# Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$?

Given $$b$$ and $$c$$ with $$b,c>1$$, is it possible to construct a polynomial $$p(x)$$, whose degree is $$n$$ for all $$c$$ and $$b$$, such that:

• $$|p|$$ is strictly increasing on $$[1,c]$$

• and $$|b \cdot p(c)| < |p(0)|$$?

This might be satisfied by an interpolating polynomial, but how to actually construct it is beyond me.

• It doesn't make sense to say that $n$ depends on neither $c$ nor $b$. If you were to word it as "Does there exist an $n$ such that for all $b,c$, there exist a polynomial of degree $n$ such that ..." Also, your first condition can almost be simplified to "the absolute value is strictly increasing". Dec 17 '20 at 1:43
• @Acccumulation Noted and changed accordingly. Dec 17 '20 at 2:22

No, it's not possible to construct such a $$p$$ whose degree $$n$$ is bounded independently of $$b$$ and $$c$$. In fact, it's not possible even if we fix the value of $$c$$. I'll prove this for $$c=2$$ below, but the same argument works in general.

Suppose to the contrary that it were possible. Then for every $$b>1$$ we could choose a polynomial $$p_b$$ such that

• $$p_b$$ has degree $$n$$;
• $$p_b$$ takes non-negative values and is strictly increasing on $$[1,2]$$;
• $$b\cdot|p_b(2)|<|p_b(0)|=1$$.

(We take a polynomial satisfying the desired conditions and multiply by an appropriate scalar.)

Key claim: The coefficients of a polynomial $$p_b$$ satisfying these three conditions are bounded independently of $$b$$. That is, there is a constant $$B$$ such that the absolute value of every coefficient of every polynomial $$p_b$$ is at most $$B$$.

Proof of claim: Fix any $$n+1$$ distinct real numbers $$x_0,x_1,\dots,x_n$$ between $$1$$ and $$2$$ inclusive, and for $$i=0,1,\dots,n$$ define the polynomial $$f_i(x)$$ by$$f_i(x)=\prod_{j\neq i}\frac{x-x_j}{x_i-x_j} \,,$$where the product is taken over all indices $$i=0,1,\dots,n$$ except $$i=j$$. Thus, $$f_i$$ is the unique degree $$n$$ polynomial such that $$f_i(x_i)=1$$ and $$f_i(x_j)=0$$ for $$j\neq i$$.

Now the theory of Lagrange interpolation says that for any polynomial $$p$$ of degree at most $$n$$, we have $$p(x) = \sum_{i=0}^np(x_i)f_i(x) \,.$$ But in our case, we know that we have $$|p_b(x_i)|\leq|p_b(2)|\leq b^{-1}\cdot|p(0)|<1$$ for every $$b$$, since $$p_b$$ is strictly increasing on $$[1,2]$$. Thus, the absolute value of every coefficient of every polynomial $$p_b=\sum_{i=0}^np(x_i)f_i$$ is at most $$(n+1)B'$$, where $$B'$$ is the largest absolute value of any coefficient of any of the $$f_i$$. This gives our bound independently of $$b$$, and proves the key claim.

Now by a compactness argument (a.k.a. the Bolzano--Weierstraß Theorem), our key claim implies that we may choose an increasing sequence of integers $$b_1 such that the polynomials $$p_{b_i}$$ converge coefficientwise to a polynomial $$p$$. What can we say about this limiting polynomial $$p$$? Well, by taking an appropriate limit of the above properties of the $$p_{b_i}$$, we find:

• $$p$$ has degree $$n$$;
• $$p$$ takes non-negative values and is weakly increasing on $$[1,2]$$;
• $$|p(0)|=1$$; and
• $$|p(2)|\leq b_i^{-1}$$ for every $$i$$.

Since the integers $$b_i$$ increase without bound, this final condition implies that actually $$|p(2)|=0$$. Since $$p$$ is non-negatively valued and weakly increasing on $$[1,2]$$, we find that $$p$$ actually has to be equal to $$0$$ on all of $$[1,2]$$. This implies that $$p$$ must be the zero polynomial. But this contradicts the assumption that $$|p(0)|=1$$.

Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$ All linear functionals on a finite-dimensional space are bounded, so if $$\deg p\leqslant n$$, we get $$|p(0)|\leqslant C_n \|p(x)\|$$ for certain $$C_n$$. Thus, if $$b>C_n$$, the second condition is not achievable.

• I think this answer really gets to the point of what's going on here. I think it might be helpful to remark that both Fedor's solution and the one I gave revolve around similar ideas: using the first condition to bound the polynomial $p$, and deriving a contradiction by playing off this bound against the value of $p(0)$ using the second condition. However, in the proof I gave, these two steps are bound up together in the choosing to normalise everything such that $|p(0)|=1$, so I think Fedor's proof makes this underlying structure clearer. Sep 18 '20 at 10:45
• @AlexanderBetts Fedor is a genius. Your answer is good too. Sep 18 '20 at 15:35

Let me present a more explicit version of Fedor’s argument.

Choose distinct $$x_0,\dots,x_n\in[1,c]$$. By Lagrange’s interpolation formula, there exist constants $$a_0,\dots,a_n$$ such that $$p(0)=\sum_{I=0}^n a_ip(x_i)$$ for each polynomial $$p$$ of degree not exceeding $$n$$. Therefore, $$|p(0)|\leq \sum_{I=0}^n |a_i|\cdot |p(x_i)|\leq |p(c)|\cdot \sum_{I=0}^n |a_i|.$$