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

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Theorem

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

Let $\map \CC I$ be the real-valued, continuous on $I$ function space.

Let $\map {\CC^1} I$ be the continuously differentiable function space.

Let $x \in \map {\CC^1} I$ be a continuoulsly differentiable real-valued function.

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

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

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


Then $D$ is not continuous.


Proof

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

By definition:

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

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:

\(\ds \norm{Dx}_\infty\) \(=\) \(\ds \norm{x'}_\infty\)
\(\ds \) \(=\) \(\ds n\)
\(\ds \) \(\le\) \(\ds M \norm {x}_\infty\)
\(\ds \) \(=\) \(\ds M \cdot 1\)

In other words:

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

But $M$ is finite.

This is a contradiction.

We have that the derivative operator is linear mapping.

By definition and continuity of linear transformations, $D$ is not continuous.

$\blacksquare$


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