Timeline for Eigenvalue bounds and triple (and quadruple, etc.) products
Current License: CC BY-SA 4.0
15 events
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| Aug 19, 2020 at 7:35 | comment | added | Mikael de la Salle | ... on the other hand, if you make the additional assumption that the $L^\infty \to L^\infty$ norm of $A$ is $\leq 1$, by interpolation and Hoelder you get the bound $|\frac{1}{X} \sum_x \prod_{i=0}^{n-1} A^i f(x) | \leq \epsilon^{2n-3}$. Same bound for $n=3$, but better for $n>3$ in the regime $\log X>>-\log \epsilon$. | |
| Aug 18, 2020 at 21:50 | comment | added | Will Sawin | 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. | |
| Aug 18, 2020 at 11:31 | comment | added | H A Helfgott | Well, I meant to define the inner product with a factor of $1/X$, and then bounded just means bounded. | |
| Aug 18, 2020 at 10:17 | comment | added | Mikael de la Salle | Just to be sure, when you write "$|f|_\infty$ is bounded", do you mean that $\sup_{x \in X} |f(x)| \leq C/\sqrt{|X|}$? | |
| Aug 16, 2020 at 23:25 | comment | added | Will Sawin | 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. | |
| Aug 16, 2020 at 23:25 | comment | added | Will Sawin | 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). | |
| Aug 16, 2020 at 23:22 | comment | added | Will Sawin | @HenrikRüping Why use the $L^2$ - $L^1$ norm estimate and not just multiply in two batches and use Cauchy-Schwarz? If $f_1,f_2, f_3$ have $L_2$ norms at most $a_1,a_2,a_3$ then $\left| \sum_{x \in X} f_1(x)f_2(x) f_3(x) \right| \leq a_1 a_2 a_3$. (But maybe the $L^2$ norm is supposed to be normalized so $X$ has total mass one and not point mass one?) | |
| Aug 16, 2020 at 20:46 | comment | added | Fedor Petrov | I would start with writing $A=\sum \lambda_i P_i$ for projectors $P_i$ and changing the order of summation. We get $\sum_{i,j} \lambda_i \lambda_j^2 \sum_x f(x) f_i(x)f_j(x)$ where $f_i(x)=(P_i f)(x)$. Do you see anything more clever than bounding this as $\varepsilon^3 \sum_{i,j}|\sum_x f(x)f_i(x)f_j(x)|$? | |
| Aug 16, 2020 at 20:46 | comment | added | H A Helfgott | Ah, well, yes, I was thinking that one might be able to afford factors of $\log X$ or $\log \log X$. A factor of $\sqrt{X}$ is too large. | |
| Aug 16, 2020 at 20:31 | comment | added | HenrikRüping | WEll maybe a bit too simple, but we have for pointwise multiplication of functions that $ |f\cdot g|_2 \le |f|_2\cdot |g|_2$, and thus we ould estimate the L_2-Norm of $f \cdot Af\cdot A^2f$ by $\varepsilon^3 |f|_2^3$. And similarly the L_2-norm of $f\cdot Af\cdot A^2 f\cdot A^3f$ by $\varepsilon^6 |f|_2^4$ etc. Then $|f|_1$ and $|f|_2$ differ at most by a factor of $\sqrt{|X|}$. | |
| Aug 16, 2020 at 20:23 | comment | added | H A Helfgott | Well, probably, but I'm looking for anything, really. What sort of bound would depend on $|X|$? | |
| Aug 16, 2020 at 20:19 | comment | added | Fedor Petrov | You look for bounds which do not depend on $|X|$, yes? | |
| Aug 16, 2020 at 19:02 | history | edited | H A Helfgott | CC BY-SA 4.0 |
edited title
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| Aug 16, 2020 at 18:48 | history | edited | H A Helfgott | CC BY-SA 4.0 |
added 1 character in body
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| Aug 16, 2020 at 6:58 | history | asked | H A Helfgott | CC BY-SA 4.0 |