Subspace of Real Functions of Differentiability Class
Theorem
Let $\mathbb J = \set {x \in \R: a < x < b}$ be an open interval of the real number line $\R$.
Let $\map {C^m} {\mathbb J}$ be the set of all continuous real functions on $\mathbb J$ in differentiability class $m$.
Then $\struct {\map {C^m} {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
Corollary
Let $\map {C^\infty} {\mathbb J}$ denote the set of all continuous real functions on $\mathbb J$ which are differentiable on $\mathbb J$ at all orders.
Then $\struct {\map {C^\infty} {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
Proof
Note that by definition, $\map {C^m} {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map {C^m} {\mathbb J}$.
Let $\lambda \in \R$.
Applying Linear Combination of Derivatives $m$ times we have:
- $f + \lambda g$ is $m$-times differentiable on $\mathbb J$ with $m$th derivative $f^{\paren m} + \lambda g^{\paren m}$.
Since both $f$ and $g$ are of differentiability class $m$:
- $f^{\paren m}$ and $g^{\paren m}$ are continuous on $\mathbb J$.
From Combined Sum Rule for Continuous Real Functions:
- $f^{\paren m} + \lambda g^{\paren m} = \paren {f + \lambda g}^{\paren m}$ is continuous on $\mathbb J$.
So:
- $f + \lambda g \in \map {C^m} {\mathbb J}$
Therefore, by One-Step Vector Subspace Test:
- $\struct {\map {C^m} {\mathbb J}, +, \times}_\R$ is a subspace of $\struct {\R^{\mathbb J}, +, \times}_\R$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.5$