2
$\begingroup$

Can we find a function $f:\mathbb R^n\to\mathbb R$ that such that $f$ is continuous and $\partial_v f(p)$ exists for all $p\in\mathbb R^n$ and $v\in\mathbb R^n$. But $f$ is not differentiable at $0$?

Is such function $f$ exists?

Here give a example that has directional derivative everywhere, but it's not continuous at the origin.

$\endgroup$
3
$\begingroup$

Consider the polar coordinates $\ (r\,\ \Phi)\ $ in $\ \mathbb R^2.\ $ Function $ f:\mathbb R^2\rightarrow\mathbb R,\ $ which in polar coordinates is given by:

$$ f(r\,\ \Phi)\,\ :=\,\ r\cdot\sin(3\cdot\Phi) $$

is continuous everywhere, is infinitely differentiable outside the origin $\ O,\ $ and $\ f\ $ has all directional derivative at $\ O,\ $ but $\ f\ $ is not differentiable at $\ O.$

$\endgroup$
  • 1
    $\begingroup$ It clearly doesn’t have a derivative in direction $(x,y) = (1,1)$ at $O$. $\endgroup$ – arseniiv Dec 16 '17 at 4:10
  • $\begingroup$ @arseniiv, I've fixed it. Thank you. I was indeed careless. $\endgroup$ – Wlod AA Dec 16 '17 at 4:43
  • 1
    $\begingroup$ Yeah, with 3 it’s great now! $\endgroup$ – arseniiv Dec 16 '17 at 5:01
  • $\begingroup$ @arseniiv, thank you. Thank you TWICE. :) $\endgroup$ – Wlod AA Dec 16 '17 at 5:52
0
$\begingroup$

On $\mathbb R^2$ define

$$ f(x,y) = \begin{cases} \dfrac{xy^2}{x^2+y^2}& (x,y)\ne (0,0) \\ 0 & (x,y)= (0,0)\end{cases}$$

Then $f\in C^\infty(\mathbb R^2 \setminus \{(0,0)\}),$ and $f$ is continuous at $(0,0).$ For any unit vector $u= (a,b),$ we have

$$\tag 1 \frac{f(tu)}{t} = ab^2.$$

Thus $f$ has directional derivatives at $(0,0)$ in all directions.

Suppose $Df(0,0)$ exists. Because $f$ is $0$ on the axes, all partial derivatives of $f$ at $(0,0)$ vanish, so $Df(0,0)$ must be the $0$ linear transformation. Thus $f(x,y) = o((x^2+y^2)^{1/2}).$ But that violates $(1),$ so we have a contradiction. This proves $f$ is not differentiable at $(0,0).$

$\endgroup$

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.