Maximal Inequalities Beyond the Boolean Cube
Maximal Inequalities Beyond the Boolean Cube Alexandra Kolla University of Illinois Urbana-Champaign
March 13, 2014
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property:
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property: Any radius around your house doesn’t have many cops.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property: Any radius around your house doesn’t have many cops. Does such block exist?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property: Any radius around your house doesn’t have many cops. Does such block exist?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube. For a collection A of linear mappings of V to itself, we define the maximal operator MA : V → V as MA f (x) = supA∈A Af (x)
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube. For a collection A of linear mappings of V to itself, we define the maximal operator MA : V → V as MA f (x) = supA∈A Af (x) Of interest is the family S = {Sk }nk=0 of spherical means, the stochastic P linear operators Sk : V → V given by Sk f (x) = {y :d(x,y )=k} f (x)/ kn
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube. For a collection A of linear mappings of V to itself, we define the maximal operator MA : V → V as MA f (x) = supA∈A Af (x) Of interest is the family S = {Sk }nk=0 of spherical means, the stochastic P linear operators Sk : V → V given by Sk f (x) = {y :d(x,y )=k} f (x)/ kn Applying M, we have the spherical maximal operator MS : V → V defined by MS f (x) = max0≤k≤n Sk f (x)
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Inequality
Theorem There is a constant AH such that for all n, kMS k2→2 < AH .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Inequality
Theorem There is a constant AH such that for all n, kMS k2→2 < AH . Equivalently, for all n and f , kMS f k2 ≤ AH kf k2 .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Inequality
Theorem There is a constant AH such that for all n, kMS k2→2 < AH . Equivalently, for all n and f , kMS f k2 ≤ AH kf k2 . AH is dimension- independent.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance. Fixing any nonnegative integer P P N let f be the indicator function of {x : xi = 0, |xi | ≤ 2N}.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance. Fixing any nonnegative integer P P N let f be the indicator function of {x : xi = 0, |xi | ≤ 2N}. Then kf k22 = 2N + 1 while kMS f k22 ∈ Ω(N 2 )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance. Fixing any nonnegative integer P P N let f be the indicator function of {x : xi = 0, |xi | ≤ 2N}. Then kf k22 = 2N + 1 while kMS f k22 ∈ Ω(N 2 )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk This has the following interpretation. Nt f (x) is the expectation of f (y ) where y is obtained by running n independent Poisson processes with parameter 1 from time 0 to time t, and re-randomizing the ith bit of x as many times as there are events in the ith Poisson process.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk This has the following interpretation. Nt f (x) is the expectation of f (y ) where y is obtained by running n independent Poisson processes with parameter 1 from time 0 to time t, and re-randomizing the ith bit of x as many times as there are events in the ith Poisson process. The Nt ’s form a semigroup: Nt1 Nt2 = Nt1 +t2 . The process is equivalent to a Poisson-clocked random walk on the hypercube.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk This has the following interpretation. Nt f (x) is the expectation of f (y ) where y is obtained by running n independent Poisson processes with parameter 1 from time 0 to time t, and re-randomizing the ith bit of x as many times as there are events in the ith Poisson process. The Nt ’s form a semigroup: Nt1 Nt2 = Nt1 +t2 . The process is equivalent to a Poisson-clocked random walk on the hypercube. Maximal inequalities are easily proved for semi-groups.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t).
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t). While direct comparison is difficult (e.g. writing Sk as a linear combination of Nt necessarily entails large coefficients), we can argue that the spectra of these operators should be qualitatively similar.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t). While direct comparison is difficult (e.g. writing Sk as a linear combination of Nt necessarily entails large coefficients), we can argue that the spectra of these operators should be qualitatively similar. Note that the (orthonormal) √ eigenvectors of the Sk are the characters χy (x) = (−1)x·y / 2n .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t). While direct comparison is difficult (e.g. writing Sk as a linear combination of Nt necessarily entails large coefficients), we can argue that the spectra of these operators should be qualitatively similar. Note that the (orthonormal) √ eigenvectors of the Sk are the characters χy (x) = (−1)x·y / 2n . And the corresponding eigenvalues are the (normalized) P (x )(n−x ) (n) Krawtchouk polynomials: κk (x) = kj=0 (−1)j j nk−j (k )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x . Goal will be to show that κk (x) has similar behavior to (1 − 2k/n)x .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x . Goal will be to show that κk (x) has similar behavior to (1 − 2k/n)x . More precisely, we show Lemma There is a constant c > 0 such that for all n and integer (n) 0 ≤ x, k ≤ n/2, |κk (x)| ≤ e −ckx/n
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x . Goal will be to show that κk (x) has similar behavior to (1 − 2k/n)x . More precisely, we show Lemma There is a constant c > 0 such that for all n and integer (n) 0 ≤ x, k ≤ n/2, |κk (x)| ≤ e −ckx/n It is enough to consider values up to n/2 due to symmetry.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n . Underlying graph is the cartesian product of m + 1-cliques.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n . Underlying graph is the cartesian product of m + 1-cliques. Challenge was to prove similar bounds on the eigenvalues, which are again the Krawtchouk polynomials P (x )(n−x ) (n) κk (x) = kj=0 (− m1 )j j nk−j . (k )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n . Underlying graph is the cartesian product of m + 1-cliques. Challenge was to prove similar bounds on the eigenvalues, which are again the Krawtchouk polynomials P (x )(n−x ) (n) κk (x) = kj=0 (− m1 )j j nk−j . (k ) Results hold for all p-norms p > 1.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products? What about products of other graphs (e.g. G (n, p)).
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products? What about products of other graphs (e.g. G (n, p)). Our bounds on Krawtchouks depend on detailed case-by case analysis and/or very tedious calculations and bounds. Is there a more “canonical” way of proving bounds for orthogonal polynomials?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products? What about products of other graphs (e.g. G (n, p)). Our bounds on Krawtchouks depend on detailed case-by case analysis and/or very tedious calculations and bounds. Is there a more “canonical” way of proving bounds for orthogonal polynomials? Try to use the three term reccurence relation and prove bounds on the matrix norms that correspond to those relations? Can we make this approach work for hermite polynomials (which are limits of Krawtcouks?). (see blackboard)
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Thank you
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
March 13, 2014
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property:
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property: Any radius around your house doesn’t have many cops.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property: Any radius around your house doesn’t have many cops. Does such block exist?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
The Drug Dealer Problem Assume you are a drug-dealer (or any criminal) in a city where a few city blocks are occupied by cops. You want to move in a house at a block with the following property: Any radius around your house doesn’t have many cops. Does such block exist?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube. For a collection A of linear mappings of V to itself, we define the maximal operator MA : V → V as MA f (x) = supA∈A Af (x)
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube. For a collection A of linear mappings of V to itself, we define the maximal operator MA : V → V as MA f (x) = supA∈A Af (x) Of interest is the family S = {Sk }nk=0 of spherical means, the stochastic P linear operators Sk : V → V given by Sk f (x) = {y :d(x,y )=k} f (x)/ kn
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Function Let H n be the n-dimensional hypercube equipped with Hamming metric, and V the vector space of real-valued functions on the hypercube. For a collection A of linear mappings of V to itself, we define the maximal operator MA : V → V as MA f (x) = supA∈A Af (x) Of interest is the family S = {Sk }nk=0 of spherical means, the stochastic P linear operators Sk : V → V given by Sk f (x) = {y :d(x,y )=k} f (x)/ kn Applying M, we have the spherical maximal operator MS : V → V defined by MS f (x) = max0≤k≤n Sk f (x)
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Inequality
Theorem There is a constant AH such that for all n, kMS k2→2 < AH .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Inequality
Theorem There is a constant AH such that for all n, kMS k2→2 < AH . Equivalently, for all n and f , kMS f k2 ≤ AH kf k2 .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Maximal Inequality
Theorem There is a constant AH such that for all n, kMS k2→2 < AH . Equivalently, for all n and f , kMS f k2 ≤ AH kf k2 . AH is dimension- independent.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance. Fixing any nonnegative integer P P N let f be the indicator function of {x : xi = 0, |xi | ≤ 2N}.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance. Fixing any nonnegative integer P P N let f be the indicator function of {x : xi = 0, |xi | ≤ 2N}. Then kf k22 = 2N + 1 while kMS f k22 ∈ Ω(N 2 )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Why is This Non Trivial? Consider the 2D Torus with the L1 distance. Fixing any nonnegative integer P P N let f be the indicator function of {x : xi = 0, |xi | ≤ 2N}. Then kf k22 = 2N + 1 while kMS f k22 ∈ Ω(N 2 )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk This has the following interpretation. Nt f (x) is the expectation of f (y ) where y is obtained by running n independent Poisson processes with parameter 1 from time 0 to time t, and re-randomizing the ith bit of x as many times as there are events in the ith Poisson process.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk This has the following interpretation. Nt f (x) is the expectation of f (y ) where y is obtained by running n independent Poisson processes with parameter 1 from time 0 to time t, and re-randomizing the ith bit of x as many times as there are events in the ith Poisson process. The Nt ’s form a semigroup: Nt1 Nt2 = Nt1 +t2 . The process is equivalent to a Poisson-clocked random walk on the hypercube.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Let’s look at the somewhat related stochastic noise operators N = {Nt }P by real t. Letting p = (1 − e −t )/2, we t≥0 indexed n n k set Nt = k=0 k p (1 − p)n−k Sk This has the following interpretation. Nt f (x) is the expectation of f (y ) where y is obtained by running n independent Poisson processes with parameter 1 from time 0 to time t, and re-randomizing the ith bit of x as many times as there are events in the ith Poisson process. The Nt ’s form a semigroup: Nt1 Nt2 = Nt1 +t2 . The process is equivalent to a Poisson-clocked random walk on the hypercube. Maximal inequalities are easily proved for semi-groups.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t).
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t). While direct comparison is difficult (e.g. writing Sk as a linear combination of Nt necessarily entails large coefficients), we can argue that the spectra of these operators should be qualitatively similar.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t). While direct comparison is difficult (e.g. writing Sk as a linear combination of Nt necessarily entails large coefficients), we can argue that the spectra of these operators should be qualitatively similar. Note that the (orthonormal) √ eigenvectors of the Sk are the characters χy (x) = (−1)x·y / 2n .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea: Spectral Approximation
In some ways Sk resembles Nk/n , since Nt is approximately an p average of Sk for k = nt ± nt(1 − t). While direct comparison is difficult (e.g. writing Sk as a linear combination of Nt necessarily entails large coefficients), we can argue that the spectra of these operators should be qualitatively similar. Note that the (orthonormal) √ eigenvectors of the Sk are the characters χy (x) = (−1)x·y / 2n . And the corresponding eigenvalues are the (normalized) P (x )(n−x ) (n) Krawtchouk polynomials: κk (x) = kj=0 (−1)j j nk−j (k )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x . Goal will be to show that κk (x) has similar behavior to (1 − 2k/n)x .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x . Goal will be to show that κk (x) has similar behavior to (1 − 2k/n)x . More precisely, we show Lemma There is a constant c > 0 such that for all n and integer (n) 0 ≤ x, k ≤ n/2, |κk (x)| ≤ e −ckx/n
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Proof Idea
Indeed, the Nt are also diagonal in the χy basis, and for |y | = x, their eigenvalue for χy is (1 − 2t)x . Goal will be to show that κk (x) has similar behavior to (1 − 2k/n)x . More precisely, we show Lemma There is a constant c > 0 such that for all n and integer (n) 0 ≤ x, k ≤ n/2, |κk (x)| ≤ e −ckx/n It is enough to consider values up to n/2 due to symmetry.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n .
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n . Underlying graph is the cartesian product of m + 1-cliques.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n . Underlying graph is the cartesian product of m + 1-cliques. Challenge was to prove similar bounds on the eigenvalues, which are again the Krawtchouk polynomials P (x )(n−x ) (n) κk (x) = kj=0 (− m1 )j j nk−j . (k )
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Other Spaces?
Greenbalt,K,Kraus,Schulman showed a Maximal Inequality for BALLS (not spheres) for Zm+1 n . Underlying graph is the cartesian product of m + 1-cliques. Challenge was to prove similar bounds on the eigenvalues, which are again the Krawtchouk polynomials P (x )(n−x ) (n) κk (x) = kj=0 (− m1 )j j nk−j . (k ) Results hold for all p-norms p > 1.
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products? What about products of other graphs (e.g. G (n, p)).
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products? What about products of other graphs (e.g. G (n, p)). Our bounds on Krawtchouks depend on detailed case-by case analysis and/or very tedious calculations and bounds. Is there a more “canonical” way of proving bounds for orthogonal polynomials?
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Open Questions Our proof rely on spectral analysis. Is there a combinatorial proof? Spheres and not balls for clique products? What about products of other graphs (e.g. G (n, p)). Our bounds on Krawtchouks depend on detailed case-by case analysis and/or very tedious calculations and bounds. Is there a more “canonical” way of proving bounds for orthogonal polynomials? Try to use the three term reccurence relation and prove bounds on the matrix norms that correspond to those relations? Can we make this approach work for hermite polynomials (which are limits of Krawtcouks?). (see blackboard)
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
Thank you
Alexandra Kolla
Maximal Inequalities Beyond the Boolean Cube
