Why does the order not matter? Partial D

Question

When taking partial derivatives, why does the order not matter as long as the function is continuous?

Any proof, intuitive or rigorous?

Answer 1

Let $f:\mathbb{R}^2\to\mathbb{R}$ have continuous second derivatives on an open $U\subset\mathbb{R}^2$. Fix $y_0,y_1$ and let $$g(x):=\int_{y_0}^{y_1}\frac{\partial f}{\partial y}(x,y)dy=f(x,y_1)-f(x,y_0).$$By assumption, $\partial f/\partial y$ is continuously differentiable, thus one can differentiate inside the integral and obtain $$\frac{dg}{dx}=\int_{y_0}^{y_1}\frac{\partial^2f}{\partial x\partial y}(x,y)dy,$$or$$\frac{\partial f}{\partial x}(x,y_1)-\frac{\partial f}{\partial x}(x,y_0)=\int_{y_0}^{y_1}\frac{\partial^2f}{\partial x\partial y}(x,y)dy.$$Fix $x,y_0$ and differentiate with respect to $y_1$ to obtain $$\frac{\partial^2f}{\partial y\partial x}(x,y_1)=\frac{\partial^2f}{\partial x\partial y}(x,y_1).$$Note that actually, for this proof it is sufficient to assume continuity of $\partial^2f/\partial x\partial y$, and existence of $\partial f/\partial y$. To begin with, one doesn't even have to know that $\partial^2f/\partial y\partial x$ exists.