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Question:

Let $P$ denote the set of polynomials

$p(x) = \sum_{i=0}^n a_{i}x^i$

for $n=0, 1, 2, ....$ and real numbers $a_{i}$. Let $D:P \to P$ be the differential operator defined by

$(Dp)(x) = \sum_{i=1}^n ia_{i}x^{i-1}$

and let $I:P \to P$ be the integration operator

$(Ip)(x) = \sum_{i=0}^n \frac{a_{i}}{i+1}x^{i+1}$

Show that $D$ is surjective, but not injective.

Attempted solution:

$D$ is surjective if $\forall$ $q$ $\in$ $P$ there exists at least one $p$ $\in$ $P$ such that $D(p) = q$. To show this I choose an arbitrary $q$ in the codomain and find a corresponding $p$ in the domain such that $D(p) = q$. Since the domain consists of the set of polynomials $p(x)$ and $p$ is a polynomial in the domain, I express $p'(x) = q$ and integrate on both sides yielding

$p(x) = \int (\sum_{i=1}^n ia_{i}x^{i-1}) dx = \sum_{i=0}^n a_{i}x^i $

But I'm quite unsure what I have achieved at this point, or if what I've done is correct. Any help will be appreciated. .....

Community
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Joey
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    For any polynomial you can find an anti-derivative polynomial, then surjective. The anti-derivatives are many ($+C$), then not injective. – A.Γ. Aug 27 '15 at 16:44

2 Answers2

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In order to show that $D$ is surjective, you want to see that, for each polynomial $p$, there exists a polynomial $q$ such that $Dq=p$ or, in functional notation, $q'(x)=p(x)$.

What you've done is proving that $p$ is an antiderivative of $p'$, which is obvious. But your idea is good: let $$ q(x)=\int_0^x p(t)\,dt=\sum_{i=0}^n \frac{a_i}{i+1}x^{i+1}=(Ip)(x) $$ The fundamental theorem of calculus tells you that $q'=p$. In other words, $DI$ is the identity, so $D$ is surjective.

As regards to injectivity, what's the derivative of a constant polynomial?

egreg
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Your wording at the beginning is off. You say

$D$ is surjective if $\forall p\in P$ there exists at least one $q$ such that $D(p)=q$.

It should say

$D$ is surjective if $\forall q\in P$ there exists at least one $p$ such that $D(p)=q$.

What you then acchieved was, well, not much.

Then, you took an arbitrary $q$. Now, you need to find some polynomial $p$ such that $p' = q$. To do that, write down

$$q=\sum_{i=0}^n a_i x^i$$

Now, try to find combination of $b_0, b_1,\dots, b_m$ such that

$$p=\sum_{i=0}^m b_i x^i$$

and $p' = q$. To do that, you can help yourself with the integration operator.

5xum
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