An extension of the Hardy-Littlewood-Polya inequality? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:35:33Z http://mathoverflow.net/feeds/question/53346 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53346/an-extension-of-the-hardy-littlewood-polya-inequality An extension of the Hardy-Littlewood-Polya inequality? Aryeh Kontorovich 2011-01-26T12:17:03Z 2011-01-26T14:32:54Z <p>Let $x,y$ be vectors in $\mathbb{R}^n$ and let's use the notation $\hat x$ for the vector $x$ with its components sorted in increasing order. The Hardy-Littlewood-Polya inequality states that $$x\cdot y \leq \hat x\cdot \hat y.$$ Let us also use the notation $xy\in\mathbb{R}^n$ to denote the coordinate-wise product of $x$ and $y$. I conjecture that $$\frac{ ||xy||_p ||xy||_r}{||xy||_q} \le \frac{ ||\hat x\hat y||_p ||\hat x\hat y||_r}{||\hat x\hat y||_q}$$ for all $1\le p\le q\le r$. For $q=p$ and $q=r$, my conjectured inequality is true by the HLP inequality. Any ideas for a proof?</p> <p>UPDATE: thank you for the quick answers. The counterexamples indeed work when negative coordinates for x and y are allowed. However, when all the coordinates of x and y are required to be positive, the conjecture seems to hold.</p> <p>UPDATE 2: so the conjecture is totally false; see below for counterexamples.</p> http://mathoverflow.net/questions/53346/an-extension-of-the-hardy-littlewood-polya-inequality/53348#53348 Answer by fedja for An extension of the Hardy-Littlewood-Polya inequality? fedja 2011-01-26T12:50:46Z 2011-01-26T12:50:46Z <p>Rather for a counterexample. Let's say all coordinates are positive.</p> <p>The inequality is equivalent to the claim that $f(t)=\frac{\|xy\|_t}{\|\hat x\hat y\|_t}$ satisfies $f(s)f(t)\le f(1)$ for $s\le 1\le t$ ($1\le p$ is not really a restriction due to the possibility to raise to positive powers inside and outside, so only the ratios $p/q$ and $r/q$ really matter). Also sums can be replaced by averages. Now, as $s\to 0$, we have the geometric means in the limit, which do not feel the rearrangements, so $f(0+)=1$. Also, $f(\infty)=1$ if only the maxima match in the original arrangements. But $f(1)&lt;1$ unless the orderings are exactly the same. </p> http://mathoverflow.net/questions/53346/an-extension-of-the-hardy-littlewood-polya-inequality/53349#53349 Answer by suVRit for An extension of the Hardy-Littlewood-Polya inequality? suVRit 2011-01-26T13:12:16Z 2011-01-26T13:12:16Z <p>Here is a Matlab script that will generate a quick counterexample for you:</p> <pre><code>function [x,y]=testIneq(n, p, q, r) % x and y are length n vectors % Try: [x,y]=testIneq(2,1,2,3) to get a counterexample! flag = 1; iter = 0; while (flag) iter = iter + 1; x = randn(n,1); y = randn(n,1); xh = sort(x); yh = sort(y); xy = x .* y; xyh = xh .* yh; lhs = norm(xy,p) * norm(xy,r) / norm(xy,q); rhs = norm(xyh,p) * norm(xyh,r) / norm(xyh,q); if (rhs &lt; lhs) flag = 0; fprintf('Found countex after %d tries\n', iter); end end </code></pre> <p>end</p> <p>Example: $x =[-2.1384,-0.8396]$, $y =[1.3546,-1.0722]$, with $p=1$, $q=2$, $r=3$.</p>