Why is $f_n(x) = \int_{0}^{1}f_{n-1}(t)\sin(x-t)dt $ uniformly convergent? The function of the sequence defined on the $\mathbb{R}$ by  $f_n(x) = \int_{0}^{1}f_{n-1}(t)\sin(x-t)dt $, $n\geq 2$, plus $f_1(x) = \cos x$.
The first step to show the uniform convergence, I'm trying the find the limit function.
Let the  $f(x)$ be the limit function of the sequence. (Under the hypothesis that $f_n$ is u.c to $f$ )
Then $f(x) = \int_{0}^{1}f(t)\sin(x-t)dt =\int_{x}^{x+1}f(u-x)\sin u du $ by putting the $u =x -t$
The only thing left to do is just find the $f(x)$. But From this step, I was totally stuck for finding the $f$.
What should I do next?
 A: straightforward method
Let's look at the sequence:
$$
f_1=\cos(x), \ f_2=\frac{1}{4}(\cos(2-x)-\cos(x)+2\sin(x)), \ f_3=-\frac{1}{4}\cos^2(1)\cos(x)
$$
From $f_3$ the sequence repeats itself with the factor $-\frac{1}{4}\cos^2(1)$ which is smaller than $1$ (by absolute value). Absolute values of the first two functions are bounded over $\mathbb{R}$ with some positive $A$. Thus we have
$$
\max_{x\in \mathbb{R}} f_n<(\frac{1}{4}\cos^2(1))^k A, \ n > 2k
$$
Thus we have a uniform limit $f=0$.
(similar to) prove with the hint
$$
\|f_n\|=\max_{x\in \mathbb{R}} |f_n(x)|=\max_{x\in \mathbb{R}} |\int_0^1 f_{n-1}(t)\sin(x-t) dt |\leq \\
\leq\max_{x\in \mathbb{R}} \int_0^1 |f_{n-1}(t)\sin(x-t)| dt\leq\|f_{n-1}\|\max_{x\in \mathbb{R}}\int_0^1|\sin(x-t)|dt =\\
\|f_{n-1}\|\max_{x\in \mathbb{R}}\int_x^{x+1}|\sin(t)|dt 
$$
The integral is obviously $<1-\delta$ for some $\delta\in(0,1)$. Thus we have
$$
\|f_n\|\leq \|f_{n-1}\|(1-\delta).
$$
A: Define the map $A:C([0,1])\rightarrow C([0,1])$ as
$$ Af(x)=\int^1_0 f(t)\sin(x-t)\,dt,\qquad 0\leq x\leq 1$$
For any $f,g\in C([0,1])$
$$|Af(x)-Ag(x)|\leq \int^1_0 |f(t)-g(t)||\sin(x-t)|\,dx\leq\sin(1)\|f-g\|_u$$
where $\|\;\|_u$ is the uniform norm on $C([0,1])$. A direct application of the Banach fixed point theorem there is a unique fixed point $f_*\in C([0,1])$ and that for any $f_0\in C([0,1]$
$f_n=Af_{n-1}$ converges (in the uniform norm) to $f_*$.
Further inspection of $A$ shows that $f(t)\equiv0$ is a fixed point of $A$; hence the only one. To summarize: for any $f_0\in C([0,1])$
$f_n=Af_{n-1}\xrightarrow{n\rightarrow\infty}0$ uniformly.
Observation: As mention in several comments, one can do without a full application of the Banach fixed point theorem.  For any $f_0\in C([0,1])$
$$\|Af_0\|_u\leq \sin1\|f\|_u$$
Setting $A^{\circ n}=A\circ\ldots \circ A$ $n$-times yields
$$\|A^{\circ n}\|_u\leq (\sin 1)^n\|f\|_u\xrightarrow{n\rightarrow\infty}0$$
(in the OP's notation, $f_n=A^{\circ n}f_0$)
