Surjectivity of a linear transformation from the space of all formal power series to itself

Question

Today, one of my friends asked that if we take the linear transformation $T : P(\mathbb{R}) \to P(\mathbb{R})$ s.t. $T(p(x))=p(x)+p''(x)$, then $T$ is onto or not. $P(\mathbb{R})$ is the vector space of all formal power series.

My attempt : If we take a polynomial $q(x) \in P(\mathbb{R})$, then we can form a differential equation $p''(x)+p(x)=q(x)$. But is it solvable ?? The confusion rises since $p(x)$ and $q(x)$ are all formal power series.

I've tried to check the coefficients and compare them, but it doesn't help in any way. Help would be appreciated.

Answer 1

It's trivial that this map is surjective.

We need to show this: Given any sequence $(a_0,a_1,\dots)$ there exists a sequence $(b_0,b_1,\dots)$ such that $$2b_2+b_0=a_0,$$ $$6 b_3+b_1=a_1,$$ $$12b_4+b_2=a_2,$$etc.

Let $b_0=b_1=0$. Solve the first equation for $b_2$. Solve the second equation for $b_3$. Since we have determined $b_2$ we can solve the third equation for $b_4$. Etc.

Answer 2

Please be careful with the difference between a polynomial and the function associated with it. You should write $T(P)=P+P´´$ The idea would be to consider the matrix of your endomorphism in $R_{n}[X]$ in the usual basis to see if it is invertible or not.

Edit : remember that derivation is a linear operation.

Answer 3

Yes, it is surjective.

Given $$ q(x) = b_0 + b_1x + b_2x^2 + ....$$

You like to find a power series $P(x)$ such $$P''(x)+P(x)=q(x)$$

Let $$P(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3+....$$

$$ P''(x) +P(x) = (a_0+2a_2) + (a_1+6a_3)x +(a_2+12a_4)x^2 +....$$

$$ q(x) = b_0 + b_1x + b_2x^2 + ....$$ We need to solve $$ a_0+2a_2=b_0$$

$$a_1+6a_3=b_1$$

$$a_2+12a_4=b_2$$

You will find many solutions for $P(x)$