Polynomials bounded by $[-1, 1]$ iff argument is in $[-1, 1]$ Problem: $f(x)$ is a polynomial with complex coefficients, such that $-1 \leq f(x) \leq 1$ iff $-1 \leq x \leq 1$. Find all such $f(x)$.
My observations: Now, its easy to see that coefficients are real, as we can get the coefficients as solutions of system of linear equations with real numbers. Now, using continuity (sign preserving property) to conclude that $f(1)=1 \text{or} -1$ and $f(-1)=1 \text{or} -1$.
With this information, it looks good to consider $f(x)^2$ as we are able to say something concrete about it at 1 and -1. So, I got  $f(x)^2-1=(x^2-1)g(x)$. For some polynomial $g(x)$.
Also, clearly Chebyshev polynomials of the first kind definitely work (I just learnt about Chebyshev polynomials). But I don't know how commment on existence of more solutions.
My intuition says there should be infinitely many polynomials as the graph outside the $[-1, 1]$ zone, is free to be anything provided its more than one in magnitude. But I don't know if this is correct as graphs of polynomials have other restrictions on them.
Also, I am in high school so please don't stuff like complex analysis (if its possible to avoid it, just a request). I know calculus of real variable. 
Someone please help!!
 A: As I said above, given any polynomial $f$ that's bound correctly, we can construct infinitely more by writing $x^n f(x)$.
Moreover, given $f,g$ that are both bound correctly, their product $fg$ will also be bound correctly.
Now, to see that there isn't some finite set $\{f_0,f_1,...,f_n\}$ such that all other polynomials bound this way are a product of $f_i's$, consider the following:
It's well known that given $n$ points a polynomial of degree $n-1$ can be "fit" to them (constructed so it passes through all of them).  (See Lagrange Interpolation).
Consider a $2$nd degree polynomial (a parabola).  Let's try fitting it through three points: $(-1,1),(1,1)$, and $(0,a)$, where $a$ isn't chosen yet (but $|a|<1$ will be true).  Via WolframAlpha, this gives us the polynomial $$f(x)=(1-a)x^2+a$$
Note that for $\|x\|\leq 1$ (and $0\leq a<1$), we have $\|f(x)\|=\|(1-a)x^2+a\|\underbrace{\leq}_{\Delta}\|1-a\|\|x^2\|+\|a\|\leq \|1-a\|+\|a\|\leq 1$
Here, the $\Delta$ is used to denote an application of the Triangle Inequality, which says that $\|x+y\|\leq\|x\|+\|y\|$.  It's called the "Triangle Inequality" because if $x,y$ are the lengths of two sides of a triangle, then $x+y$ is the length of the third (for some triangle), and that third side must be shorter than the sum of the lengths of the other two sides.
So, we have that $\|x\|\leq 1\implies \|f(x)\|\leq 1$.  Now, assume that $\|f(x)\|\leq 1$, so $\|(1-a)x^2+a\|\leq 1$.  This just means that $-1\leq (1-a)x^2+a\leq 1$, so we have that $-1-a\leq (1-a)x^2\leq 1-a$, or that $\frac{a+1}{a-1}\leq x^2\leq 1$.  Note that this upper bound is good, but $\frac{a+1}{a-1}\leq -1$, with equality when $a=0$, so we need to find a lower bound still.  Fortunately, we know that $f:\mathbb{R}\to\mathbb{R}$, so $x\in\mathbb{R}$, so it follows that $x^2\in\mathbb{R}_{\geq 0}$, so we know that $0\leq x^2\leq 1$, and thus it follows that  $-1\leq x\leq 1$.
So, only quadratic functions, we've shown that given any $a\in [0,1)$, there's a unique quadratic function such that $f(0)=a$, and $f$ is bounded how you want it to be.  As there aren't finitely many $a$, we don't have finitely many solutions, so describing them all in terms of some finite set of functions $\{f_0,f_1,...,f_n\}$ seems impossible.  I'd imagine you could show similar things for polynomials of degree other than 2, and possibly for $a\in (-1,0)$ as well.
Essentially, there are for sure infinitely many polynomials, and to get a better answer you'll need to pose a more specific question.
