Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
Theorem
Let $I := \closedint a b$ be a closed real interval.
Let $\map C I$ be a space of real-valued functions continuous on $I$.
Let $\map {C^1} I$ be a space of continuously differentiable functions on $I$.
Let $\struct {\R, +_\R, \times_\R}$ be the field of real numbers.
Let $\paren +$ be the pointwise addition of real-valued functions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of real-valued functions.
Then $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ is a vector space.
Proof
- $\struct {\map C I, +, \, \cdot \,}_\R$ is a vector space.
By Differentiable Function is Continuous:
- $\map {C^1} I \subset \map C I$
Let $f, g \in \map {C^1} I$.
Let $\alpha \in \R$.
Let $\map 0 x$ be a real-valued function such that:
- $\map 0 x : I \to 0$
Restrict $\paren +$ to $\map {C^1} I \times \map {C^1} I$.
Restrict $\paren {\, \cdot \,}$ to $\R \times \map {C^1} I$.
Closure under vector addition
- $f + g \in \map {C^1} I$
$\Box$
Closure under scalar multiplication
By Derivative of Constant Multiple:
- $\alpha \cdot f \in \map {C^1} I$
$\Box$
Nonemptiness
By Derivative of Constant, a constant mapping is differentiable.
Hence, $\map 0 x \in \map {C^1} I$.
$\Box$
We have that $\map {C^1} I$ is closed under restrictions of $\paren +$ and $\paren {\, \cdot \,}$ to $\map {C^1} I$.
Also, $\map {C^1} I$ is non-empty.
By definition, $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ is a vector subspace of $\struct {\map C I, +, \, \cdot \,}_\R$.
Since $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ satisfies vector space axioms under given restrictions, it is a vector space.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.1$: Vector Space