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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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I am sure you are asking about injectivity and surjectivity of $D$, not $f$.

Injectivity: Use Zhen Lin's comment and show that the kernel of $D$ consists only of the function that is constantly zero. I.e., the only function in $A_0$ whose derivative is constantly $0$ is constantly zero. This follows easily from the MVT.

The Fundamental Theorem of Calculus (in the appropriate formulation) pretty explicitly says that $D$ is surjective.

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    If you mean Ker$(D)$ is all $f \in A_0$ such that $D(f) = 0$, then it follows $f = c$ and since $f(0) = 0$ then f = 0. Not sure if using the MVT makes this much easier. And I don't see how showing this shows injectivity.2011-09-20
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    $D$ is a linear map and one can show linear maps are injective if and only if $\ker D = 0$.2011-09-20
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    @mahin: Let $f,g\in A_0$. Then $f-g\in A_0$. If $D(f)=D(g)$, then $D(f-g)=0$. As you pointed out, this implies $f-g=0$ and thus $f=g$. Now, in order to show that for a function $h$, if $D(h)$ is constantly $0$, then $h$ is constant you would usually use the MVT, if I am not mistaken.2011-09-20
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    In the sense that $D^{-1} : A \rightarrow A_0, D^{-1}(f)=\int_0^x f(t) \, {\rm d}t$ and thus $D^{-1}(D(f)) = \int_0^x \frac{\rm d}{{\rm d}t} f(t) \, {\rm d}t = f(x)-f(0)=f(x)$ ??2018-08-07