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!
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.