$f_1 \in L^1_{loc}(\mathbb{R})$ and $f_{n+1} (x)= \int_0^x f_n(t) dt$, What is $\sum_n f_n$?

Question

$f_1 \in L^1_{loc}(\mathbb{R})$ and $f_{n+1}(x) = \int_0^x f_n(t) dt$, What is $\sum_n f_n$? (and converges in what sense?)

My attempt:

Suppose $f_1$ is bounded, define the continuous linear operator from Banach space $(X,\|\cdot\|_{max})$ $$X:= \{\text{ bounded continuous functions with } f(0) = 0\}$$ to itself by $$A[f](x) : = \int_0^x f(t) dt,$$ we have $\|A\|_{op} <1$.

The summation above can be rewritten as $$\sum_{n=1}^\infty f_n(x) = f_1(x) + \bigg(\sum_{n=1}^\infty A^n\bigg)[f_1](x) = f_1(x) + (I-A)^{-1}[f_1](x). $$

Questions:

1. My space $X$ is too small for $f_1\in L^1_{loc}$, clearly there exists unbounded $f_1$ such that $A[f_1]\not\in X$. If I require $X$ to be a (closed) subspace of continuous functions instead of bounded continuous functions, is there a norm for the space $C(\mathbb{R})$?

2. I can not workout explicitly what $(I-A)^{-1}$ is.

3. Is there a better way to solve this problem? Using basic real analysis and calculus?

Thank you very much!

Edit: My partial solution is incorrect.

Answer 1

HINT:

Use basic analysis to check that \begin{equation} f_n(x)= \int_{0}^x \frac{\ \ (x-t)^{n-2}}{(n-2)!} f_1(t)\, d t \end{equation}

for $n\ge 2$.