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$F (x ,y) = |x| + |y|$ 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?

cmi
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  • The question in the title is completely different from the question in the post – Jack M Jul 24 '18 at 07:40
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    Let $f$ be differentiable in a point $x$ and $Df$ denote ist derivative which is a linear map. Then all directional derivatives $D_vf$ exist and we have $D_vf = Df(v)$. Therefore the following condition is necessary for $f$ being differentiable in $x$: All directional derivatives $D_vf$ exist and $D_{v+w}f = D_vf + D_wf$ for all $v,w$ (note that $D_{\lambda v}f = \lambda D_vf$ is trivial). I do not know whether this condition is sufficient, but I am sure you will find an answer somewhere in the literature. – Paul Frost Jul 24 '18 at 08:03

3 Answers3

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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.

Theo Bendit
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  • But if I can show one counter example then I will be done. Here can I not consider the case when $hk$ not equal to $0$?@Theo Bendit – cmi Jul 24 '18 at 06:19
  • @cmi You absolutely can. If you put in the caveat about $hk \neq 0$, then the proof is perfect. I'm just nit-picking, because your formula for the directional derivative is only almost always true rather than always true. – Theo Bendit Jul 24 '18 at 06:22
  • I have edited..Is it now okay?@Theo Bendit – cmi Jul 24 '18 at 06:47
  • @cmi Yes, it's fine. Like I said, there wasn't anything terribly wrong before; I was just nit-picking. :-) – Theo Bendit Jul 24 '18 at 06:49
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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.

Calvin Khor
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  • Yea..I got you. But if I get directional dervative in any direction is a non linear function , then I can tell that the function is not differentiable at that point.@Calvin Khor – cmi Jul 24 '18 at 06:40
  • @cmi yes. Nonexistence of directional derivative implies nonexistence of derivative. And not the other way round (which is the point I'm making) – Calvin Khor Jul 24 '18 at 06:41
  • No directional derivative may exist but if it is not linear then only I think we can say that the function is not differentiable at that point.If directional derivative itself does not exist, then it must not be differentiable at that point.. Calvin Khor – cmi Jul 24 '18 at 06:43
  • yes, there is only one candidate map as a (Frechet) derivative, and this is the map you are claiming is not linear – Calvin Khor Jul 24 '18 at 06:48
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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.

Jens Schwaiger
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