Given $f_1(x)$ continuous function in $[0,1]$ and differentiable in $(0,1)$. The series {$f_n(x)$} is defined by $f_{n+1}(x)=\int_{0}^{x}f_n(t)dt$ for $x\in [0,1]$.

I need to prove that $\sum_{1}^{\infty}f_n(x)$ converges and is differentiable in $(0,1)$.

I tried to bound $f_n(x)$ by some other series and to use Weierstrass M-Test to prove uniform convergence which will lead to regular one, but I didn't manage to find a series.

I'd really love your help with this one.

Thanks a lot!


Since $f_1$ is continuous on $[0,1]$, there exists $M$ such that $|f_1(x)|\leq M$ for all $x\in [0,1]$. Then $$|f_{2}(x)|=\left|\int_{0}^{x}f_1(t)dt\right|\leq\int_{0}^{x}|f_1(t)|dt\leq Mx.$$ Similarly, $$|f_{3}(x)|=\left|\int_{0}^{x}f_2(t)dt\right|\leq\int_{0}^{x}|f_2(t)|dt\leq Mx^2/2,$$ and by induction, $$|f_{n+1}(x)|=\left|\int_{0}^{x}f_n(t)dt\right|\leq\int_{0}^{x}|f_n(t)|dt\leq Mx^n/n!.$$ Therefore, we have $$\left|\sum_{n=1}^{\infty}f_n(x)\right|\leq\sum_{n=1}^{\infty}|f_n(x)|\leq\sum_{n=1}^{\infty}Mx^n/n!$$ which converges for all $x\in(0,1)$. By Weierstrass_M-test, $\displaystyle \sum_{n=1}^{\infty}f_n(x)$ converges on $(0,1)$.

On the other hand, Since $f_1$ is differentiable on $(0,1)$ and $f_{n+1}(x)=\int_{0}^{x}f_n(t)dt$ for $n\geq 2$, by induction and the fundamental theorem of calculus, $f_n$ is differentiable on $(0,1)$ for all $n$. This implies that $\displaystyle \sum_{n=1}^{\infty}f_n(x)$ is differentiable on $(0,1)$.

  • 5
    $\begingroup$ More is needed than the differentiability of some functions and their summability to deduce that the sum of a series of functions is differentiable. $\endgroup$ – Did Feb 15 '12 at 9:06
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    $\begingroup$ @Paul, I suggest you edit this into your answer to make it more complete, since as Didier mentioned your current results are not strong enough to ensure differentiability of the series: Theorem: Suppose $ f_n:(0,1) \to \mathbb{R}$ is a sequence of differentiable functions and $f$ is such that $f_n \to f $ uniformly. Also assume that $ f'_n $ converges uniformly to some function $g.$ Then $f$ is differentiable and $f'=g.$ Since the uniform convergence of $ \sum f'_n$ follows from that of $ \sum f_n $ we reach the desired conclusion. $\endgroup$ – Ragib Zaman Feb 15 '12 at 10:06

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