Given $-1<a<b<0$, I want to find a polynomial $f(x)\in\Bbb R[x]$ such that $f(x)\in[a,b]$ at every $x\in[b^2,a^2]$ and $f(0)=0$. What is minimum degree that is needed and maximum degree that will suffice?


What is degree as function of $a,b$ that is necessary and sufficient using chebyshev polynomials?


Is $O(1)$ degree sufficient?


Quadratic will not work. Min of quadratic goes below $[a,b]$ at some point in middle of $[b^2,a^2]$. Example take $b=−1/9,a=−1/3$. Quadratic is $3x(−13+81x)/4$ which takes min value at $x=13/162\in[1/81,1/9]$ with value $-169/432<-1/3$. Although $-((111 x)/10) + (729 x^2)/4 - (19683 x^3)/20$ seems to work. Is answer cubic always?


I think theory of Tchebyshev polynomials or Jackson's theorem in approximation theory should be useful here.


The question is about finding a polynomial that hits zero one place, and then is squeezed in a rectangle in some other region with rectangle $\mathsf{Height} = |a-b|$ here while $\mathsf{Width}=|a^2-b^2|=|(a-b)(a+b)|\leq |a-b|=\mathsf{Height}$ if $a+b\leq1$, $\mathsf{Width}=|a^2-b^2|=|(a-b)(a+b)|\geq |a-b|=\mathsf{Height}$ if $a+b\geq1$.



Degree necessary of a polynomial?
**Update**: The solution has been edited in order to answer a more general question.


We can always rescale to obtain the following problem: find a polynomial $f(x)$ of minimal degree such that $f(0)=0$ and $f([\beta,1])\subseteq [\alpha,1]$. Here $1>\alpha=\sqrt\beta=b/a>0$ in the original problem. 


Suppose we have a polynomial $f\_0(x)$ satisfying $f([\beta,1])\subseteq [\alpha,1]$, but $f\_0(0)\leq 0$. Then it is possible to construct a solution to the original problem of degree $\deg f\_0$ by the formula 
$$f(x)=\beta(f\_0(x)),\quad \beta(y)=\frac{y-1}{1-f\_0(0)}+1,$$
since $\beta([\alpha,1])\subseteq[\alpha,1]$.


Therefore the following is true



> 
> Lemma: The problem can be solved with a polynomial of degree $n$ if and only if $\inf\_{f\in P\_n} f(0)\leq 0$, where $P\_n$ is the space of polynomials $f$ of degree $n$ having the property $f([\beta,1])\subseteq [\alpha,1]$. 
> 
> 
> 


Now, this $\inf$ is known and given by Chebyshev polynomials. Indeed, let $Q\_n$ denote the space of polynomials $g$ of degree $n$ satisfying $g([-1,1])\in [-1,1]$. Then we have
$$
g\in Q\_n\Longrightarrow 1-(1-\alpha)\frac{g\left(1-2\frac{x-\beta}{1-\beta}\right)+1}{2}\in P\_n
$$
and this correspondence is bijective. Therefore, the $\inf$ in question is given by
$$
\inf\_{f\in P\_n} f(0)=1-(1-\alpha)\frac{1+\sup\_{g\in Q\_n} g\left(\frac{1+\beta}{1-\beta}\right)}{2},
$$
and since $\frac{1+\beta}{1-\beta}>1$, we have
$$
\sup\_{g\in Q\_n} g\left(\frac{1+\beta}{1-\beta}\right)=T\_n\left(\frac{1+\beta}{1-\beta}\right).
$$


Therefore we get the following statement



> 
> The problem can be solved in degree $n$ if and only if $T\_n\left(\frac{1+\beta}{1-\beta}\right)\geq\frac{1+\alpha}{1-\alpha}$, where $0<\alpha,\beta<1$, and in the original problem $\alpha=\sqrt\beta=b/a$
> 
> 
> 


Alternatively, defining $\gamma=\frac{1+\sqrt\beta}{1-\sqrt\beta}=\frac{a+b}{a-b}$, $\delta=\frac{1+\alpha}{1-\alpha}$ we get the inequality
$$
\cosh (n\log\gamma)\geq\delta.
$$
Looking for the threshold value, replace this by equality and obtain
$$
n=\left\lceil \frac{\cosh^{-1}\delta}{\log\gamma}\right\rceil=\left\lceil \sqrt{\frac{a}{b}}\left(1+\frac{1}{3}\frac{b}{a}+O(b^2/a^2)\right)\right\rceil,
$$
where the last equality sign is for the original problem.
Keeping just the leading term gives a pretty good approximation.


