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Let $C^{\infty}[a,b]$ be the space of all infinitely differentiable functions on [a,b] with norm $$ || f || = \max _{[0,1]} | f(x) | , f \in C^{\infty}[a,b]$$

Is the differentiation operator $\frac{d}{dx}$ a contraction mapping on $C^{\infty}[a,b]$?

I'm confused. Operators are not until later in the textbook, and the contraction mappings I've worked with are using the Mean Value Theorem which are referencing a derivative? Ah!

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Questions for you: (1) What is the definition of contraction mapping? That is, what exactly is the property you're trying to see whether or not $\frac{d}{dx}$ has? (2) Have you tried some examples of functions $f$ in the space, computing the norms of $f$ and $\frac{df}{dx}$, to gather evidence as to whether or not it is plausible? –  Jonas Meyer May 14 '13 at 5:45
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Don't worry about the word "operator". It is just asking about the map $f\mapsto f'$ from $C^\infty[a,b]$ to itself. –  Jonas Meyer May 14 '13 at 5:47
    
I know intuitively it is a mapping that when we're measuring distance in one space (a norm) brings the things closer together for all elements in the interval (so we say it is a contraction mapping on this interval or subset?). The actual definition from my textbook is $||Ax, Ay|| \leq \alpha ||x, y||$ for $\alpha \in (0,1)$ then A is a contraction. should I just try for any smooth function? –  Questioneer May 14 '13 at 5:51
    
Questioner: You mean $\|Ax-Ay\|\leq\alpha\|x-y\|$? I would suggest trying a variety of simple examples of functions, or drawing pictures, to see what is going on, how $\|f\|$ and $\|\frac{df}{dx}\|$ are related. A rougher version of the question is: If $f$ is small, must $\frac{df}{dx}$ be small? –  Jonas Meyer May 14 '13 at 5:53
    
Thank you for your help, Jonas. I first thought of $sin(x)$ and $cos(x)$, the magnitude or amplitude of these functions will not decrease after a derivative. I would think that would be enough just in the Euclidean space with the standard distance, how do I incorporate this norm? –  Questioneer May 14 '13 at 5:58
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up vote 1 down vote accepted

It's not a contraction.

Let $f(x)=\sin nx$, $g(x)=\cos nx$ ($n$ is a number we need to find). Then $f,g\in C^{\infty}[a,b]$. And \begin{align*} \left\|f-g\right\|=&\underset{[a,b]}{\max}|\sin nx-\cos nx|=\sqrt{2}\underset{[a,b]}{\max}|\sin(nx-\frac{\pi}{4})| \\ \left\|\frac{df}{dx}-\frac{dg}{dx}\right\|=&\left\|n\cos nx+n\sin nx\right\|=n\underset{[a,b]}{\max}|\cos nx+\sin nx|=\sqrt{2}n\underset{[a,b]}{\max}|\sin(nx+\frac{\pi}{4})| \end{align*} If we take $n$ large enough that $n\ge\frac{2\pi}{b-a}$, then \begin{align*} \left\|f-g\right\|=\sqrt{2},\qquad\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|=\sqrt{2}n. \end{align*} If we also make $n\ge1$, $$\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|\ge\left\|f-g\right\|.$$ Therefore the operator $d/dx$ is not contraction.

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