3
$\begingroup$

I'm having a hard time trying to understand a theorem of multivariable calculus.

Let:

  • Statement A = "The partial derivatives of $f$ are continuous in an open set containing (a,b)"
  • Statement B = "$f$ is differentiable at (a,b)"

I was wondering why the theorem: A$\implies$B was only true in that way. Why is A$\iff$B false ?

How can a function be differentiable without continuous partial derivatives ? Do you have an example of such a function ?

Thanks a lot for your help and happy new year ;)

Diego, student from Belgium

$\endgroup$
3
$\begingroup$

An example for such a function is: $f(x,y)=\begin{cases} (x^2+y^2)sin(\frac{1}{\sqrt{x^2+y^2}}) & if\;(x,y)\neq (0,0)\\ 0 & if\;(x,y)=(0,0). \end{cases}$

It is differentiable in $(0,0)$, but the partial derivatives are not continous in $(0,0)$.

More explicitely: The partial derivative $\frac{\partial f}{\partial x}$ is: (analogous for $y$)

$\frac{\partial f}{\partial x}(0,0)=\lim\limits_{h \rightarrow 0}{\frac{f(h,0)-f(0,0)}{h}}=\lim\limits_{h \rightarrow 0}{hsin(\frac{1}{|h|})}=0$.

$\frac{\partial f}{\partial x}(x,y)=2xsin(\frac{1}{\sqrt{x^2+y^2}})-\frac{xcos(\frac{1}{\sqrt{x^2+y^2}})}{\sqrt{x^2+y^2}}$.

If you consider this along the x-axis (i. e. y=0), you get $\frac{\partial f}{\partial x}(x,0)=2xsin(\frac{1}{|x|})-sgn(x)cos(\frac{1}{|x|})$, which is oscillating around $(0,0)$ (hence we don't have $\lim\limits_{x \rightarrow 0}\frac{\partial f}{\partial x}(x,0)=0)$, hence the partial derivative $\frac{\partial f}{\partial x}(x,y)$ is not continous.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ +1. $f(x)=x^2\sin(\frac{1}{x}),x\neq 0,f(0)=0$ works in one dimension: $\lim\sup_{x\rightarrow 0}f'(x)=1\neq \lim\inf_{x\rightarrow 0}f'(x)=-1$ $\endgroup$ – Peter Melech Jan 7 '19 at 13:13
  • $\begingroup$ You're right, but since Diego asked for multivariable and spoke of $(a,b)$, I thought he'd need an example in two dimensions. ^^ $\endgroup$ – Student7 Jan 7 '19 at 13:15
  • $\begingroup$ Of course! Your answer is excellent, my comment was just supplementary! $\endgroup$ – Peter Melech Jan 7 '19 at 13:18
  • $\begingroup$ Thank you for your answer ;) Have a very nice day ! $\endgroup$ – Diego H Jan 7 '19 at 13:36
  • $\begingroup$ Thank you, i wish you happy new year and a nice day, too. ^^ $\endgroup$ – Student7 Jan 7 '19 at 13:41

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.