# Equivalent definitions of $C^r(\Omega)$?

The question is motivated from the definition of $C^r(\Omega)$ I learned from S.S.Chern's Lectures on Differential Geometry:

Suppose $f$ is a real-valued function defined on an open set $\Omega\subset{\bf R}^m$. If all the $k$-th order partial derivatives of $f$ exist and are continuous for $k\leq r$, then we say $f\in C^r(\Omega)$. Here $r$ is some positive integer.

While in Folland's Introduction to Partial Differential Equations, $C^r(\Omega)$ denotes the space of functions possessing continuous derivatives up to order $r$ on $\Omega$, where $\Omega$ is an open subset of ${\bf R}^m$ and $k$ is a positive integer.

It's trivial to show that these two definitions are equivalent when $m=1$. So here is my question:

Are these two definitions equivalent in the higher dimensions? How to prove it?

Edit: The question was partially answered by Didier here. At least for me, I do not think this is a trivial question. For answering the question above, the key point, I think, is to answer the following question:

In the higher dimension case, say, $f:{\bf R}^n\to {\bf R}$($n\geq 2$), what's the relationship between the higher order partial derivative of $f$ and the high order derivative $f^{(k)}$?

When $1\leq k\leq 2$, $f^{(k)}$ is the gradient and the Hessian matrix respectively. The answer to the above questions in these two cases is clear. I have no idea for $k>2$. I don't know much about tensor, I'm not sure if the question is related to the topic of multilinear algebra.

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I think something went wrong here. Look at the exponents of your ${\mathbb R}$'s. – Christian Blatter Jun 10 '11 at 14:31
@Christian: Notation corrected. Thanks. – Jack Jun 10 '11 at 16:41

## 2 Answers

The $r$-th derivative $d^r f(x)$ of a function $f:\ {\mathbb R}^n\to{\mathbb R}$ at a point $x\in {\mathbb R}^n$ is a homogeneous polynomial of degree $r$ in the $n$ variables $X_1$, $\ldots$, $X_n$, where $X$ is a tangent vector "attached to the point $x$". The coefficients of this polynomial are the $r$-th partial derivatives of $f$ at the point $x$, and things are set up such that $x \to d^r f(x)$ is continuous as a "polynomial-valued function" iff all these partial derivatives are continuous. So the two definitions alluded to are actually the same.

In the case $r=1$ this means that $f\in C^1$ iff $\nabla f(x)$ is a continuous vector-valued function of $x$ iff the $n$ partial derivatives ${\partial f\over \partial x_k}$ are continuous.

If $f$ is vector-valued then the above applies to every component $f_i$ of $f$; but of course which enough sophistication we can forget about coordinates altogether.

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Yes they are, see here.

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@Jack: I wonder how many times you will ask/erase/re-ask somewhere else the same (trivial) question. This is an odd way of showing your appreciation of the answers you receive. – Did Jun 10 '11 at 14:37