Multilinear generalization of Cauchy-Schwarz inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:52:08Z http://mathoverflow.net/feeds/question/73923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73923/multilinear-generalization-of-cauchy-schwarz-inequality Multilinear generalization of Cauchy-Schwarz inequality Nate Eldredge 2011-08-28T23:07:41Z 2011-08-29T21:32:37Z <p>Let $V$ be a real vector space, and let $(\cdot,\cdot;\cdot,\cdot) : V^4 \to \mathbb{R}$ be a multilinear form with the following properties:</p> <ol> <li>$(x,y;z,w) = (y,x;z,w) = (x,y;w,z)$ (symmetry in the first and second pairs)</li> <li>$(x,x;z,z) \ge 0$ (positive semidefiniteness in the first and second pairs).</li> </ol> <blockquote> <p>Must such a form satisfy the inequality $$|(x,y;z,w)| \le \sqrt{(x,x;z,z)(y,y;w,w)}?$$</p> </blockquote> <p>The prototype I have in mind is something like $V = C_c^\infty(\mathbb{R}^n)$, with $$(f,g;h,k) = \int f g \nabla h \cdot \nabla k$$ in which case the inequality follows by using Cauchy-Schwarz twice (first in $\mathbb{R}^n$, and then in $L^2(\mathbb{R}^n)$).</p> <p>I'd settle for the inequality $$|(x,y;z,w)| \le C({\epsilon}(x,x;z,z) + \epsilon^{-1}(y,y;w,w))$$ which follows from the above by AM-GM (with $C = 1/2$). I'd also settle for the special case $x=w, y=z$ where it reads $|(x,z;x,z)| \le \sqrt{(x,x;z,z)(z,z;x,x)}$.</p> <p><s>Simply using Cauchy-Schwarz in each pair gives the inequality $$|(x,y;z,w)| \le [(x,x;z,z)(x,x;w,w)(y,y;z,z)(y,y;w,w)]^{1/4}$$ which has cross terms that I don't want.</s> Edit: Of course, as Willie Wong points out and zeb's counterexample confirms, this doesn't work.</p> <p>Thanks!</p> http://mathoverflow.net/questions/73923/multilinear-generalization-of-cauchy-schwarz-inequality/73934#73934 Answer by zeb for Multilinear generalization of Cauchy-Schwarz inequality zeb 2011-08-29T01:41:54Z 2011-08-29T15:22:50Z <p>Even the inequality $(x,z;x,z)^2 \le (x,x;z,z)(z,z;x,x)$ is false:</p> <p>Let $V = \mathbb{R}^2$, with basis $x,z$. Take $(x,x;x,x) = 100$, $(x,z;x,x)=0$, $(z,z;x,x)=1$, $(x,x;x,z)=0$, $(x,z;x,z)=50$, $(z,z;x,z)=0$, $(x,x;z,z) = 1$, $(x,z;z,z)=0$, $(z,z;z,z)=100$, and extend to all of $V^4$ by symmetry and multilinearity.</p> <p>To check that positive semi-definiteness holds, note that we just need to check that</p> <p>$(x+az,x+az;x+bz,x+bz) = 100 + a^2 + 200ab + b^2 + 100a^2b^2 \ge 0$,</p> <p>which easily follows from AM-GM.</p> <p>Now note that we have $2500 = (x,z;x,z)^2 > (x,x;z,z)(z,z;x,x) = 1$.</p> <p>In fact, we even have $6250000 = (x,z;x,z)^4 > (x,x;x,x)(x,x;z,z)(z,z;x,x)(z,z;z,z) = 10000$.</p> <p>Edit: On the other hand, we can prove the following inequality:</p> <p>$4(x,y;z,w)^2 \le ((x,x;z,z)+(x,x;w,w))((y,y;z,z)+(y,y;w,w))$.</p> <p>To see this, note that by positive semi-definiteness we have</p> <p>$0 \le (x+ay,x+ay;z+w,z+w) + (x-ay,x-ay;z-w,z-w)$ $= 2((x,x;z,z)+(x,x;w,w)) + 8a(x,y;z,w) + 2a^2((y,y;z,z)+(y,y;w,w))$</p> <p>for any $a$, and plugging in $a = -\frac{2(x,y;z,w)}{(y,y;z,z)+(y,y;w,w)}$ we get the desired inequality.</p>