Twice Continuous Differentiability on $\mathbb{R}^n$ ¶

Definition¶

Let $E\subseteq \mathbb{R}^n$ be open and let $\boldsymbol{f}\colon E\to\mathbb{R}^k$ be a function.

The function $\boldsymbol{f}$ is continuously differentiable, or $C^1$ , on $E$ if all first-order partial derivatives of $\boldsymbol{f}$ exist on $E$ and are continuous on $E$ .

The function $\boldsymbol{f}$ is twice continuously differentiable, or $C^2$ , on $E$ if all second-order partial derivatives of $\boldsymbol{f}$ exist on $E$ and are continuous on $E$ .

Interpretation¶

A $C^1$ function has first-order partial derivatives that change without jumps, so its slopes in the coordinate directions at nearby points are close to each other. $C^2$ asks the same one level deeper, of the second-order partial derivatives.

Three graphs: |x|, continuous but not differentiable at 0; x|x|, continuously differentiable but not twice; and x^2, twice continuously differentiable.

The three graphs show the classes for $n=1$ $1$ . $|x|$ is continuous but has a corner at $0$ , so it is not differentiable there; $x|x|$ has derivative $2|x|$ , continuous but again with a corner at $0$ , so it is $C^1$ but not $C^2$ . Each class lies inside the previous one: $C^2\subseteq C^1\subseteq C^0$ $C^{0}$ .

Continuity of the partial derivatives is required, not just their existence: the function $f(x,y)=xy/(x^2+y^2)$ with $f(0,0)=0$ has both partial derivatives at every point, yet it equals $\tfrac{1}{2}$ on the line $y=x$ away from the origin, so it is not even continuous at the origin. Continuous partial derivatives rule this out: a $C^1$ function is totally differentiable, as the Key Results state.

This is the $n$ -variable form of the differentiability class from Math I, with partial derivatives in place of $f'$ and $f''$ . $C^2$ is the assumption behind the symmetry of mixed partial derivatives, hence of a symmetric Hessian, and of the second-order Taylor approximation.

A common slip is to read "differentiable everywhere" as $C^1$ . In one variable, $f(x)=x^2\sin(1/x)$ with $f(0)=0$ is differentiable at every point, but $f'(x)=2x\sin(1/x)-\cos(1/x)$ for $x\neq0$ has no limit at $0$ , so $f$ is not $C^1$ .

Key Results¶

A $C^1$ function on an open set is totally differentiable at every point of that set.
A $C^2$ scalar function has symmetric mixed second derivatives.
Polynomials and the usual elementary smooth functions are $C^2$ on the interior of their domains.

Examples¶

The function $f(x,y)=x^2+\sin y$ is $C^2$ on $\mathbb{R}^2$ .
The function $f(x)=|x|$ is not $C^1$ on $\mathbb{R}$ , because it is not differentiable at $0$ .

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 1, Lecture 1

Aliases: twice continuously differentiable Keywords: twice continuously differentiable, C2 function on Rn, C1 function on Rn Symbols: real numbers, subset

Last updated: 20 October 2026 03:02

Twice Continuous Differentiability on Rn\mathbb{R}^nRn¶