Subspace of Real Functions of Differentiability Class

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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 $\struct {\R^{\mathbb J}, +, \times}_\R$.

$\blacksquare$


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