Can the non-differentiability of a function $f:R^n \to R$ always be proved by using directional derivative?

Question

$F (x ,y) = |x| + |y|$ when $xy \neq 0$ and $F(x,y) =0 $ elsewhere.

How can I prove or disprove this function is differentiable at $(0,0)$?

My Try : The directional derivative at $(0,0)$ in the direction $(h,k)$ is $|h| + |k|$ (where $hk \neq 0$ )which is not a linear function of $(h,k)$. So it can not have differentiation at $(0,0)$.

Can anyone please tell me if I am wrong?

Answer 1

That's a fine approach. In order to be differentiable, the directional derivative must be linear, which is not the case for this function.

The only nitpick I have is with the directional derivative you've computed. It's right when $h$ and $k$ is non-zero, but the function is constant along the axes, so the directional derivatives will be $0$ whenever $(h, k)$ points along the axes.

But, either way, it's definitely not linear.

Answer 2

For the question in the title, no. If $f(x)=1$ if $0<x=y^2$ and zero otherwise, the directional derivative at 0 in any direction is 0, but the function is not even continuous at zero.

Answer 3

The question in the title asks, at least in my understandig, also whether there are functions for which no directional derivatives exist at all. In dead such (continuous) functions exist. Take any continuous nowhere differentiable function $g\colon\mathbb{R}\to\mathbb{R}$. Then $f\colon\mathbb{R}^2\to\mathbb{R}$ with $f(x,y):=g(x)+g(y)$ has the properties to be continuous and to have no directional derivatives at all.

Edit: I'm rather sure that my claim is true, but it seems to be not so easy to proof it, as the fact that two limits do not exist does not always imply that the limit of the sum also does not exist.