The Problem is : Let $C^{\infty}(0,1)$ denote the vector space of all smooth functions on $(0,1)$, and let $\phi : C^{\infty}(0,1) \to C^{\infty}(0,1)$ be a linear map such that, $\phi(f)= (f + f')$, then verify ontoness of $\phi$ .
Obviously, here $\phi$ is not one-one as for any $f$, $f$ and $(f+ \exp^{-x})$ maps into same thing . But, I am not sure how to use linearity of the map $\phi$ for verifying ontoness, as the vector space is infinite dimensional.
For the functions $f$ in the vector space which are integrable on $(0,1)$, if $F(x)=\int_0^1 f$, then $(F+f)'$= $\phi(f)$, but $g(x)=1/x$ is not integrable on $(0,1)$ , then what's the pre-image of $g$ under $\phi$ ???
Let $g$ be any smooth function. Let $c \in (0,1)$. Consider the equation $f'+f=g$. We can write this as $e^{x}f'(x)+f(x)e^{x}=g(x)e^{x}$ or $(e^{x}f(x))'=g(x)e^{x}$. This gives $e^{x}f(x)=e^{c}f(c)+\int_c^{x} e^{t}g(t)dt + C$. Hence $f(x)=e^{-x}(e^{c}f(c)+\int_c^{x} e^{t}g(t)dt+C) $. We can take $C=0$ and verify that this $f$ indeed satisfies the equation $f'+f=g$. Also this function is smooth and its image under the given map is $g$.