# Abstract Nonsense

## Functions of Class C^k

Point of Post: In this post we define the notion of classes of differentiability and discuss what the membership in some of these classes implies about the function.

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Motivation

Since differentiability and related notions are our focus for right now, it seems prudent that we should define some kind of notation which shortens saying things like “$f$ has partial derivatives of all types of order $7$“, ” $f$ has partial derivatives of all types of order $1$ and each partial derivative is continuous”, or ” $f$ has partial derivatives of all orders and all types”. This is taken up by the notion of $C^k$ classes which, put simply tells you how well-behaved the function is.

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$C^k$ Classes

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Let $f:E\to\mathbb{R}$ where $E\subseteq\mathbb{R}^n$. We say that $f\in C^k(U)$ ( with $k>1$) if $U\subseteq E$ is open in $\mathbb{R}^n$ and $f$ has partial derivatives of all types of order $k$ and that $D_{(i_1,\cdots,i_k)}f$ is continuous on $U$ for every possible $(i_1,\cdots,i_k)$. We say that $f\in C^0(U)$ if $f$ is continuous on $U$. We often say $f\in C^k(U)$ without specifying any larger domain $f$ may be defined on. We first note the following facts:

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Theorem: Let $f\in C^1(U)$, with $U\subseteq\mathbb{R}^n$, then $f$ is total differentiable everywhere on $U$.

Proof: This follows immediately from our previous theorem pertaining to this. $\blacksquare$

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Consequently, recalling that total differentiability implies continuity we have the following corollary:

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Corollary: For any open $U\subseteq\mathbb{R}^n$ we have that $C^0(U)\supseteq C^1(U)$.

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In fact, taking this a step further it’s easy to see that

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Theorem: Let $U\subseteq\mathbb{R}^n$ be open. Then,

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$C^0(U)\supseteq C^1(U)\supseteq C^2(U)\supseteq C^3(U)\supseteq\cdots$

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What we next note is that order of derivatives doesn’t really matter for $C^k(U)$ functions in the following sense:

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Theorem: Let $f\in C^k(U)$. Then, for every $(i_1,\cdots,i_p)$ and $(j_1,\cdots,j_p)$ with $p\leqslant k$ we have

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$D_{(i_1,\cdots,i_p)}f=D_{(j_1,\cdots,j_p)}f$

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Proof: We know that $f\in C^p(U)$ by the prior theorem and thus we may use Clairaut’s theorem to make successive switches, and so the conclusion follows. $\blacksquare$

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If $U\subseteq\mathbb{R}^n$ is open we say that $f\in C^\infty(U)$ if $f\in C^k(U)$ for every $k\in\mathbb{N}$. We call such functions smooth.

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We extend the notion of $C^k$ functions to vector valued functions in the obvious way (same definition). Or, we could alternatively say that a mapping $f:U\to\mathbb{R}^m$ ,where $U\subseteq\mathbb{R}^n$ is open, is $C^k(U,\mathbb{R}^m)$ (with $k\in\mathbb{N}\cup\{\infty\}$)  if $f_1,\cdots,f_m\in C^k(U)$. Clearly then the theorems from the above apply now to this extended notion.

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We last note that if the domain and/or codomain of a mapping is either irrelevant or implies we are apt to just say that $f$ is $C^k$.

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References:

1.  Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.

June 4, 2011 -

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