Derivative Operator on Continuously Differentiable Function Space with C^1 Norm is Continuous
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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 $\map {CL} {X,Y}$