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Problem: Let $\mathcal P $ bet the set off all polynomials in one variable with rational coefficients. For $P\in\mathcal P$ define $I(P)(x) = \int_0^xP(t)\,dt$. Show that the map $I:\mathcal P\rightarrow \mathcal P$ ist well defined (i.e., maps the set $\mathcal P$ to itself), injective, but not surjective. Determine the image $I(\mathcal P)$ of this map and the inverse function $I^{-1}(\mathcal P)\rightarrow\mathcal P$.

First off, I was trying to think of a way to define P: as it's written, there doesn't seem to be a restriction on the degree of the polynomials in $\mathcal P$.

Also, I don't.. well, I don't really understand much about this problem. The function given just takes an antiderivative of the entire polynomial, correct? It's clearly injective because no two distinct functions have the same antiderivative but I have no idea how to prove that. Finally, how can I find the inverse function of I if it isn't even bijective? (it says so right in the problem that it will not be surjective. and intuitively it won't be, $0$ is an element of $\mathcal P$ and there is no way to integrate and get $0$.)

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  • $\begingroup$ No two distinct functions have the same antiderivative. Luckily that is what you need. Note that if $Q(x)= I(P)(x)$, then $Q(0)=0$. That shows the map is not surjective. Note that the $0$ function is in the image, $\int_0^x 0\,dt=0$. $\endgroup$ – André Nicolas Jan 31 '14 at 1:05
  • $\begingroup$ Sorry, that's what I meant to say! I will edit that now. Also, thank you for pointing out that zero gets mapped to, I was thinking only of the antiderivative of 0.. i'm a bit slow right now apparently. $\endgroup$ – furashu Jan 31 '14 at 1:09
  • $\begingroup$ @AndréNicolas i am a bit confused on what you gave about surjectivity (perhaps it's the Q(x) notation but I believe that is the same as the P my question gives): are you saying that, given this function, if i plug in x and send that value to 0 I am effectively "skipping" what the antiderivative of P is and thus I am not surjective? $\endgroup$ – furashu Jan 31 '14 at 1:39
  • $\begingroup$ also, what is a sufficient argument for injectivity? $\endgroup$ – furashu Jan 31 '14 at 3:29
  • $\begingroup$ Use the Fundamental Theorem of Calculus. Suppose that $\int_0^x f(t)\,dt=\int_0^x g(t)\,dt$ (identically as a function of $x$). Differentiate, using FTC. We get $f(x)=g(x)$. $\endgroup$ – André Nicolas Jan 31 '14 at 17:37
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First, it's not true that two distinct functions have the same derivative; it's only true up to addition of constants. Second, being bijective isn't strictly necessary to have an inverse. Note that it's asking for $I^{-1} \colon I(P) \to P$, not $I^{-1} \colon P \to P$. It's injective, so it's trivially bijective onto its image. So the question is then asking, if some polynomial is in the range of $I$, what is its (unique, because $I$ is injective) inverse? Third, the lack of a degree bound doesn't matter. $P$ is already defined for you, so there's no need to define it. Experiment by seeing what $I$ does to a basis of $P$, which will tell you (because $I$ is linear) what $I$ does to an arbitrary element of $P$. The most obvious basis is $\{1, t, t^2, \ldots\}$. After experimenting a little it should be clear why the map isn't surjective.

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  • $\begingroup$ Thanks for this starting point, and I realized a bit ago that the lack of a degree bound doesn't matter. I forgot about the addition of constants somehow. It seems like it may make injectivity harder to prove, but I will think a bit with these hints you have given me. $\endgroup$ – furashu Jan 31 '14 at 1:06
  • $\begingroup$ Once you get a symbolic representation for what $I$ does to an arbitrary element of $P$ (which you get from the basis), it shouldn't be terribly difficult to prove injectivity. By all means ask for more help if you get stuck. $\endgroup$ – Julien Clancy Feb 1 '14 at 2:13

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