derivative operator is unbounded in the sup norm

Consider $C^{\mathrm{\infty }}([-1,1])$ the vector space of functions with derivatives or arbitrary order on the set $[-1,1]$.

This space admits a norm called the supremum norm given by

$$|f|=sup\{|f(x)|:x\in [-1,1]\}$$

This norm makes this vector space into a metric space.

We claim that the derivative operator $D:(Df)(x)=f^{\prime }(x)$ is an unbounded operator.

All we need to prove is that there exists a succession of functions $f_n\in C^{\mathrm{\infty }}([-1,1])$ such that $\frac{|D(f_n)|}{|f_n|}$ is divergent as $n\to \mathrm{\infty }$

consider

$$f_n(x)=\exp (-n^4{x}^2)$$

$$(D{f}_n)(x)=-2x{n}^4\exp (-n^4{x}^2)$$

Clearly $|f_n|=f_n(0)=1$

To find $|D{f}_n|$ we need to find the extrema of the derivative of $f_n$, to do that calculate the second derivative and equal it to zero. However for the task at hand a crude estimate will be enough.

$$|D{f}_n|\ge |(D{f}_n)(\frac{1}{n^2})|=\frac{2n^2}{e}$$

So we finally get

$$\frac{|D{f}_n|}{|f_n|}\ge \frac{2n^2}{e}$$

showing that the derivative operator is indeed unbounded since $\frac{2n^2}{e}$ is divergent as $n\to \mathrm{\infty }$.

Title	derivative operator is unbounded in the sup norm
Canonical name	DerivativeOperatorIsUnboundedInTheSupNorm
Date of creation	2013-07-14 20:22:55
Last modified on	2013-07-14 20:22:55
Owner	cvalente (11260)
Last modified by	cvalente (11260)
Numerical id	8
Author	cvalente (11260)
Entry type	Proof
Classification	msc 47L25