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Let $A$ be the set of all infinitely differentiable functions $f:[0,1] \rightarrow \mathbb{R}$, and let $A_0 \subset A$ be the set of all such functions for which the condition $f(0) = 0$ holds. Define the function $D:A_0 \rightarrow A$, $D(f) = df/dx$.

Use the Mean Value Theorem to show that $D$ is injective.

Use the Fundamental Theorem of Calculus to show that $D$ is surjective.

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    I guess you mean $D$ injective/surjective. What did you try?2011-09-20
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    @mahin: I'm quite sure you mean $D$ injective and $D$ surjective. The proof is indeed ‘rather basic’. It may be easier to think of it from a linear algebra point of view, since $A_0$ and $A$ are vector spaces and $D$ is linear.2011-09-20
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    Thanks for spotting the typo. I corrected it.2011-09-20
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    I'm not exactly sure why $A$ and $A_0$ are vector spaces. Can you elaborate?2011-09-20
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    Define the sum of $f$ and $g$ by letting $(f+g)(x)=f(x)+g(x)$. Similarly, for $c\in\mathbb R$ let $(c\cdot f)(x)=c\cdot(f(x))$. Both $A$ and $A_0$ are closed with respect to $+$ and under multiplication with constants from $\mathbb R$. The usual axioms for a vector space are satisfied.2011-09-20

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