Regularity conditions needed for divergence to equal "flux density"? (I.e. for "coordinate-free definition" to be valid?)

Question

Background:
Let $\vec{V} : \mathbb{R}^n \to \mathbb{R}^n$ be treated as a vector field, $V_i$ denote the corresponding scalar coordinate functions $\mathbb{R}^n \to \mathbb{R}$, and then if these partial derivatives exist define $\operatorname{div}(\vec{V}) = \sum_{i=1}^n \partial_{x_i} V_i$. For the sake of simplicity below feel free to assume $n=2$.

Either by first proving the divergence theorem and then deriving this as a consequence, or first proving this (and then deriving the divergence theorem from it) something to the effect of the following is stated in many sources:

• "the divergence at a point is equal to the limit of the ratio of the flux of $\vec{V}$ through the boundary of an enclosing region as the volume of that region approaches zero", and
• "it doesn't matter what the precise sequence of enclosing regions are, as long as their volumes approach zero".

Questions:

1. Is it necessary, but not sufficient, for the volumes of the enclosing regions to approach zero?

   Consider for example enclosing regions for which the diameters (maximum distance between any two points) do not approach zero even as the volumes approaches zero -- presumably such regions could continue to "catch flux away from the point of interest" (via "a long, skinny part") even as the volumes technically approach zero.

2. Does $\vec{V}$ need to be continuously differentiable? (Equivalently totally/Frechet differentiable with continuous partial derivatives?)

   Or does it suffice merely to be totally differentiable? Gateaux differentiable? Or even only the $\partial_{x_i} V_i$ need to exist and not necessarily be continuous (which would suffice for the divergence as defined above and in most sources to exist)?

Additional context:
To the extent that this result is equivalent to the divergence theorem (but is it equivalent, or is one stronger than the other?), the regularity conditions needed of $\vec{V}$ for the divergence theorem would suffice to answer the second question.

However the conditions of the divergence theorem of course would not address the first question about regularity conditions on the limiting sequence of enclosing regions.

In the case of $n=2$ I have tried to prove the result using square contours parallel to the $x$ and $y$ axes, centered at the point in question, and with area $\epsilon^2$ as $\epsilon \to 0$. I haven't been able to make the proof work even assuming that $\vec{V}$ is totally differentiable, so I suspect that the continuous differentiability assumption is necessary (although I am bad at proofs).

I am also not familiar with many examples of totally differentiable functions that aren't also continuously differentiable, so am not sure what to check for potential counterexamples. I suspect that for the Gateaux differentable vector field $\vec{V}(x,y) = |y|$ that the limit of the flux surrounding $(x_0,0)$ depends on the shape of the enclosing contour but haven't tried yet to prove or disprove it. (Because I'm more interested in the totally differentiable case.)

In Kreyszig, Advanced Engineering Mathematics, in the statement of the divergence theorem for $n=3$ (chapter 10.7) it is assumed that $\vec{V}$ is continuously differentiable (in order to allow turning a multiple integral of one of the partial derivatives into an iterated integral inside of the proof). In formula (2) of chapter 10.8 of the same book, the result about "divergence = flux density" is stated (for $n=3$) with the requirement that the diameters of the enclosing regions approach zero, not merely that their volumes approach zero.

When $n=1$, the analogous statement that the derivative equals the limit of the difference quotient only requires the function in question to be differentiable, not necessarily continuously differentiable. So it would be interesting to me if for $n \ge 2$ suddenly additional regularity assumptions (beyond being totally differentiable) are needed.

This question is related but the answers to that question do not address what is asked here: Conditions for existence of divergence of a vector field Specifically I already understand the distinction between Gateaux differentiability, Frechet differentiability, and continuous differentiability.

The comments on that question do seem to strongly suggest (if not outright state) that continuous differentiability is not only sufficient, but necessary. However, no explicit counterexamples of a totally differentiable but not continuously differentiable vector field for which the result fails are given. Moreover, the discussion for that question does not address the concern about "diameters approaching zero" vs. "merely volumes approaching zero" that are the first part of this question.

Answer 1

This attempted proof outline suggests that (1) totally differentiability is sufficient (continuous partial derivatives are not necessary) and (2) the diameter of the enclosing curves/hypersurfaces does need to go to zero. This should be considered circumstantial evidence at best and not an answer to the question.

I've only verified the lemma used in the case of $\mathbb{R}^2$ though, nor have I yet set aside the time to try to do the proof from start to finish with rigorous $\epsilon$/$\delta$ accounting.

1. If $f$ is totally differentiable, then for enclosing hypersurfaces whose diameter is going to zero, the error from using the total derivative at the point in question in place of $f$ itself when computing the flux can be bounded / controlled.

2. The total derivative is a linear function, so we can apply the Lemma that its flux through the hypersurface equals its trace times the hypervolume enclosed by the hypersurface.

Because the trace of the total derivative equals the divergence, it follows that the flux of $f$ divided by the enclosed hypervolume approaches the divergence as the diameters of the enclosing hypersurfaces approaches zero. If the diameters did not approach zero, even if the enclosing hypervolumes approached zero, there would be no guarantee that the total derivative would become a more accurate approximation in the limit.

Lemma: If $f$ is linear $\mathbb{R}^2 \to \mathbb{R}^2$, the flux of $f$ through a positively oriented triangle with vertices $(v_1, w_1)$, $(v_2, w_2)$, $(v_3, w_3)$ equals the trace of $f$ multiplied by the area of the triangle.

The above lemma can be verified using the "shoelace formula" for the area of a triangle and a tedious calculation.

Because every simple, closed, rectifiable curve can be arbitarily closely approximated by polygons, and every polygon can be subdivided into triangles, it follows after taking limits that

Lemma: If $f$ is linear $\mathbb{R}^2 \to \mathbb{R}^2$, the flux of $f$ through any simple, closed, rectifiable curve equals the area of the curve times the trace of $f$

using the standard trick that the flux of adjacent positively oriented curves (in this case the triangles in the triangular subdivision of the approximating polygon) will have the contributions cancel out from adjacent portions of the curves, i.e. both the total flux and the area enclosed is additive.

Now, I don't know whether the above lemma generalizes to $\mathbb{R}^n$ using a suitable analogue of "rectifiable hypersurface". So even if the result is true in $\mathbb{R}^2$, I don't know whether it generalizes to $\mathbb{R}^n$.

I think the error in using the total derivative at $p$ in place of $f(p)$ to approximate the flux of the hypersurface enclosing $p$ can be bounded using the definition of total derivative and the inequality $||Df(p) v || \le |Df(p)| ||v||$, but again I haven't checked this rigorously.