When is a function of the type $\mathbb{R}^2 \rightarrow \mathbb{R}$ $C^2$?

Question

A function $f$ of the type $\mathbb{R}^2 \rightarrow \mathbb{R}$ is $C^1$ if $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous. What does it mean for a $f$ to be $C^2$? Is it enough that $\frac{\partial^2 f}{\partial x^2}$ and $\frac{\partial^2 f}{\partial y^2}$ are continuous or does $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$ have to be continuous as well?

Answer 1

The definition of $C^2$ are functions such that its derivatives are in $C^1$. Therefore, in particular, $\frac{\partial f}{\partial x}$ must be in $C^1$. So, both its partial derivatives $\frac{\partial^2f}{\partial y\partial x}$ and $\frac{\partial^2f}{\partial x^2}$ exist and are continuous. The same condition is required of $\frac{\partial f}{\partial y}$.

Answer 2

A function is $C^{k+1}$ when its partial derivatives exist and are $C^k$. So let $f \in C^{2}(\mathbb R^2)$: this means that $\frac{\partial f}{\partial x}$ and the analogous partial w.r.t. $y$ exist and must both be $C^1(\mathbb R^2)$, that is their partials must exist and be continuous (all four). This can easily be extended to functions over $\mathbb R^n$, you just get more partials to deal with.

Functions like this enjoy many properties. Since $\mathbb R^2$ is open, Schwarz’s Lemma ensures you that the two mixed second-order partials are the same continuous function over $\mathbb R^2$. Also, the first order partials turn out to be differentiable functions (which is a slightly stronger condition that the existence of partials).