# Can symmetrizing a contraction increase the speed of convergence?

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

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 $\tilde{d}=0$.