Mixed successive partial derivatives: On the statement of Schwarz's theorem

Question

I have seen many "variations" of Schwarz's theorem on mixed partial derivatives. I found one in a lecture note that states that:

Given f: $A \subseteq \mathbb{R}^n \longrightarrow \mathbb{R}$. A is an open set. $x \in A$

If the following statements are true:

1. f is differentiable at x.
2. Given $i, j = 1,2,...,n$ (i,j are fixed now). These two derivatives: $\frac{\partial f^2}{\partial x_i\partial x_j}$ and $\frac{\partial f^2}{\partial x_j\partial x_i}$ exist at any point in a neighborhood of x, and they are continuous at x.

then $\frac{\partial f^2}{\partial x_i\partial x_j} = \frac{\partial f^2}{\partial x_j\partial x_i}$

I want to know if the theorem above is correct. Is the existence at all points in a neightborhood of x and the continuity at x of all second derivatives a necessary condition for the theorem to hold?

Answer 1

This theorem is also known as Clairaut's theorem.

If you want $\frac{\partial f^2}{\partial x_i \partial x_j}(x) = \frac{\partial f^2}{\partial x_j \partial x_i}(x)$ for a fixed point $x \in A$, it is sufficient for $\frac{\partial f^2}{\partial x_i \partial x_j}$ and $\frac{\partial f^2}{\partial x_j \partial x_i}$ to exist in an open neighborhood about $x$ and for them to be continuous at $x$. Note that the mixed partial derivatives only commute at the point $x$, and not necessarily at any other point. If you want $\frac{\partial f^2}{\partial x_i \partial x_j}(y) = \frac{\partial f^2}{\partial x_j \partial x_i}(y)$ for all $y$ in an open neighborhood about $x$, then the prior reasoning says that it suffices to have $\frac{\partial f^2}{\partial x_i \partial x_j}$ and $\frac{\partial f^2}{\partial x_j \partial x_i}$ exist in that open neighborhood about $x$ and for them to be continuous at each point in the neighborhood.

Here is the proof for $\mathbb{R}^2$, although the proof generalizes. Note that the statement of the theorem in this source says "continuous near $(a, b)$" but continuity is only used at the point $(a,b)$ specifically. Continuity at $(a,b)$ is used only in the very last line of the proof.