Abstract Nonsense

Crushing one theorem at a time

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.

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June 4, 2011 - Posted by | Analysis, Uncategorized | , , , , , , , ,

9 Comments »

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