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 $\map \CC I$ be a space of continuous on closed interval real-valued functions.

Let $\map {\CC^1} I$ be a space of continuously differentiable functions on closed interval $I$.

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 $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ is a vector space.

Proof
We have that $\struct {\map \CC I, +, \, \cdot \,}_\R$ is a vector space.

By Differentiable Function is Continuous, $\map {\CC^1} I \subset \map \CC I$.

Let $f, g \in \map {\CC^1} I$.

Let $\alpha \in \R$.

Let $\map 0 x$ be a real-valued function such that:


 * $\map 0 x : I \to 0$

Restrict $\paren +$ to $\map {\CC^1} I \times \map {\CC^1} I$.

Restrict $\paren {\, \cdot \,}$ to $\R \times \map {\CC^1} I$.

Closure under vector addition
By Sum Rule for Derivatives:
 * $f + g \in \map {\CC^1} I$

Closure under scalar multiplication
By Derivative of Constant Multiple:
 * $\alpha \cdot f \in \map {\CC^1} I$

Nonemptiness
By Derivative of Constant, a constant mapping is differentiable.

Hence, $\map 0 x \in \map {\CC^1} I$.

We have that $\map {\CC^1} I$ is closed under restrictions of $\paren +$ and $\paren {\, \cdot \,}$ to $\map {\CC^1} I$.

Also, $\map {\CC^1} I$ is non-empty.

By definition, $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ is a vector subspace of $\struct {\map \CC I, +, \, \cdot \,}_\R$.

Since $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ satisfies vector space axioms under given restrictions, it is a vector space.