TITLE: Show that $f_n(x)$ converges uniformly to a continuous function on $[0,1 ]$
QUESTION [6 upvotes]: Let $f_0(x): [0,1 ] \rightarrow \mathbb{R}$, where $f_0(x)=x$.
Define $f_n(x)=-1 +\frac{1}{3} \int_0 ^1 \cos (3x^2+y^2+1) f_{n-1}(y) dy$
Show that $f_n(x)$ converges uniformly to a continuous function on $[0,1 ]$.
I thought of ways of solving this questions.

First we want to find a function $f(x)$ that $f_n(x)$ converges to, then use the $\epsilon-\delta$ definition. But the integral part just makes it difficult to proceed this way.

We can try the Weierestrass M-test, by first proving $f_n$ is continuous. Then use the Weierestrass M-test to prove uniform convergence. Then conclude that $f$ is continuous. But I had trouble finding a bound for the M-test. I think it's the recursive part that is making it difficult. But I'm sort of guessing $|f_n|\leq 1$. But $\sum 1$ does not converge.


Any help will be appreciated!

REPLY [5 votes]: Use the contraction mapping theorem applied to the space $C[0,1]$ and the mapping $T:C[0,1]\to C[0,1]$ with
$$Tf(x)=-1+\frac{1}{3}\int_0^1\cos\left(3x^2+y^2+1\right)f(y)\,dy.$$
To see that this is a contraction, note that
$$\begin{align*}
\left|Tf(x)-Tg(x)\right|&=\frac{1}{3}\left|\int_0^1\cos\left(3x^2+y^2+1\right)\big(f(y)-g(y)\big)dy\right| \\
&\leq\frac{1}{3}\int_0^1\left|\cos\left(3x^2+y^2+1\right)\right|\left|f(y)-g(y)\right|\,dy \\
&\leq\frac{1}{3}\sup_{x\in[0,1]}\left|f(x)-g(x)\right|,
\end{align*}$$
and thus
$$\left\Vert Tf-Tg\right\Vert=\sup_{x\in[0,1]}\left|Tf(x)-Tg(x)\right|\leq\frac{1}{3}\sup_{x\in[0,1]}\left|f(x)-g(x)\right|=\frac{1}{3}\left\Vert f-g\right\Vert.$$