# Polynomial Length upper bound

Define the length of a polynomial $$A = a_0 + a_1x+\cdots +a_nx^n \in \mathbb{R}[x]$$ to be $$L(A) = \sum_{i=0}^n |a_i|$$, and let $$\lambda (r,s) : = \inf \frac{L(AB)}{L(A)L(B)}$$ where the degree of $$A$$ is at most $$r$$, and the degree of $$B$$ is at most $$s$$.

Prove that $$\lambda(r,r) \leq \frac{1}{2^r}$$.

I've tried two ideas, neither of which I could really get to work. The first was an induction argument, in which I was able to successfully show the base case, but couldn't get the inductive step.

The second thought I had was to see if I could just explicitly find polynomials $$A$$ and $$B$$ that have length $$(1/2)^r$$ for any $$r$$, that would also work to show the infimum is less then that value.

Any hints or thoughts would be greatly appreciated.

• Source of this problem, please? – Gerry Myerson Sep 26 '18 at 7:11
• This is from Dr. David Masser's book titled Auxilliary polynomials, and it's problem 5.5 – JonHales Sep 26 '18 at 13:10

For a fixed $$r\in\mathbb{Z}_{\geq0}$$, let $$A(x):=(1+x)^r$$ and $$B(x):=(1-x)^r$$. Then, $$L(A)=L(B)=L(AB)=2^r\,.$$ That is, $$\frac{L(AB)}{L(A)\,L(B)}=\frac{1}{2^r}\,.$$ Hence, $$\lambda(r,r)\leq \frac{1}{2^r}\,.$$
There are some easy-to-prove results. First, $$\lambda(r,0)=\lambda(0,r)=1$$ for all $$r\in\mathbb{Z}_{\geq 0}$$. Next, $$\lambda(r,1)=\lambda(1,r)\leq \dfrac{1}{r+1}$$ for all $$r\in\mathbb{Z}_{>0}$$, where the equality holds, at least, for $$r=1$$. Finally, for all positive integers $$r$$ and $$s$$, we get $$\lambda(r,rs)=\lambda(rs,r)\leq \dfrac{1}{(s+1)^r}\,.$$
• @JonHales Does the book mention any specific formula for $\lambda(r,s)$? It would be interesting to know. – Batominovski Sep 26 '18 at 14:53
• I asked my professor today during lecture, and he said as far as he was concerned, currently in the literature, there is no known solution for how to calculate $\lambda(r,s)$ for a given $r$ and $s$. Sounds like an interesting research problem! – JonHales Sep 28 '18 at 16:50