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.
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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.
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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!
Best Answer
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 $$\left\Vert Tf-Tg\right\Vert=\sup_{x\in[0,1]}\left|Tf(x)-Tg(x)\right|\leq\frac{1}{3}\sup_{x\in[0,1]}\left|f(x)-g(x)\right|=\frac{1}{3}\left\Vert f-g\right\Vert.$$