# derivative operator is unbounded in the sup norm

Consider $C^{\infty}([-1,1])$ 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

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

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 $f_{n}\in C^{\infty}([-1,1])$ such that $\frac{|D(f_{n})|}{|f_{n}|}$ is divergent as $n\to\infty$

consider

 $f_{n}(x)=\exp(-n^{4}x^{2})$
 $(Df_{n})(x)=-2xn^{4}\exp(-n^{4}x^{2})$

Clearly $|f_{n}|=f_{n}(0)=1$

To find $|Df_{n}|$ 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.

 $|Df_{n}|\geq|(Df_{n})(\frac{1}{n^{2}})|=\frac{2n^{2}}{e}$

So we finally get

 $\frac{|Df_{n}|}{|f_{n}|}\geq\frac{2n^{2}}{e}$

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

Title derivative operator is unbounded in the sup norm DerivativeOperatorIsUnboundedInTheSupNorm 2013-07-14 20:22:55 2013-07-14 20:22:55 cvalente (11260) cvalente (11260) 8 cvalente (11260) Proof msc 47L25