Limit of a sequence of functions defined recursively Let $\{f_n\}$ be a sequence of polynomials with real coefficients defined by $f_0 = 0$ and for $n = 0, 1, 2,\ldots ,$
$$f_{n+1}(x) = f_n(x) + \frac{x^2 − f_n\,^2(x)}{2}$$
.Find $\lim_{n\to\infty} f_n$ on $[−1, 1]$, where the limit is taken in the supremum norm of $f_n$ over the interval $[−1, 1]$.
Should we use induction here? Straightforward substitution seems not to simplify the problem. Any hints? Thanks beforehand.
 A: If we consider for a given $x\in[-1,1]$ that the sequence $f_n(x)$ converges to a limit $l(x)$ we have that
$$l(x)=l(x)+{x^2-l^2(x)\over 2}$$
Which means $l(x)=|x|$
Now consider
$$f_{n}(x)-|x|=f_{n-1}(x)-|x|+{x^2-f^2_{n-1}(x)\over 2}=\left(f_{n-1}(x)-|x|\right)\left(1-{f_{n-1}(x)+|x|\over 2}\right)$$
Let’s prove by induction that $\forall n \,\,0\leq f_n(x)\leq |x|$. It is true for $f_0$ and assume it is true for $f_{n-1}$. One has
$$|x|-f_n(x)|=\left(|x|-f_{n-1}(x)\right)\left(1-{|x|+f_{n-1}(x)\over 2}\right)$$
We have $0\leq f_{n-1}(x)\leq |x|\leq 1$ so $|x| \leq f_{n-1}(x)+|x|\leq 2|x|$ by assumption. This means
$$0\leq \left(|x|-f_{n-1}(x)\right)\left(1-{|x|}\right)\leq |x|-f_n(x)\leq \left(|x|-f_{n-1}(x)\right)\left(1-{|x|\over 2}\right)$$
And this gives us more than we expected because it proves the convergence to $l(x)=|x|$
A: The recursion $f_{n+1}(x) = f_n(x) + \frac{x^2 − f_n^2(x)}{2}$ makes $f_n(x)$ an even function. Using the identity 
$$(|x|-f_{n+1}(x)) = (|x|-f_n(x))(1-(f_n(x)+|x|)/2)$$
and induction we can prove that $\;f_n(x)\le|x|,\;f_n(x)\le f_{n+1}(x),\;$ and $\;f_n(x)\to|x|\;$ as $\;n\to\infty$. The limit  is also true for all $\;|x|<2\;$ because the factor $\;|1-(f_n(x)+|x|)/2|<1\;$ in that case.
This sequence of polynomials is interesting precisely because it approaches $|x|$ in the limit and so is a constructive proof that $|x|$ is the limit of a sequence of polynomials in $x$.
A: Yes I think you should use induction. Also, as a hint, consider that both $f_{n}$ and $f_{n+1}$ should have the same limit
