Derivative Operator on Continuously Differentiable Function Space with C^1 Norm is Continuous

Theorem

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

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

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

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

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

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

Suppose $\map C I$ and $\map {C^1} I$ are equipped with $C^0$ and $C^1$ norms respectively

Then $D$ is continuous.

Proof

\(\ds \norm {D x}_\infty\)

\(=\)

\(\ds \norm {x'}_\infty\)

\(\ds \)

\(\le\)

\(\ds \norm {x'}_\infty + \norm x_\infty\)

\(\ds \)

\(=\)

\(\ds \norm x_{1, \infty}\)

We have that the Derivative Operator is Linear Mapping.

By definition and Continuity of Linear Transformation between Normed Vector Spaces, $D$ is continuous.

$\blacksquare$

Sources

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