Derivative Operator on Continuously Differentiable Function Space with Supremum Norm is not Continuous

Theorem

Let $I = \closedint 0 1$ be a closed real interval.

Let $D : \map {C^1} I \to \map C I$ be the derivative operator such that:

$\forall t \in \closedint 0 1 : \map {Dx} t := \map {x'} t$

Suppose $\map C I$ and $\map {C^1} I$ are equipped with the supremum norm.

Then $D$ is not continuous.

Aiming for a contradiction, suppose $D$ is continuous.

$\exists M \in \R_{> 0} : \forall x \in \map {C^1} I : \norm {\map D x}_\infty \le M \norm x_\infty$

Suppose $x = t^n$ with $n \in \N$.

Then:

$\norm x_\infty = \norm {t^n}_\infty = 1$

$\norm {x'}_\infty = \norm {n t^{n-1}}_\infty = n$

Hence:

$\ds \norm {D x}_\infty$

$=$

$\ds \norm {x'}_\infty$

$\ds $

$=$

$\ds n$

$\ds $

$\le$

$\ds M \norm x_\infty$

$\ds $

$=$

$\ds M \cdot 1$

In other words:

$\forall n \in \N : n \le M$

But $M$ is finite.

Hence, $D$ is not continuous.

$\blacksquare$

2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$