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I've been stuck on this problem for a bit now:

Let $g$ and $f_0$ be continuous functions on $[0,1]$. Define the sequence on $[0,1]$ by $$f_n(x) = \int_0^t g(t)f_{n-1}dt.$$

I have to prove that the sequence is converges uniformly.

Here are my thoughts: I know $f_n$'s are continuous and I also tried to solve this question using the $\epsilon-\delta$ definition for continuity and uniform convergence without any luck.

Any hint?

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I suppose that the upper limit in the integral is $x$ and not $t$.

Let $G$ and $F$ be constants such that $|g(x)|\le G$ and $|f_0(x)|\le F$ for all $x\in[0,1]$. Then $$\begin{align} |f_1(x)|&\le\int_0^xG\,F\,dt=G\,F\,x\\ |f_2(x)|&\le\int_0^xG^2\,F\,t\,dt=\frac{G^2\,F}{2}\,x^2\\ |f_3(x)|&\le\int_0^x\frac{G^3\,F}{2}\,t^2\,dt=\frac{G^3\,F}{2\cdot3}\,x^3\\ \end{align}$$ Can you see a pattern?

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  • $\begingroup$ The upper limit is t, but I thought was a mistake in the mock exam too..And yes, I can to also to what is converges now! thanks! $\endgroup$ – 2410 May 11 '16 at 15:41

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