By being differentiable, I mean a function of several real variables $f: \mathbb{R^m}\rightarrow \mathbb{R}$ is said to be differentiable at a point $x_0$ if there exists a linear map $J: \mathbb{R}^m → \mathbb{R}$ such that $\lim_{\mathbf{h}\to \mathbf{0}} \frac{\|\mathbf{f}(\mathbf{x_0}+\mathbf{h}) - \mathbf{f}(\mathbf{x_0}) - \mathbf{J}\mathbf{(h)}\|}{\| \mathbf{h} \|} = 0.$

By having a gradient at $x_0$, I mean there exists a vector denoted as $\nabla f$ such that its dot product with any unit vector $v$ is the directional derivative of $f$ along $v$ at the point $x_0$. That is, $$\nabla f \cdot v = D_v f(x_0)$$

We assume all the directional derivative of $f$ at $x_0$ exists, that being said $f$ may or may not be differentiable at $x_0$, if in addition we assume the gradient exists at $x_0$, then $f$ should be differentiable at $x_0$, right?

I think this is simple or well known question, I can visualize it in the following way: if $f$ has a gradient at $x_0$ then all its directional derivative should lie in the same plane defined by $\nabla f$, therefore it should be differentiable using the definition above. But surprisingly I couldn't find any rigorous proof or material discussing this. Is this too simple or too well-known?

  • $\begingroup$ That’s not the usual meaning that I’ve seen for “having a gradient at $x_0$.” What I’ve always seen it mean in textbooks is only that $\nabla f(x_0)$ exists, i.e., that all of the partial derivatives at $x_0$ exist. That’s not enough to guarantee differentiability, nor is existence of all directional derivatives. $\endgroup$ – amd Apr 18 '19 at 2:08
  • $\begingroup$ I know the existence of all the partial derivatives does NOT guarantee differentiability. My definition of having a gradient at $x_0$ is from wikipedia. en.wikipedia.org/wiki/Gradient#Definition. I don't know if I should assume the uniqueness of the gradient, but I think it should be the same, you can not have two different gradient vectors which are consistent with all the directional derivatives at the same point. $\endgroup$ – Pew Apr 18 '19 at 2:13
  • $\begingroup$ If it exists, then how can it not be unique? If $u\cdot v=u\cdot w$ for all $u$, then $v=w$. $\endgroup$ – amd Apr 18 '19 at 2:15
  • $\begingroup$ Yes, that is the point I realized the same thing. $\endgroup$ – Pew Apr 18 '19 at 2:18

Consider the function $$ f(x,y) = \begin{cases} 1, & y=x^2 \wedge x \ne 0; \\ 0, & \mathrm{otherwise}. \end{cases} $$ This function $f$ satisfies your gradient condition at the origin with $\nabla f(0, 0) = (0, 0)$, yet it is not even continuous at $(0,0)$.

If you want a function which is continuous at $(0,0)$ and has a gradient in your sense, but is still not differentiable, instead let $f(x,y) = \sqrt{|x|}$ along the parabola $y=x^2$.

  • $\begingroup$ I found this wiki subpage calculus.subwiki.org/wiki/…. It seems that existence of gradient vector implies the differentiablity according to this page $\endgroup$ – Pew Apr 18 '19 at 2:20
  • $\begingroup$ The definition of gradient linked from that page essentially requires that the function satisfies the definition of differentiability with $J(\mathbf{h}) = \nabla f(\mathbf{x}) \cdot \mathbf{h}$, which is stronger from your definition of gradient. $\endgroup$ – Daniel Schepler Apr 18 '19 at 2:38
  • $\begingroup$ Ok, thanks, but I didn't find the requirement $J(h) = \nabla f(x) \cdot h$, where is the link for that requirement? $\endgroup$ – Pew Apr 18 '19 at 2:46
  • $\begingroup$ The "Formal epsilon-delta definition" section of calculus.subwiki.org/wiki/Gradient_vector is equivalent to $\lim_{\bar h\to 0} \frac{\lVert f(\bar c + \bar h) - f(\bar c) - \bar v \cdot \bar h \rVert}{\lVert \bar h \rVert} = 0$, if you expand that limit in terms of an epsilon-delta definition and then "substitute" $\bar h = \bar x - \bar c$. $\endgroup$ – Daniel Schepler Apr 18 '19 at 2:58

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