$f_n:=\int _I h(x,y)f_{n-1}(y)dy$ uniformly convergent Proposition:$h(x,y)$ is $C^1$ function on $[0,1]^2$ and $f_0(x)$ is continuous on $I:=[0,1]$.
Let $f_n:=\int _I h(x,y)f_{n-1}(y)dy$ $(n=1,2,\cdots)$
Suppose  $M:=\sup_n \max_x |f_n(x)| <\infty $ and for all continuous $g(x)$ on $I$, $\int_I f_n(y)g(y)dy$ convergent.
Then, $f_n(x)$ uniformly convergent.
My idea: I proved $f_n$ pointwise convergent by $g(y):=h(x,y)$ and there exist subsequence $f_{n_k}$ uniformly convergent by Ascoli-Arzela.
But I can't prove $f_n$ convergent uniformly.

In the metric space $C[0,1]$ with the supremum metric the sequence $\{f_n\}$ is relatively compact. If a subsequence $\{f_{n_{k}}\}$ converges to a function $h$ then $\int f_{n_{k}} g \to \int hg$. If $h'$ is the limit of another subsequence then we get $\int hg =\int h'g$ for all $g \in C[0,1]$ and this implies $h=h'$. Thus all subsequential limits of $\{f_n\}$ are the same. This implies that the entire sequence converges in the metric of $C[0,1]$,i.e. uniformly. 
