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