Classes of continuously differentiable functions

Note: this page covers the use of $C^k$ on $U$, $C^k(U)$ for $U\subseteq\mathbb{R}^n$ and $C^k(\mathbb{R}^n)$ such.

Definition

Given $U\subseteq\mathbb{R}^n$ (where $U$ is open) and some $k\ge 0$, a function of the form:

• $f:U\rightarrow\mathbb{R}^m$ is given.

We may say $f$ is:

• Of class $C^k$ or
• $k$-times differentiable

if $f$ has the following two properties:

1. All partial derivatives of $f$ of order $\le k$ exist and
2. All the partial derivatives are continuous on $U$

Explicit cases

Class	Property	Meaning
$C^0$	All continuous functions on $U$	Here we take it as the class of functions whose zeroth-order partial derivatives exist and are continuous This is simply the function itself.
$C^\infty$	Contains functions that are of class $C^k$ for all $k\ge 0$	This is essentially a limit definition (see also smooth and diffeomorphism)

Notations

Of class $C^k$ on $U$

To say a function is of class $C^k$ on $U$ we require:

• Here we must have $U\mathop\subseteq_{\text{open} }\mathbb{R}^n$ (with the standard topology (see Euclidean space))
• For an $f$ of class $C^k$ on $U$ we know:
  • All of the partial derivatives of $f:U\rightarrow\mathbb{R}^m$ (of order $\le k$) exist and are continuous on $U$
  • We do not know even what $m$ is.

Of class $C^k(U)$

This is a set. It can be constructed and one can (sensibly) write $f\in C^k(U)$ (where as $f\in C^k\text{ on }U$ wouldn't be suitable and doesn't tell us the co-domain of $f$)

• Here $U\subseteq\mathbb{R}^n$ must be open (as before).
• By definition we denote:
  • The set of all real valued functions of class $C^k$ on $U$ as $C^k(U)$
• All members are real-valued functions - these are functions with co-domain $\mathbb{R}$

This means that for:

• A given open $U\subseteq\mathbb{R}^n$

To say

• $f\in C^k(U)$ means that $f:U\rightarrow\mathbb{R}$ has continuous partial derivatives of all orders up to or equal to $k$ on $U$

Unresolved issues

Warning: this section contains conflicts or ambiguities I am trying to resolve

1. According to smooth on $A$ works for an arbitrary $A\subseteq\mathbb{R}^n$ however:

   • He defines smooth as being of class $C^k$ for all $k\ge 0$

   So to be smooth on $A$ is to be of class $C^k$ on $A$ forall $k$ - but we only define being of class $C^k$ for open subsets.

2. Other content on page 645

See also

• Index of notation