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:

  1. $(x,y;z,w) = (y,x;z,w) = (x,y;w,z)$ (symmetry in the first and second pairs)
  2. $(x,x;z,z) \ge 0$ (positive semidefiniteness in the first and second pairs).

Must such a form satisfy the inequality $$|(x,y;z,w)| \le \sqrt{(x,x;z,z)(y,y;w,w)}?$$

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 $L^2(\mathbb{R}^n)$).

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 $|(x,z;x,z)| \le \sqrt{(x,x;z,z)(z,z;x,x)}$.

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. Edit: Of course, as Willie Wong points out and zeb's counterexample confirms, this doesn't work.


  • $\begingroup$ I wonder how far CS can be generalized... $\endgroup$ – M.G. Aug 28 '11 at 23:34
  • $\begingroup$ Without multilinearity this breaks down; also without the said symmetry it breaks down easily. But otherwise, not sure yet. $\endgroup$ – Suvrit Aug 28 '11 at 23:40
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    $\begingroup$ I think these notes by Terence Tao might interest you: terrytao.wordpress.com/2010/05/19/higher-order-hilbert-spaces $\endgroup$ – Mark Aug 29 '11 at 8:18
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    $\begingroup$ I don't think "Cauchy-Schwarz in each pair" is legal. Fixing $z,w$ you don't know that $(\cdot,\cdot;z,w)$ is a positive semi-definite bilinear form. $\endgroup$ – Willie Wong Aug 29 '11 at 16:37

Even the inequality $(x,z;x,z)^2 \le (x,x;z,z)(z,z;x,x)$ is false:

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.

To check that positive semi-definiteness holds, note that we just need to check that

$(x+az,x+az;x+bz,x+bz) = 100 + a^2 + 200ab + b^2 + 100a^2b^2 \ge 0$,

which easily follows from AM-GM.

Now note that we have $2500 = (x,z;x,z)^2 > (x,x;z,z)(z,z;x,x) = 1$.

In fact, we even have $6250000 = (x,z;x,z)^4 > (x,x;x,x)(x,x;z,z)(z,z;x,x)(z,z;z,z) = 10000$.

Edit: On the other hand, we can prove the following inequality:

$4(x,y;z,w)^2 \le ((x,x;z,z)+(x,x;w,w))((y,y;z,z)+(y,y;w,w))$.

To see this, note that by positive semi-definiteness we have

$0 \le (x+ay,x+ay;z+w,z+w) + (x-ay,x-ay;z-w,z-w)$ $ = 2((x,x;z,z)+(x,x;w,w)) + 8a(x,y;z,w) + 2a^2((y,y;z,z)+(y,y;w,w))$

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.

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  • $\begingroup$ Thanks for the counterexample! This was helpful in finding out what extra structure I needed to use to solve my actual problem. I'll try to post some more info soon. $\endgroup$ – Nate Eldredge Aug 29 '11 at 21:34

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