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Let $A$ be a symmetric linear operator of norm $\leq 1$ on the space of functions $f:S\to \mathbb{R}$, where $S$ is a set with $N$ elements. Define the inner product $\langle \cdot,\cdot\rangle$ by letting $\{x\}$ have measure $1/N$ for each $x\in S$.

Assume that, for every function $f$ such that $|f|_2=1$ and $|f|_\infty\leq K$, we have $|\langle f,A f\rangle|\leq \alpha$.

Does it follow that there must be a $Y\subset S$ of measure $O(1/K)$ (say) such that, for $X = S\setminus Y$ and any $f:S\to \mathbb{R}$ with $|f|_2=1$, $$|\langle f|_X, A(f|_X)\rangle|\leq 10 \alpha,$$ say?

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  • $\begingroup$ Why doesn't this hold just by homoeneity for $Y=\emptyset $? And how do you define $A (f|_X) $? Do you put the restriction as $0$ outside $X $? $\endgroup$
    – Jochen Wengenroth
    Commented Aug 4, 2020 at 17:23
  • $\begingroup$ I'm not sure how homogeneity would be enough - $A$ could have eigenfunctions with enormous peaks. Sure, define $A(f|_X)$ that way if you wish -- it doesn't matter, since you are going to take the inner product with $f|_X$. $\endgroup$
    – H A Helfgott
    Commented Aug 4, 2020 at 18:03
  • $\begingroup$ I must misunderstand something. For $t=K/|f|_\infty$ one has $|tf|_\infty\le K$ and hence $$\langle f, A(f)\rangle= t^{-2} \langle tf, A(tf)\rangle \le t^{-2}\alpha |tf|_2=\alpha|f|_2.$$ $\endgroup$
    – Jochen Wengenroth
    Commented Aug 5, 2020 at 6:17
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    $\begingroup$ Yes, I stated the question wrongly. Just fixed it. Thanks! $\endgroup$
    – H A Helfgott
    Commented Aug 5, 2020 at 6:32

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