Symmetry of second derivatives

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$, and let $a \in U$. Suppose that $f$ is differentiable at each point of $U$, and that the derivative $Df \colon U \to L(\mathbb{R}^n, \mathbb{R}^m)$ is differentiable at $a$. Then the second derivative $D^2 f(a)$ is a symmetric bilinear map: $D^2 f(a)(u, v) = D^2 f(a)(v, u)$ for all $u, v \in \mathbb{R}^n$. In particular, $\dfrac{\partial^2 f}{\partial x_i \, \partial x_j}(a) = \dfrac{\partial^2 f}{\partial x_j \, \partial x_i}(a)$ for all $i, j$.

Remarks

Here $D^2 f(a)$ is the iterated derivative of order $2$, and $D^2 f(a)(e_i, e_j)$ is the mixed partial derivative $\partial_i \partial_j f(a)$. The hypothesis is that $f$ is twice differentiable at $a$. It holds when the second partial derivatives are continuous near $a$, which gives Schwarz's theorem. The existence of the second partial derivatives at $a$ is not enough: there is a function with unequal mixed partial derivatives.

History

Euler published a proposed proof in 1740, and many incomplete proofs followed over more than a century. In 1867 Lindelöf analysed the earlier proofs and gave a counterexample to the general claim. Schwarz gave the first rigorous proof six years later, and Jordan gave in 1883 the general version that most textbooks use.

Explanations

L1 human

The theorem says that the order of two differentiations does not matter, when $f$ is regular enough. For $f(x, y) = x^3 y^2$, differentiating first in $x$ and then in $y$ gives $6 x^2 y$, and the other order gives $6 x^2 y$ too.