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## Reccurent function sequence SEEMOUS 2010

Suppose $f_0:[0,1]\to \mathbb{R}$ is a continuous function, and define the sequence $(f_n)_n, f_n:[0,1]\to \mathbb{R}$ in the following way:

$f_n(x)=\int_0^x f_{n-1}(t)dt,\ \forall x \in [o,1]$.

a) Prove that the series $\sum_{n\geq 0}f_n(x)$ converges for any $x \in [0,1]$.

b) Find an explicit formula in terms of $x$ for the above series.

Seemous 2010, Problem 1

a) It is obvious that $f_n'=f_{n-1}$. The function $f_0$ is continuous defined on a compact set, therefore is bounded. This means we can find $M>0$ such that $|f_0(x)|\leq M$ on $[0,1]$. But then

$\displaystyle |f_1(x)|=|\int_0^x f_0(t)dt|\leq \int_0^x |f_0(t)|dt \leq Mx\leq M$ on $[0,1]$.

Continuing we get

$\displaystyle |f_2(x)|=|\int_0^x f_1(t)dt|\leq \int_0^x |f_1(t)|dt \leq M\frac{x^2}{2} \leq M/2$ on $[0,1]$.

By induction we have

$\displaystyle |f_n(x)| \leq M\frac{x^n}{n!}\leq \frac{M}{n!}$ on $[0,1]$.

We are needed to prove that the series $\sum_{n=0}^\infty f_n(x)$ is convergent. For this it is enough to prove that it is absolutely convergent, and for that we have

$\displaystyle \sum_{n=0}^\infty |f_n(x)|\leq \sum_{n=0}^\infty \frac{1}{n!}<\infty$.

This proves by using the Weierstrass test that the series converges (moreover it converges uniformly).

b) In the second part denote by $F(x)=\displaystyle \sum_{n=1}^\infty f_n(x)$, and note that by differentiating term by term (this is possible due to the uniform convergence) we obtain

$F'(x)=f_0+F(x)$,

a first order linear differential equation which can be immediatley solved.