1. ## Show uniform convergence

Hello,

I want to show that the family of functions defined by $f_n(x)=\int_0^x f_{n-1}(t) dt$ converges uniformly on $[0,a]$.

Since I don't need a particular limit I've been fishing around with proof by contradiction.

I know a fair amount about the family:

Every member is continuous, so uniformly continuous as they are defined on a compact set (this should mean that the family is equicontinuous under Rudins definition in chapter 7.)

Every member is bounded, so the family is uniformly bounded.

Each $f_n$ is integrable so the $\int_0^x f_{n-1}(t) dt$ is finite at every $x\in [0,a]$.

I can use the ArzeląAscoli theorem to show the existence of a uniformly convergent subsequence (I wanted to use this condition to get a contradiction)

I can use the mean value theorem to show that $|f_n(x)-f_m(x)|=|f_n(c)x-f_m(c')(x)|\leq |f_n(c)-f_m(c')||a|= |\int_0^c f_{n-1}(t) dt - \int_0^{c'} f_{m-1}(t)dt|$

I should have $f_n(x) = \int_0^x f_{n-1}(t)dt=F(x)-F(0)$ as well no?

I'm not so sure what else to do at the moment. I'll try again in the morning.

2. ## Re: Show uniform convergence

Since the constant function c is continuous on $[0,a]$, if $c=f_0$ then you can derive $f_1=\int_0^x c dt = c t \big|_0^x = c x$. So then $f_2 = \int_0^x c t dt = 1/2 c t^2 \big|_0^x = 1/2 c x^2$. Continuing in this way until $f_n= \frac{1}{(n-1)!} \int_0^x f_0 \cdot (t)^{n-1} dt = \frac{1}{(n-1)!} f_0 \frac{1}{(n)} t^n \big|_0^x = \frac{1}{(n-1)!} f_0 \frac{1}{(n)} x^n$. So you would get the family of functions $\{c, c x, c 1/2 x^2,\dots, \frac{1}{(n-1)!} f_0 \frac{1}{(n)} x^n \}$.

3. ## Re: Show uniform convergence

Here,

Let $f_0(t)=c$ and write the terms of the sequence out as:

\begin{equation*}\begin{split}f_1(x)&=\int_0^x c \: dt = x c \\ f_2(x)&=\int_0^x t c \: dt = \frac{1}{2}x^2 c \\ f_3(x)&=\int_0^x \frac{1}{2} t^2 c \: dt = \frac{1}{2 \cdot 3} x^3 c \\ f_4(x) & = \int_0^x \frac{1}{2\cdot 3} t^3 c = \frac{1}{2\cdot 3 \cdot 4} x^4 c \\ \vdots \\ f_n(x)&=\int_0^x \frac{1}{(n-1)!} t^{n} c \: dt = \frac{1}{(n)!} x^{n+1} c \end{split}\end{equation*}

Now argue that the $n!$ factor dominates the $x^n$ factor, this implies there exists an $N$ such that for all $n \geq N$ you have $|f_n(x)|<\varepsilon$.

This proof I think is not very good.

4. ## Re: Show uniform convergence

First, you did not specify the function $f_0$; by inference, it is the constant c. A small modification to your "proof" shows that the conclusion remains true for any continuous $f_0$.

Now argue that the $n!$ factor dominates the $x_n$ factor, this implies there exists an $N$ such that for all $n \geq N$ you have $|f_n(x)|<\epsilon$.
I agree. This is not a good proof. In fact, in the above, the $N$ will depend on the size of $x$; exactly what you don't want for uniform convergence. You must use the fact that the convergence is to be uniform on the closed interval $[0,a]$.