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Let us consider some function in $\mathbb{R}^2$ of the form $$u(x,y) = \begin{cases} \frac{x^a y^b}{x^c + y^d}, & (x,y)\neq(0,0)\\ 0, & (x,y)=(0,0)\end{cases}$$

for some $a,b,c,d\in\mathbb{N}$. I know there are functions of this sort that are continuous everywhere, especially in $(0,0)$, as well as directionally differentiable in any direction $v\in\mathbb{R}^2$ but not (totally) differentiable in $(0,0)$. If the directional derivative in $(0,0)$ is some expression that is an expression that is non-linear in the direction $v$, then can one already conclude that it cannot be differentiable in $(0,0)$? As otherwise we would have $u'((0,0))(v+w) = u'((0,0))(v) + u'((0,0))(w)$ for any $v,w\in\mathbb{R}^2$ as the differential must be a linear map, right? This would be a pretty easy way to show the the function is not differentiable.

Thanks!

HelloEveryone
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    How are you defining "linear in the direction v" ? – Dan Asimov May 16 '23 at 17:01
  • Oh sorry for the confusion, I just meant that $f'((0,0))(\cdot)$ is not a linear map (which maps in this case from $\mathbb{R}^2 \to \mathbb{R}$, so the argument would e.g. be the direction vector $v$), then it cannot be differentiable in this point. – HelloEveryone May 16 '23 at 17:06

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Yes, if you show that one of the directional derivatives does not exist or that the directional derivative is not linear in $v$ then your function will not be differentiable at the considered point.

Something devious however is that those are not necessary conditions for non-differentiability: there exist functions (of the type you described too!) for which all directional derivatives exist and for which the directional derivative is a linear function of the direction, yet the functions are still non-differentiable, see this post: Continuous function with linear directional derivatives=>Total differentiability?

Bruno B
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