Does the Existence of Derivatives along All Smooth Paths Guarantee Multivariable Differentiability?

Question

This question is inspired by other relevant questions on MSE regarding continuity of partial derivatives and differentiability, existence of partial derivatives and differentiability, and limits along arbitrary smooth curves.

I'm trying to get a handle on the extra bit of weirdness introduced by multivariable functions when it comes to differentiability and continuity. The applicable definition here is:

Definition (Cain & Herod, 1997): Let $f: D \rightarrow \mathbb{R}^p$, where $D \subset \mathbb{R}^n$ and let $\mathbf{x}$ be an interior point of $D$. Then $f$ is differentiable if there exists a linear function $L$ such that $\lim\limits_{\mathbf{h} \rightarrow \mathbf{0}} \dfrac{f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - L(\mathbf{h})}{\|\mathbf{h}\|} = \mathbf{0}$

So, given that the multivariable limit in the definition above requires that the limit exists (and is zero) independent of the path, can I interpret this as translating to the requirement that the derivative along any smooth path through $\mathbf{x}$ must exist?

That is, if we define a so-called path derivative $D_\mathbf{p}f$:

$D_{\mathbf{p}}f : \mathbf{x} \mapsto \lim\limits_{t \rightarrow t_0} \dfrac{f(\mathbf{x} + \mathbf{p}(t)) - f(\mathbf{x})}{\|\mathbf{p}(t)\|}$

where $\mathbf{p}$ defines any path with $t_0$ such that $\lim\limits_{t \rightarrow t_0} \mathbf{p}(t) = \mathbf{0}$,

1. If a path is defined as smooth if it is representable by an infinitely differentiable vector function, is differentiability equivalent to the existence of all path derivatives where $\mathbf{p}$ defines a smooth path?

Reading this answer, I initially suspected that this path derivative condition was identical to the total derivative which defines differentiability, but I wonder what kind of wrinkle the requirement that $\mathbf{p}$ defines a smooth path introduces. Proving that differentiability implies the existence of all such path derivatives is straightforward enough. I can't see if there's a way to prove differentiability assuming the existence of all path derivatives, however...

2. How does this condition compare to the condition of the existence of all partial derivatives and $n-1$ continuous partial derivatives in some $n$-ball containing $\mathbf{x}$ (quoted here)? Is it stronger/weaker?

At the very least, I think that this condition is not useless. For example, if you can find a smooth path for which the path derivative does not exist, then $f$ is not differentiable. In reference to this question–where one partial derivative is not defined along some line with $\mathbf{x}$ deleted–the notion of path derivatives implies that such a function is automatically not differentiable.

Answer 1

Remark. The original version of this answer contained an error which was pointed out by Vercassivelaunos. I corrected the error, but the price is that the question remains open. However, I think the answer is still useful to some extent because it corrects the definition of the path-derivative $D_{\mathbf x}f$ given in the question.

Your definition of the path derivative $D_\mathbf pf$ does not make much sense. You define

$$D_{\mathbf{p}}f(\mathbf x) = \lim\limits_{t \rightarrow t_0} \dfrac{f(\mathbf{x} + \mathbf{p}(t)) - f(\mathbf{x})}{\|\mathbf{p}(t)\|}$$

where $\mathbf{p}$ is a path defined on an open interval containing $t_0$ such that $\lim\limits_{t \rightarrow t_0} \mathbf{p}(t) = \mathbf{0}$.

The minimal requirement for a path is continuity, thus $\lim\limits_{t \rightarrow t_0} \mathbf{p}(t) = \mathbf{0}$ simply means that $\mathbf{p}(t_0) = \mathbf{0}$. Also there is no reason to work with an arbitrary $t_0$. It suffices to consider $t_0 = 0$, i.e. paths $\mathbf{p}$ defined on an open interval containing $0$ such that $\mathbf{p}(0) = \mathbf{0}$. But these are only formal technical issues. Here are more serious defects.

1. In order that $D_{\mathbf{p}}f(\mathbf x)$ be well-defined, you need to assume $\mathbf{p}(t) \ne \mathbf 0$ for $t \ne 0$.

2. Even for very nice paths $\lim\limits_{t \rightarrow 0} \dfrac{f(\mathbf{x} + \mathbf{p}(t)) - f(\mathbf{x})}{\|\mathbf{p}(t)\|}$ usually does not exist. For example, take $\mathbf p(t) = t\mathbf e_i$, where $\mathbf e_i$ is the $i$-th coordinate vector. Then $\lim\limits_{t \rightarrow 0} \dfrac{f(\mathbf{x} + t\mathbf e_i) - f(\mathbf{x})}{\lvert t \rvert}$ resembles the partial derivative $\frac{\partial f}{\partial x_i}(\mathbf x) = \lim\limits_{t \rightarrow 0} \dfrac{f(\mathbf{x} + t\mathbf e_i) - f(\mathbf{x})}{t}$. If $\frac{\partial f}{\partial x_i}(\mathbf x)$ exists and is non-zero, then $\lim\limits_{t \rightarrow 0} \dfrac{f(\mathbf{x} + t\mathbf e_i) - f(\mathbf{x})}{\lvert t \rvert}$ clearly does not exist.

Therefore a better definition is $$D_{\mathbf{p}}f(\mathbf x) = \lim\limits_{t \rightarrow 0} \dfrac{f(\mathbf{x} + \mathbf{p}(t)) - f(\mathbf{x})}{t}$$ which can be considered for any path with $\mathbf{p}(0) = \mathbf{0}$. This generalizes the directional derivatives $$D_{\mathbf v}f(\mathbf x) = \lim\limits_{t \rightarrow 0} \dfrac{f(\mathbf{x} + t\mathbf v) - f(\mathbf{x})}{t}$$ with $\mathbf v \in \mathbb R^n$: Just take $\mathbf p_\mathbf v (t) = t \mathbf v$.

Now have a look at Generalized directional derivatives and consider $f$ which is differentiable at $\mathbf x$.

You will see that if $\mathbf{p}$ is not differentiable at $0$, then in general $D_{\mathbf{p}}f(\mathbf x)$ does not exist. See the answers to the the linked question.

We conclude that considering paths which are non-differentiable at $0$ does not make sense as a test for $f$ being differentiable at $\mathbf x$.

In contrast, if $\mathbf p$ is differentiable at $0$, then $D_{\mathbf{p}}f(\mathbf x)$ exists and agrees with the directional derivative $D_{\mathbf{p}'(0)}f(\mathbf x)$.

We conclude that a necessary condition for $f$ being differentiable at $\mathbf x$ is the existence of all $D_{\mathbf{p}}f(\mathbf x)$ with $\mathbf p$ differentiable at $0$.

An example due to Vercassivelaunos shows that is a stronger condition than the existence of all directional derivatives (the latter is not sufficient to conclude that $f$ is differentiable at $\mathbf x$, see e.g. directional derivatives implication):

Consider the function $f : \mathbb R^2 \to \mathbb R$ which is $0$ everywhere except the parabola $y=x^2$, where it is $0$ for rational and $1$ for irrational $x$. The path derivative at $0$ along the parabola does not exist, but every directional derivative does exist.

Answer 2

1. The simplest answer is yes. Without getting too deep into the analysis, you can somewhat easily show that if it is not the case for some smooth path, then it is not the case for the derivative. That is, if the gradient is defined you get a derivative, and if not, not. Try the contrapositive for the limit of the derivative at the point if you are curious.

2. These are equivalent, specifically because you can have a derivative undefined along a certain direction UNLESS we define it there. If I define or find that the derivative along one of the axes exists, and the others are all continuous in the ball, they are continuous at the point and we have all the derivatives defined. This is neither weaker nor stronger, but directly equivalent. If the derivatives are discontinuous along other axes toward the point, we have an entire subspace approaching the point with possible discts or undefined derivative curves. (If this is incorrect, please someone else correct me!)