Let $f\to f+ f'$ be a map from $C^\infty (0,1) \to C^\infty (0,1)$ Let $f\to f+ f'$  be a map from $C^\infty (0,1) \to C^\infty (0,1)$  is this map injective and surjective?
It's not injective since take $f=e^{-x},$ 
I could not think a counterexample to disprove surjectivity? Is the map surjective?
I can see that all the polynomials are in the image since $x^n$ is in the image for all n. Can I use this to pro that the map is surjective?
Any help is appreciated
 A: Surjectivity is equivalent to the existence of the solution $f\in C^\infty(0,1)$ of the first order linear differential equation
$$
f'+f = g
$$ where $g\in C^\infty(0,1)$. It can be solved explicitly using integrating factor $e^x$. This gives
$$
\left(e^xf(x)\right)' = e^x(f'(x)+f(x)) = e^x g(x),
$$ which implies
$$
f(x) = e^{-x}\left(\int_a^x e^tg(t) \mathrm dt +C\right).
$$ for some constant $C$ and some $a\in (0,1)$. Note that $f$ is smooth, hence we proved surjectivity of the given map.
A: Pick any
$g(x) \in C^\infty(0, \infty); \tag 1$
then
$f'(x) + f(x) = g(x) \tag 2$
is a simple, first order, linear differential equation for $f(x)$; it has constant coefficients and is inhomogeneous.  As such, there is a very well-known formula for it solution, viz:
$f(x) = \exp \left ( -\displaystyle \int_1^x ds \right) \left (f(1) + \displaystyle \int_1^x \exp \left (  \int_1^u du \right ) g(u) \; du \right )$
$e^{1 - x} \left ( f(1) + \displaystyle \int_1^x e^{u - 1} g(u) \; du \right ); \tag 3$
we note that this equation invokes $f(1)$; we may in fact choose $f(1) \in \Bbb R$ arbitrarily and we will obtain a different solution for each such value.  Thus there is an entire family of functions $f(x) \in C^\infty(0, \infty)$ satisfying (2); the solutions are all clearly $C^\infty$ by virtue of (1).
It follows that the mapping
$f \to f' + f \tag 4$
is surective.
Nota Bene:  Taking $g(x) = x^n$, the above argument shows the existence of an $f_n(x)$ with
$f_n'(x) + f_n(x) = x^n; \tag 5$
this does indeed allow the conclusion that all polynomials $g(x)$ are in the image of $f' + f$; but it is difficult to extend this to all $g(x) \in C^\infty(0, \infty)$ due to convergence issues.  Fortunately, the formula (3) resolves this difficulty, since it binds for any $g(x)$ as in (1).  End of Note.
