Dear community,


I have a problem which is very simple to state but seems to be hard to answer. 


Statement of the problem
------------------------


Let $f$ and $g$ be two symmetric, real functions in $n$ and $m$ variables, both of them square integrable over $T^n$ and $T^m$, respectively, where $T$ is some subset of $\mathbb{R}$. Fix $1 \leq r \leq \min \{p,q \}$ and define the *contraction* $f \otimes\_r g$ to be the function of $p+q-2r$ variables obtained by identifying $r$ variables of $f$ and $g$ and integrating them out, i.e.
$$(f \otimes\_r g)(x\_1,\ldots,x\_{p+q-2r}) := \int\_{T^r} f(x\_1,\ldots,x\_p,t\_1,\ldots,t\_r) \ g(t\_1,\ldots,t\_r,x\_{p+1},\ldots,x\_{q}) \ \mathrm{d} (t\_1,\ldots,t\_r)$$


The contraction $f \otimes\_r g$ is not necessarily symmetric, but we can always canonically symmetrize it by averaging over all permutations of its variables. Denote this symmetrization by $\widetilde{f \otimes\_r g}$.


Suppose now that you have two sequences $(f\_n)$ and $(g\_n)$ of functions such that  

$$ d(n) := \lVert f\_n \otimes\_r g\_n \rVert\_2$$
is positive and converges to zero. 
The triangle and Cauchy-Schwarz inequalities show, that $\widetilde{d}(n):= \lVert \widetilde{f\_n \otimes\_r g\_n} \rVert\_2$ is bounded by $d(n)$ and therefore also goes to zero.


I would like to know if it is possible that symmetrizing increases the speed of convergence, i.e. that $$\widetilde{d}(n) / d(n) \to 0$$.


All the examples I did so far suggest that this can'thappen but I'm not able to find a rigorous proof.


An example
----------


Let's take the simple case where $n=m=2$ and therefore $r=1$. 


We have $$(f \otimes\_1 g) (x,y) = \int\_T f(x,t) \ g(t,y) \ \mathrm{d}t,$$
$$(\widetilde{f \otimes\_1 g})(x,y) = \frac{1}{2} \int\_T \big( f(x,t) \ g(t,y) + f(y,t) \ g(t,x) \big) \ \mathrm{d}t$$
and therefore
$$\lVert f \otimes\_1 g \rVert\_2^2 = \int\_{T^4} f(t\_1,t\_2)g(t\_2,t\_3)g(t\_3,t\_4)f(t\_4,t\_1) \ \mathrm{d} (t\_1,t\_2,t\_3,t\_4),$$
$$\begin{aligned} \lVert \widetilde{f \otimes\_1 g} \rVert\_2^2 &= \frac{1}{2} \int\_{T^4} 
\big( f(t\_1,t\_2)g(t\_2,t\_3)g(t\_3,t\_4)f(t\_4,t\_1) \\ &\qquad\qquad\qquad + f(t\_1,t\_2)g(t\_2,t\_3)f(t\_3,t\_4)g(t\_4,t\_1)\big) \ \mathrm{d} (t\_1,t\_2,t\_3,t\_4)
\\ &=
\frac{1}{2} \lVert f \otimes\_1 g \rVert\_2^2
+
\frac{1}{2} \int\_{T^4} f(t\_1,t\_2)g(t\_2,t\_3)f(t\_3,t\_4)g(t\_4,t\_1) \mathrm{d} (t\_1,t\_2,t\_3,t\_4)
\end{aligned}.
$$


This gives, in terms of sequences,


$$2 \frac{\widetilde{d}(n)^2}{d(n)^2} = 1
+ 
\frac{
\int\_{T^4} f\_n(t\_1,t\_2)g\_n(t\_2,t\_3)f\_n(t\_3,t\_4)g\_n(t\_4,t\_1) \ \mathrm{d} (t\_1,t\_2,t\_3,t\_4)
}{
\int\_{T^4} f\_n(t\_1,t\_2)g\_n(t\_2,t\_3)g\_n(t\_3,t\_4)f\_n(t\_4,t\_1) \ \mathrm{d} (t\_1,t\_2,t\_3,t\_4)
}
$$


Is it possible for the fraction of integrals on the right to converge to -1 in the limit?


Thanks for your help,


Simon



Can symmetrizing a contraction increase the speed of convergence?
Actually, it is even possible that $\widetilde{f \otimes\_1 g}\equiv0$ without $f \otimes\_1 g$ being zero. Just consider a domain $T$ consisting into two subdomains of unit volume, and functions $f$ and $g$ that are constant on each of these. Then it is the same as dealing with symmetric $2\times2$ matrices $F$ and $G$. Your $f \otimes\_1 g$ is nothing but the ordinary product $FG$ of matrices, and we know that the product of symmetric matrices need not be symmetric. Your $\widetilde{f \otimes\_1 g}\equiv0$ is therefore $\frac12(FG+GF)$. Now, let us take
$$F=\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \end{pmatrix},\qquad
G=\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}.$$
We have $(FG)^2=I\_2$, whereas $FG+GF=0\_2$. Therefore $d=2$ and $\tilde{d}=0$.


