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Very basic and somewhat open-ended question:

Let $A$ be a symmetric operator on functions $f:X\to \mathbb{R}$, where $X$ is a finite set. Assume that the $L^2\to L^2$ norm of $A$ is $\leq \epsilon$, i.e., every eigenvalue $\lambda$ of $A$ has $|\lambda|\leq \epsilon$, or, what is the same, $|\langle f,Af\rangle|\leq \epsilon |f|_2^2$ for all $f:X\to \mathbb{R}$, where $\langle f,Af\rangle:= \sum_{x\in X} f(x) Af(x)$.

Can we conclude anything about sums of the type $$\sum_{x\in X} f(x) Af(x) A^2f(x),\;\;\;\;\; \sum_{x\in X} f(x) Af(x) A^2f(x) A^3f(x),\;\;\;\;\text{etc.?}$$ Assume $|f|_2=1$ and $|f|_\infty$ bounded, if needed.

(I'm interested both in counterexamples that show that one cannot conclude anything and in cases where one can say something, possibly given some auxiliary information.)

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  • $\begingroup$ You look for bounds which do not depend on $|X|$, yes? $\endgroup$
    – Fedor Petrov
    Commented Aug 16, 2020 at 20:19
  • $\begingroup$ Well, probably, but I'm looking for anything, really. What sort of bound would depend on $|X|$? $\endgroup$
    – H A Helfgott
    Commented Aug 16, 2020 at 20:23
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    $\begingroup$ You can do a little better using the boundedness of $f_1$ to bound some higher norm. The worst case for these simple estimates is when the functions $f_1,f_2,f_3$ are both highly concentrated and highly correlated. I don't see any obstruction to this happening except the boundedness of $f_1$, which is mild. Take $A$ to be an $n\times n$ symmetric matrix where the first row and column are all $\epsilon/\sqrt{|X|}$ except for the first entry which is $\epsilon$, and all other entries are $0$. Take $f_1$ to be the all $1/\sqrt{X}$ vector (or all $1$s depending on normalization). $\endgroup$
    – Will Sawin
    Commented Aug 16, 2020 at 23:25
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    $\begingroup$ In this setting $f_2,f_3,\dots$ all attain the maximal $L^2$ norm, or close to it, and a positive proportion of their $L^2$ mass is concentrated at the point $0$, so these types of bounds will basically be sharp. $\endgroup$
    – Will Sawin
    Commented Aug 16, 2020 at 23:25
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    $\begingroup$ So in this normalization the inequality is $||fg||_2 \leq \sqrt{X} ||f||_2 ||g||_2$ and we can get a bound of $|\sum_{x \in X} \prod_{i=0}^{n-1} A^i f(x) | \leq X ||\prod_{i=0}^{n-2} A^i f ||_2 ||A^{n-1} f ||_2 \ll X ||\prod_{i=1}^{n-2} A^i f ||_2 ||A^{n-1} f ||_2 \leq X^{1 + \frac{n-3}{2}} \prod_{i=1}^{n-1} || A^i f||_2 \leq X^{ \frac{n-1}{2}} \epsilon^{ n (n-1)/2}$ which, if I calculated correctly, my counterexample shows is sharp to within a constant factor. $\endgroup$
    – Will Sawin
    Commented Aug 18, 2020 at 21:50

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