derivative operator is unbounded in the sup norm

Consider C([-1,1]) the vector space of functions with derivativesPlanetmathPlanetmath or arbitrary order on the set [-1,1].

This space admits a norm called the supremum normMathworldPlanetmath given by


This norm makes this vector space into a metric space.

We claim that the derivative operatorMathworldPlanetmath D:(Df)(x)=f(x) is an unbounded operator.

All we need to prove is that there exists a succession of functions fnC([-1,1]) such that |D(fn)||fn| is divergent as n



Clearly |fn|=fn(0)=1

To find |Dfn| we need to find the extrema of the derivative of fn, to do that calculate the second derivative and equal it to zero. However for the task at hand a crude estimate will be enough.


So we finally get


showing that the derivative operator is indeed unboundedPlanetmathPlanetmath since 2n2e is divergent as n.

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