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.

  1. 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.

  2. 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!

  • 1
    $\begingroup$ The $M$-test tells you about series of functions, not sequences - so you should be trying to bound $|f_{n + 1} - f_n|$. You can show $\lVert f_{n + 1} - f_n \rVert_\infty \le \tfrac 13 \lVert f_n - f_{n - 1} \rVert_\infty$ using the calculation from the answer below, and then the $M$-test actually applies, as you can bound the differences by a geometric series. This is essentially re-proving part of the contraction mapping theorem. $\endgroup$ May 27, 2022 at 22:01

1 Answer 1


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


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