Functions in Vector Space of Real-Valued Functions Continuously Differentiable on Closed Interval vanish at Endpoints

Theorem
Let $I := \closedint a b$ be a closed real interval.

Let $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ be the continuously differentiable on closed interval real function vector space.

Let $S := \set {x \in \map {C^1} I : \map x a = y_a, \map x b = y_b}$.

Then $S$ is a vector subspace of $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ $y_a = y_b = 0$.

Necessary Condition
Suppose $y_a = y_b = 0$.

Closure under Vector Addition
Let $x_1, x_2 \in \map {C^1} I$.

By sum rule for derivatives, $x_1 + x_2 \in \map {C^1} I$

Evaluate the sum at both endpoints:

Hence, if $x_1, x_2 \in S$ then $x_1 + x_2 \in S$.

Closure under Scalar Multiplication
Let $x \in \map {C^1} I$ and $\alpha \in \R$.

By derivative of constant multiple, $\alpha \cdot x \in \map {C^1} I$.

Evaluation at both endpoint yields:

Hence, if $x \in S$ and $\alpha \in \R$, then $\alpha \cdot x \in S$.

Nonemptiness
Let $\map 0 x \in \map {C^1} I$ be such that:


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

Then:

Hence, $S$ is a subspace of $\struct {\map {C^1} I, +, \, \cdot \,}_\R$

Sufficient Condition
Suppose $S$ is a subspace of $\struct {\map {C^1} I, +, \, \cdot \,}_\R$.

Let $x \in S$.

Then $2 \cdot x \in S$ and $\map {\paren {2 \cdot x} } a = y_a$.

However, by Pointwise Scalar Multiplication of Real-Valued Functions we have that:


 * $\map {\paren {2 \cdot x} } a = 2 \map x a = 2 y_a$

Hence, $2 y_a = y_a$, or $y_a = 0$.

Analogously, $y_b = 0$.