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, $|\langle f,Af\rangle|\leq \epsilon |f|_2^2$, 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.)

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.)

Very basic and somewhat open-ended question:

Let $A$ be a symmetric operator on functions $f:X\to \mathbb{R}$, where $X$ is a bounded 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, $|\langle f,Af\rangle|\leq \epsilon |f|_2^2$, 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.)

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, $|\langle f,Af\rangle|\leq \epsilon |f|_2^2$, 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.)