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

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Theorem

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

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

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


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


Proof

Necessary Condition

Suppose $y_a = y_b = 0$.


Closure under Vector Addition

Let $x_1, x_2 \in \map {\CC^1} I$.

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

Evaluate the sum at both endpoints:

\(\ds \map {\paren {x_1 + x_2} } a\) \(=\) \(\ds \map {x_1} a + \map {x_2} a\) Definition of Pointwise Addition of Real-Valued Functions
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds y_a\) Assumption
\(\ds \map {\paren {x_1 + x_2} } b\) \(=\) \(\ds \map {x_1} a + \map {x_2} b\) Definition of Pointwise Addition of Real-Valued Functions
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds y_b\) Assumption

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


Closure under Scalar Multiplication

Let $x \in \map {\CC^1} I$ and $\alpha \in \R$.

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

Evaluation at both endpoint yields:

\(\ds \map {\paren {\alpha \cdot x} } a\) \(=\) \(\ds \alpha \cdot \map x a\) Definition of Pointwise Scalar Multiplication of Real-Valued Functions
\(\ds \) \(=\) \(\ds \alpha \cdot 0\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds y_a\) Assumption
\(\ds \map {\paren {\alpha \cdot x} } b\) \(=\) \(\ds \alpha \cdot \map x b\) Definition of Pointwise Scalar Multiplication of Real-Valued Functions
\(\ds \) \(=\) \(\ds \alpha \cdot 0\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds y_b\) Assumption

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


Nonemptiness

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

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

Then:

\(\ds \map 0 a\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds y_a\) Assumption
\(\ds \) \(=\) \(\ds y_b\) Assumption
\(\ds \) \(=\) \(\ds \map 0 b\)

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

$\Box$

Sufficient Condition

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

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, $y_b = 0$.

$\blacksquare$


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