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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.

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  • $\begingroup$ Source of this problem, please? $\endgroup$ – Gerry Myerson Sep 26 '18 at 7:11
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    $\begingroup$ This is from Dr. David Masser's book titled Auxilliary polynomials, and it's problem 5.5 $\endgroup$ – JonHales Sep 26 '18 at 13:10
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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 $$\lambda(r,rs)=\lambda(rs,r)\leq \dfrac{1}{(s+1)^r}\,.$$

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  • $\begingroup$ I knew there was something simple, thanks a ton. $\endgroup$ – JonHales Sep 26 '18 at 14:51
  • $\begingroup$ @JonHales Does the book mention any specific formula for $\lambda(r,s)$? It would be interesting to know. $\endgroup$ – Batominovski Sep 26 '18 at 14:53
  • $\begingroup$ It does not, the concept was introduced in the exercises. It might in some later chapters. I'll let you know if it turns up. $\endgroup$ – JonHales Sep 26 '18 at 14:54
  • $\begingroup$ Please do. This is an interesting problem. So, I would greatly appreciate what you'd find out. $\endgroup$ – Batominovski Sep 26 '18 at 14:55
  • $\begingroup$ 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! $\endgroup$ – JonHales Sep 28 '18 at 16:50

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