# Series of Functions of integrals

How do you solve this question? We had that in a test and I've been staring on it for around 30 minutes without any solution.

Given $g(x)$ a differentiable and bounded function over $\mathbb{R}$. Define,

\begin{align} f_0(x) & =g(x) \\ f_n(x) &= \int _0^xdt_{n-1}\int _0^{t_{n-1}}dt_{n-2}\int _0^{t_{n-2}}dt_{n-3}...\int _0^{t_3}dt_2\int _0^{t_2}dt_1\int _0^{t_1} g(t_0)\,dt_0 \end{align}

A) Prove that the series of functions $$\sum_{n=0}^{\infty }\:f_n(x)$$ converges uniformly in every closed interval $[0,a]$ for $a > 0$.

B) We define $G(x)= \displaystyle\sum_{n=0}^{\infty} f_n(x)$ for every $x \in [0,a]$. Prove that $G'(x)=\displaystyle\sum_{n=0}^{\infty } f_n'(x)$.

C) Show that for every $x \in [0,a]$ applies: $G(x)=\displaystyle\int _0^xe^{x-t}g'\left(t\right)dt\:$

Please, solve it if you may.

And something a little bit important as well, did anyone came across this question in the past? Where is the source of this question?

• It's impossible to understand that. Please do learn the easy rules to properly type mathematics here: meta.math.stackexchange.com/questions/5020/… – Timbuc Feb 18 '15 at 10:14
• Check the edit to make sure this is what you intended. Also, please add some more context and thoughts of your own. Right now the question is just a verbatim copy of the problem statement. Such questions are generally discouraged. – mrf Feb 18 '15 at 10:23
• Taking derivatives allows you get forms that might be simpler to work with: $f_n'(x) = f_{n-1}(x)$ and $f_n^{(n)}(x) = g(x)$. – Winther Feb 18 '15 at 10:25
• Thank you mrf. Winther, Could you please elaborate? – Billy McGeen Feb 18 '15 at 10:36
• Use that $g$ is bounded to calculate a bound for $f_i$ for the first few $i$. For example: $|f_1(x)| \leq \int_0^{x}|g(t_0)|dt_0 \leq Mx$ and $|f_2(x)| \leq \frac{M}{2!}x^2$ ... See the pattern that emerges. Try to use induction using $f_n'(x) = f_{n-1}(x)$ to prove the formula guessed from the pattern. Finally use Weierstrass M-test to prove uniform convergence. (haven't checked all the steps, but it seems to work) – Winther Feb 18 '15 at 10:40

## 1 Answer

Show by induction that for $n \ge 1$,

$$f_n(x) = \frac{1}{(n-1)!}\int_0^x (x - t)^{n-1} g(t)\, dt = \frac{1}{(n-1)!}\int_0^x t^{n-1}g(x - t)\, dt.$$

Let $a > 0$, and set $M = \max\{|g(x)|: 0\le x \le a\}$. For all $x\in [0,a]$, $|f_0(x)| \le M$ and for $n\ge 1$,

$$|f_n(x)| \le \frac{M}{(n-1)!}\int_0^a t^{n-1}\, dt = \frac{Ma^n}{n!}.$$ Since $\sum_{n = 0}^\infty Ma^n/n!$ converges (to $Me^a$, in fact), by the Weierstrass $M$-test, the series $\sum_{n = 0}^\infty f_n(x)$ converges uniformly on $[0,a]$. This proves A).

To prove B), note that since $f_n'(x) = f_{n-1}(x)$ and the series $\sum_{n = 0}^\infty f_n(x)$ converges uniformly on $[0,a]$, the series $\sum_{n = 0}^\infty f_n'(x)$ converges uniformly on $[0,a]$. Therefore, $G'(x) = \sum_{n = 0}^\infty f_n'(x)$.

Part C) does not hold unless $g(0) = 0$. The formula for $G$ should be

$$G(x) = g(0)e^x + \int_0^x e^{x-t}g'(t)\, dt.$$

You can either use the formula for $f_n$ I displayed in the beginning to find $G$ directly (integrating term-wise then applying integration by parts), or by writing a first-order differential equation in $G$ and solving by method of integrating factors. If you choose the latter, then use the relation $f_n'(x) = f_{n-1}(x)$, $f_0(x) = g(x)$ and the value $G(0) = g(0)$ to develop the inital value problem

$$G' - G = g',\, G(0) = g(0).$$

The integrating factor is $e^{-x}$, so the general solution of the ODE is

$$G(x) = Ae^x + \int_0^x e^{x-t}g'(t)\, dt.$$

The initial condition $G(0) = g(0)$ yields $A = g(0)$. Hence

$$G(x) = g(0)e^x + \int_0^x e^{x-t}g'(t)\, dt.$$

If you choose the former method, then write

\begin{align} G(x) &= g(x) + \sum_{n = 1}^\infty \frac{1}{(n-1)!}\int_0^x (x - t)^{n-1}g(t)\, dt\\ &= g(x) + \int_0^x \sum_{n = 1}^\infty\frac{(x-t)^{n-1}}{(n-1)!}g(t)\, dt\\ &= g(x) + \int_0^x e^{x-t}g(t)\, dt\\ &= g(x) + \int_0^x e^{x-t}g'(t)\, dt - g(x) + g(0)e^x\\ &= g(0)e^x + \int_0^x e^{x-t}g'(t)\, dt. \end{align}

The interchange of sum and integral in second step is justified by the integrability of the terms $(x-t)^{n-1}/(n-1)!$ and the uniform convergence of the series $\sum_{n = 1}^\infty (x-t)^{n-1}/(n-1)!$ over $[0,a]$. The second to the last step is where integration by parts is used.

• A very good job. :) – PepperSausage Feb 19 '15 at 20:56