Smoothness

The smoothness of a function is the number of continuous derivatives it has. When applied to multivariate functions, the smoothness defines the number of continuous partial derivatives.

A function $f$ can belong into a differentiability class, which is a measure of its smoothness.

Definition §

Smoothness ^definition

Let $f$ be a function. Then:

$$f\in C^n⟺f^{\prime },f^{\prime \prime },\dots ,f^n\text{ exist and are continuous}$$

This is sometimes referred to as ”$f$ is of order/class $C^n$“. $n$ is the number of continuous derivatives $f$ has. In some cases a function may be $C^n$ over a certain subset of its domain.

Smoothness (Multivariate) ^definition-multivariate

Let $f(x_1,x_2,\dots ,x_k)$ be a multivariate function Then:

$$f\in C^n⟺\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots ,\frac{\partial f}{\partial x_k},\dots ,\frac{{\partial }^{2}f}{\partial x^2},\dots ,\frac{{\partial }^{n}f}{\partial x_k^n},\text{ exist and are continuous}$$

This is sometimes referred to as ”$f$ is of order/class $C^n$“. $n\cdot k$ is the number of continuous partial derivatives $f$ has.

Note, if a function is $C^n$, it is automatically also $C^{n-1},C^{n-2},\dots ,C^0$.

• All continuous function are $C^0$, even if they are not differentiable.
• All differentiable function that have a continuous derivative are $C^1$
• Functions can be infinitely smooth. They are of class $C^{\infty }$

Theorems §

T1: All $C^1$ functions are differentiable

T2: Smoothness means partial derivatives can be swapped ^t2

$$f\in C^n⟹f_{x_1,x_2,\dots ,x_n}=f_{\text{any permutation of }{x}_1\dots x_n}$$

If a function $f$ is of order $C^n$ then we can change the order of evaluation when evaluating mixed partial derivatives.

• If $f(x,y)\in C^2$ then $f_{xy}=f_{yx}$ or $\frac{\partial f}{\partial x\partial y}=\frac{\partial f}{\partial y\partial x}$

This theorem is a direct result from the continuity theorem on partial derivatives

Examples §

1: Showing a function is $C^1$

Where is the function below $C^1$?

$$f(x,y)=(x^2+y^2{)}^{3/4}$$

First, we need to check if $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist. Then we need to check where they are continuous to show $C^1$.

• $\frac{\partial f}{\partial x}=2x\cdot \frac{3}{4}\cdot (x^2+y^2{)}^{-1/4}=\frac{3x}{2\cdot (x^2+y^2{)}^{1/4}}$ (Chain Rule)
• $\frac{\partial f}{\partial y}=\frac{3y}{2\cdot (x^2+y^2{)}^{1/4}}$ (By symmetry)

Note that these functions don’t exist at $(x,y)=(0,0)$, as that would result in a division by zero. However, we can still show the derivative at the point using differentiation by first principles. First, we find $\frac{\partial f}{\partial x}$ by fixing $y$ to 0.

$$\frac{\partial f}{\partial x}{|}_{(0,0)}=\underset{h\to 0}{\lim }\left[\frac{f(h,0)-f(0,0)}{h}\right]=\underset{h\to 0}{\lim }\frac{(h^2+0{)}^{3/4}-0}{h}=\underset{h\to 0}{\lim }\sqrt{h}=0$$

Then, using the symmetry, we can also show that $\frac{\partial f}{\partial y}{|}_{(0,0)}=0$

Now, we need to check if these partial derivatives are continuous:

• When $(x,y)\ne (0,0)$, both $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are simply compositions of polynomial functions. Hence by continuity theorems 1 and 2, and because polynomials are continuous everywhere functions, we know that these partial derivatives are continuous over their domain.
• However, we still need to check and show that the limits ${\lim }_{(x,y)\to (0,0)}{f}_x=0$ and ${\lim }_{(x,y)\to (0,0)}{f}_y=0$.
• #todo Sandwich theorem with bounds $-\frac{3}{2}\sqrt{x}$ and $\frac{3}{2}\sqrt{x}$ Hence, $f$ is $C^1$ for all $(x,y)\in {\mathbb{R}}^2$.