# Polynomials with prescribed points to match prescribed bounds

Consider real polynomials on the interval $I=[-1,1]$. It is easy to see that the smallest degree for a non-negative polynomial with given zeros $x_1,\dots,x_s\in I^\circ$ is $n=2s$ (e.g. $P(x) = \prod_{i=1}^s (x-x_i)^2$ works).

My question is:

What is the smallest degree for a polynomial such that it is bounded by $\pm 1$ on $I$ and attains the value $1$ on a set $x_1^+,\dots,x_s^+$ and the value $-1$ on a set $x_1^-,\dots,x_r^-$?

Background: I know that the fact about nonnegative polynomials with presribed zeros can be generalized to "generalized polynomials" built from Tchebycheff-systems (due to a theorem by Krein). I would love to see a similar theorem on bounded generalized polynomials which attain the bounds at prescribed points.

Edit: In this question I leanerd from the answer of Gjergji Zaimi that there are bounds on the degree of increasing interpolating polyomials. How does the bounds change for monotone interpolation are described above? Are there (algebraical or numerical) methods to calculate the polynomial?

It seems to me that monotone interpolating polynomials are not treated in the current literature and are not subject of current research. Is that right, and if so is there a special reason for that?

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You understand that the answer depends heavily on the location of the points and, thereby, is by no means as simple and clean as in the non-negative case, right? –  fedja Nov 3 '11 at 13:17
That is pretty clear. I am willing to add more assumptions and not hoping for a very clean answer. –  Dirk Nov 3 '11 at 13:35
@Dirk and @fedia: why does such a polynomial exist for every choice of $x^+, x^-$? Say, what is the polynomial if $x_1^+=1/5, x_2^+=1/4, x_1^-=1/3, x_2^-=1/2$? –  Mark Sapir Nov 3 '11 at 22:13
Such a polynomial always exists: Take a Chebycheff polynomial $P$ of degree high enough (such that is attains the values 1 and -1 in the right order at point $t_i$. Then choose a monotone polynomial $Q$ which maps the points $t_i$ to the prescribed $x_i$ and take $P\circ Q$. Such monotone polynomial interpolants exit by a theorem of Young (ams.org/mathscinet-getitem?mr=0212455). –  Dirk Nov 4 '11 at 7:58
Oh, there is a simpler argument: Since there are piecewise monotone interpolating polynomials we can use them directly to interpolate the $x^+$ and $x^-$ (and possibly adding zero interpolation points inbetween). –  Dirk Nov 4 '11 at 8:56