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Source: A Course in Modern Analysis and Its Applications by Cohen page 196

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As in the image, we can see that $(P_a-P_c)(x_n)$ alternates between $\leq 0$ and $\geq 0$.

If we make $(P_a-P_c)(x_n)>0$ whenever $(P_a-P_c)(x_n)\geq 0$ in the image attached and $(P_a-P_c)(x_n)=0$ whenever $(P_a-P_c)(x_n)\leq 0$ in the image attached, we should get a polynomial equal to $0$ at $\geq \frac{r}{2}$(if even) or $\geq\frac{r}{2}+1$ (if odd) points.

Darius
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Françoise Nicolas
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    So you want to prove that if a real polynomial $P$ of degree $\leq r-1$ and $r+1$ reals $a_1 < a_2 < \cdots < a_{r+1}$ (I hope you have these strict inequalities!) satisfy $\left(-1\right)^i P\left(a_i\right) \leq 0$ for all $i$, then $P = 0$. Interesting question -- I'm quite sure that it's true, but as you correctly observe, simple root-counting in intervals does not suffice to prove it, since two consecutive intervals can share a root. I think you can instead argue that if $P$ is nonzero, then the derivative $P'$ is negative somewhere on $\left(a_1, a_2\right)$, positive somewhere ... – darij grinberg May 01 '19 at 02:26
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    ... on $\left(a_2, a_3\right)$, negative again somewhere at $\left(a_3, a_4\right)$, etc. (since otherwise $P$ would be constant on a whole interval $\left(a_i, a_{i+1}\right)$, which would make $P$ identically zero). Thus, $P'$ must have at least $r-1$ roots, which (in view of $\deg P' \leq r-2$) causes $P'$ to be $0$. But this means that $P$ is constant, hence $0$ by the conditions. Can you check this? (I would personally prefer an analysis-free proof.) – darij grinberg May 01 '19 at 02:28
  • since P=Pa-Pc(hence a polynomial function hence continuous and differentiable) is either positive/negative in an alternative pattern at points a1<....<ar+1, we'll show that if P(an)>=0 and P(an+1)<=0 then there exists a point in (an,an+1) such that P'(x)<0( there is an analogous conclusion for P(an)<=0 and P(an+1)>=0)if P'=0 on [an,an+1] then P=0 – Françoise Nicolas May 01 '19 at 03:26
  • ,so let's work on the case where that's not the case meaning P' is a polynomial different from 0,first we start by showing that there exists a point in [an,an+1] for which P(x)=/=0, since P' is also a polynomial we can only have a finite number of points for which P'(x)=0 hence there is an interval in (an,an+1) for which P'(x)>0 or <0 let's call such an interval [u,v] using the mean value theorem there is a point in c in (u,v) such that P'(c)(v-u)=P(v)-P(u), if P(u)=0 then it follows that P(v)=/=0, hence proving the existence of such a point in [an,an+1] such that P(x)=/=0, – Françoise Nicolas May 01 '19 at 03:27
  • if there is such a point in [an,an+1] then it is either >0 or <0, let's suppose P(x)>0 then since P(an+1)<=0 it follows from the mean value theorem that there exists a point c such that P'(c)=P(x)-P(an+1)/(x-an+1)<0 and if P(x)<0, P'(c)=P(an)-P(x)/(an-x)<0, hence proving what we set out to prove – Françoise Nicolas May 01 '19 at 03:27
  • that checks out the existence of such points assumed in ur proof, and the rest checks out by default – Françoise Nicolas May 01 '19 at 03:41

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