$f_{n+1}(x):= \int_a ^x f_n(t)dt$, $\sum_{m=1} ^{\infty} f_m(x)$ is uniformly convergent

Question

Let $f_1 : [a,b] \rightarrow \mathbb{R}$ be an integrable function.

Let's define a sequence $(f_n)$, $ \ \ f_n : [a,b] \rightarrow \mathbb{R}$ as $f_{n+1}(x):= \int_a ^x f_n(t)dt$.'

Prove that $\sum_{m=1} ^{\infty} f_m(x)$ is uniformly convergent on $[a,b]$.

Could you help me with that?

Thank you.

Answer 1

Denote $$M:=\int_a^b|f_1(t)|dt<\infty.\tag{1}$$ By definition, $$|f_{n+1}(x)|\le \int_a^x |f_n(t)|dt\quad \forall x\in[a,b],~\forall n\ge 1.\tag{2}$$ Using $(1)$ and $(2)$, by induction, it is easy to show that $$|f_{n+1}(x)|\le \frac{M (x-a)^{n-1}}{(n-1)!},\quad \forall x\in[a,b],~\forall n\ge 1.\tag{3}$$ The conclusion follows from $(3)$.

Answer 2

Here is an other way to consider the problem. You can consider the the operator \begin{equation} Tx(t) = \int_a^t x(t') dt' \ . \end{equation} Then $f_{n+1} = Tf_n$. Define \begin{equation} s_n(t) = \sum_{k=1}^{n+1} f_k(t) \ . \end{equation} Now $s_1-s_0 = Tf_1$. Note that $|Tf_1(t)| \leq ||f_1||_1$. Hence $Tf_1$ is bounded on $[a,b]$ and we can apply the norm $||x|| = \sup_{t \in [a,b]} |x(t)| e^{-\alpha t}$ where $\alpha = 2$. This is a norm. The proof is similar to that of the supremum norm. We will show \begin{eqnarray} ||s_{n+1}-s_n|| \leq \frac{1}{\alpha^n} ||s_1-s_0|| \ . \end{eqnarray} The basic step $n=0$ is trivial. Induction step: We estimate \begin{eqnarray} ||Ts_{n+1}-Ts_n|| & = & \sup_{t \in [a,b]} \Bigg|\int_a^t s_{n+1}(t') dt' - \int_a^t s_n(t') dt'\Bigg| e^{-\alpha t} = \sup_{t \in [a,b]} \Bigg|\int_a^t (s_{n+1}-s_n)(t') dt'\Bigg| e^{-\alpha t} \\ & \leq & \sup_{t \in [a,b]} \int_a^t |(s_{n+1}-s_n)(t')|e^{-\alpha t'} e^{-\alpha(t-t')} dt' \leq \sup_{t \in [a,b]} \int_a^t ||s_{n+1}-s_n|| e^{-\alpha (t-t')} dt' \\ & = & \sup_{t \in [a,b]} ||s_{n+1}-s_n|| \int_0^{t-a} e^{-\alpha t'} dt' = ||s_{n+1}-s_n|| \sup_{t \in [a,b]} \bigg|_0^{t-a} \frac{1}{-\alpha} e^{-\alpha t'} \\ & = & ||s_{n+1}-s_n|| \sup_{t \in [a,b]} \frac{1}{\alpha} (1-e^{-\alpha(t-a)}) \leq \frac{1}{\alpha} ||s_{n+1}-s_n|| \ , \end{eqnarray} \begin{eqnarray} ||s_{n+1+1}-s_{n+1}|| & = & ||f_{n+3}|| = ||Tf_{n+2}|| = ||T(s_{n+1}-s_n)|| = ||Ts_{n+1}-Ts_n|| \\ & \leq & \frac{1}{\alpha} ||s_{n+1}-s_n|| \leq \frac{1}{\alpha} \frac{1}{\alpha^n} ||s_1-s_0|| = \frac{1}{\alpha^{n+1}} ||s_1-s_0|| \ . \end{eqnarray} Now assume $m > n$. Otherwise the following estimate is trivial or we can intercahnge the roles of $m$ and $n$. We estimate \begin{eqnarray} ||s_m-s_n|| & = & ||\sum_{k=n}^{m-1} s_{k+1}-s_k|| \leq \sum_{k=n}^{m-1} ||s_{k+1}-s_k|| \leq \sum_{k=n}^{m-1} \frac{1}{\alpha_k}||s_1-s_0|| \\ & = & \sum_{k=0}^{m-n-1} \frac{1}{\alpha^{k+n}} ||s_1-s_0|| = \frac{1}{\alpha^n}\sum_{k=0}^{m-n-1} \frac{1}{\alpha^k} ||s_1-s_0|| \\ & = & \frac{1}{\alpha^n} \frac{1-(\frac{1}{\alpha})^{m-n}}{1-\frac{1}{\alpha}} ||s_1-s_0|| \leq \frac{1}{\alpha^n} \frac{1}{1-\frac{1}{\alpha}} ||s_1-s_0|| \ . \end{eqnarray} We can now estimate \begin{eqnarray} |s_m(t)-s_n(t)| & = & |s_m(t)-s_n(t)|e^{-\alpha t} e^{\alpha t} \leq ||s_m-s_n||e^{\alpha t} \leq \frac{1}{\alpha^n} \frac{1}{1-\frac{1}{\alpha}} ||s_1-s_0|| e^{\alpha b} \rightarrow 0 \ , \end{eqnarray} as $m,n \rightarrow \infty$. Hence there is a pointwise limit \begin{equation} s(t) = \lim_{n \rightarrow \infty} s_n(t) = \lim_{n \rightarrow \infty} \sum_{k=1}^{n+1} f_k(t) = \sum_{k=1}^\infty f_k(t) \ . \end{equation} Taking now the limit $m \rightarrow \infty$ from the pointwise estimate we obtain \begin{equation} |s(t)-s_n(t)| \leq \frac{1}{\alpha^n} \frac{1}{1-\frac{1}{\alpha}} ||s_1-s_0||e^{\alpha b} \rightarrow 0 \ , \end{equation} as $n \rightarrow \infty$. This shows that the convergence is uniform.