Difference between revisions of "Functional inequalities"
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== General == | == General == | ||
| − | Boolean functions are, as their name suggests, functions, and many types of results from functional analysis and geometry carry over to the Boolean setting. That is, it is possible to relate between various integrals, derivatives and evaluations of the functions, resulting in a functional inequality. | + | Boolean functions are, as their name suggests, functions, and many types of results from functional analysis and geometry carry over to the Boolean setting. That is, it is possible to relate between various integrals, derivatives and evaluations of the functions, resulting in a functional inequality. Recurring quantities which tend to show up are the expectation, variance, and [[influence]] of the function. |
| − | == | + | == Isoperimetric inequality == |
| − | + | === The precise statement === | |
| + | Let <math>f</math> be a Boolean function on <math>n</math> bits, and suppose that the support <math>A</math> of <math>f</math> has size <math>m</math>. Then total influence of <math>f</math> (which is exactly equal to the number of edges from <math>A</math> to <math>\{-1,1\}^n \backslash A</math>) is larger than the total influence of a function whose support is the <math>m</math>-smallest elements in lexicographical order on <math>\{-1,1\}^n</math>. In other words, among all functions with support of size <math>m</math>, the radius-<math>m</math> Hamming ball has the smallest total influence. | ||
| + | === An approximation === | ||
| + | A good approximation to the above theorem is the following '''isoperimetric inequality'''. Denoting by <math>\mu</math> the uniform measure on <math>\{-1,1\}^n</math>, | ||
| + | |||
| + | :::{| class="wikitable" | ||
| + | |- | ||
| + | |<math>\mathrm{Inf}(f) \geq 2 \mu(A) \log_2\left(\frac{1}{\mu(A)}\right) = 2\frac{|A|}{2^n}\log_2\left(\frac{2^n}{|A|}\right)</math> | ||
| + | |} | ||
| + | Equality is attained only when the support of <math>f</math> is a [[cube| subcube]]. | ||
| + | |||
| + | == Poincare inequality == | ||
| + | Let <math>f</math> be a Boolean function on <math>n</math> bits. Then | ||
| + | |||
| + | :::{| class="wikitable" | ||
| + | |- | ||
| + | |<math>\mathrm{Var}(f) \leq 2 \sum_{i=1}^n \mathrm{Inf}_i(f) </math>. | ||
| + | |} | ||
| + | |||
| + | == KKL inequality == | ||
| + | === Maximum influence formulation === | ||
| + | Let <math>f</math> be a Boolean function on <math>n</math> bits. Then there exists an input bit with large influence: | ||
| + | |||
| + | :::{| class="wikitable" | ||
| + | |- | ||
| + | |<math>\max_i \mathrm{Inf}_i(f) \geq C\cdot \mathrm{Var}(f) \frac{\log(n)}{n}</math>. | ||
| + | |} | ||
| + | |||
| + | for some universal constant <math>C > 0 </math>. | ||
| + | |||
| + | === Influence inequality formulation === | ||
| + | This inequality generalizes Poincare's inequality. Let <math>f</math> be a Boolean function on <math>n</math> bits. Then | ||
| + | |||
| + | :::{| class="wikitable" | ||
| + | |- | ||
| + | |<math>\mathrm{Var}(f) \leq C \frac{\sum_i \mathrm{Inf}_i(f)}{\log \left(1 / \max_i \mathrm{Inf}_i(f) \right)}</math> | ||
| + | |} | ||
| + | |||
| + | for some universal constant <math>C > 0 </math>. | ||
== Talagrand's influence inequality == | == Talagrand's influence inequality == | ||
| − | + | This inequality generalizes the KKL inequality. Let <math>f</math> be a Boolean function on <math>n</math> bits. Then | |
| + | |||
| + | :::{| class="wikitable" | ||
| + | |- | ||
| + | |<math>\mathrm{Var}(f) \leq C \sum_{i=1}^n\frac{\mathrm{Inf}_i(f)}{1+\log \left(1 / \mathrm{Inf}_i(f) \right)}</math> | ||
| + | |} | ||
| + | |||
| + | for some universal constant <math>C > 0 </math>. | ||
| + | |||
| + | == Talagrands surface area inequality== | ||
| + | Let <math>f</math> be a Boolean function on <math>n</math> bits, and let <math>h(x)</math> be the local [[sensitivity]] of <math>f</math> (the number of neighbors of <math>x</math> on which <math>f</math> obtains a different value than <math>f(x)</math>). Then | ||
| + | |||
| + | :::{| class="wikitable" | ||
| + | |- | ||
| + | |<math>\int \sqrt{h(x)} d\mu(x) \geq C \cdot \mathrm{Var}(f) \left(\log \left(2 + \frac{e}{\sum_i \mathrm{Inf}_i(f)^2} \right) \right) ^ {1/2}</math> | ||
| + | |} | ||
| − | + | for some universal constant <math>C > 0 </math>. | |
| − | |||
Latest revision as of 14:10, 5 April 2021
Contents
General
Boolean functions are, as their name suggests, functions, and many types of results from functional analysis and geometry carry over to the Boolean setting. That is, it is possible to relate between various integrals, derivatives and evaluations of the functions, resulting in a functional inequality. Recurring quantities which tend to show up are the expectation, variance, and influence of the function.
Isoperimetric inequality
The precise statement
Let [math]f[/math] be a Boolean function on [math]n[/math] bits, and suppose that the support [math]A[/math] of [math]f[/math] has size [math]m[/math]. Then total influence of [math]f[/math] (which is exactly equal to the number of edges from [math]A[/math] to [math]\{-1,1\}^n \backslash A[/math]) is larger than the total influence of a function whose support is the [math]m[/math]-smallest elements in lexicographical order on [math]\{-1,1\}^n[/math]. In other words, among all functions with support of size [math]m[/math], the radius-[math]m[/math] Hamming ball has the smallest total influence.
An approximation
A good approximation to the above theorem is the following isoperimetric inequality. Denoting by [math]\mu[/math] the uniform measure on [math]\{-1,1\}^n[/math],
[math]\mathrm{Inf}(f) \geq 2 \mu(A) \log_2\left(\frac{1}{\mu(A)}\right) = 2\frac{|A|}{2^n}\log_2\left(\frac{2^n}{|A|}\right)[/math]
Equality is attained only when the support of [math]f[/math] is a subcube.
Poincare inequality
Let [math]f[/math] be a Boolean function on [math]n[/math] bits. Then
[math]\mathrm{Var}(f) \leq 2 \sum_{i=1}^n \mathrm{Inf}_i(f) [/math].
KKL inequality
Maximum influence formulation
Let [math]f[/math] be a Boolean function on [math]n[/math] bits. Then there exists an input bit with large influence:
[math]\max_i \mathrm{Inf}_i(f) \geq C\cdot \mathrm{Var}(f) \frac{\log(n)}{n}[/math].
for some universal constant [math]C \gt 0 [/math].
Influence inequality formulation
This inequality generalizes Poincare's inequality. Let [math]f[/math] be a Boolean function on [math]n[/math] bits. Then
[math]\mathrm{Var}(f) \leq C \frac{\sum_i \mathrm{Inf}_i(f)}{\log \left(1 / \max_i \mathrm{Inf}_i(f) \right)}[/math]
for some universal constant [math]C \gt 0 [/math].
Talagrand's influence inequality
This inequality generalizes the KKL inequality. Let [math]f[/math] be a Boolean function on [math]n[/math] bits. Then
[math]\mathrm{Var}(f) \leq C \sum_{i=1}^n\frac{\mathrm{Inf}_i(f)}{1+\log \left(1 / \mathrm{Inf}_i(f) \right)}[/math]
for some universal constant [math]C \gt 0 [/math].
Talagrands surface area inequality
Let [math]f[/math] be a Boolean function on [math]n[/math] bits, and let [math]h(x)[/math] be the local sensitivity of [math]f[/math] (the number of neighbors of [math]x[/math] on which [math]f[/math] obtains a different value than [math]f(x)[/math]). Then
[math]\int \sqrt{h(x)} d\mu(x) \geq C \cdot \mathrm{Var}(f) \left(\log \left(2 + \frac{e}{\sum_i \mathrm{Inf}_i(f)^2} \right) \right) ^ {1/2}[/math]
for some universal constant [math]C \gt 0 [/math].
