# Entropy optimality: Bloom’s Chang’s Lemma

[Added in post: One might consult this post for a simpler/more general (and slightly more correct) version of the results presented here.]

In our last post regarding Chang’s Lemma, let us visit a version due to Thomas Bloom. We will offer a new proof using entropy maximization. In particular, we will again use only boundedness of the Fourier characters.

There are two new (and elementary) techniques here: (1) using a trade-off between entropy maximization and accuracy and (2) truncating the Taylor expansion of ${e^x}$.

We use the notation from our previous post: ${G=\mathbb F_p^n}$ for some prime ${p}$ and ${\mu}$ is the uniform measure on ${G}$. For ${\eta > 0}$ and ${f \in L^2(G)}$, we define ${\Delta_{\eta}(f) = \{ \alpha \in G : |\hat f(\alpha)| \geq \eta \|f\|_1 \}}$. We also use ${Q_{\mu} \subseteq L^2(G)}$ to denote the set of all densities with respect to ${\mu}$.

Theorem 1 (Bloom) There is a constant ${c > 0}$ such that for every ${\eta > 0}$ and every density ${f \in Q_{\mu}}$, there is a subset ${\Delta \subseteq \Delta_{\eta}(f) \subseteq G}$ such that ${|\Delta| \geq c \eta |\Delta_{\eta}(f)|}$ and ${\Delta}$ is contained in some ${\mathbb F_p}$ subspace of dimension at most

$\displaystyle \frac{1+\mathrm{Ent}_{\mu}(f)}{c\eta}\,.$

Note that we only bound the dimension of a subset of the large spectrum, but the bound on the dimension improves by a factor of ${1/\eta}$. Bloom uses this as the key step in his proof of what (at the time of writing) constitutes the best asymptotic bounds in Roth’s theorem on three-term arithmetic progressions:

Theorem 2 If a subset ${A \subseteq \{1,2,\ldots,N\}}$ contains no non-trivial three-term arithmetic progressions, then

$\displaystyle |A| \leq O(1) \frac{\left(\log \log N\right)^4}{\log N} N\,.$

This represents a modest improvement over the breakthrough of Sanders achieving ${\frac{(\log \log N)^{O(1)}}{\log N} N}$, but the proof is somewhat different.

1.1. A stronger version

In fact, we will prove a stronger theorem.

Theorem 3 For every ${\eta > 0}$ and every density ${f \in Q_{\mu}}$, there is a random subset ${\Delta \subseteq G}$ such that almost surely

$\displaystyle \dim_{\mathbb F_p}(\mathrm{span}\, \Delta) \leq 12 \frac{\mathrm{Ent}_{\mu}(f)}{\eta} + O(\log (1/\eta))\,,$

and for every ${\alpha \in \Delta_{\eta}(f)}$, it holds that

$\displaystyle \mathbb P[\alpha \in \Delta] \geq \frac{\eta}{4}\,.$

This clearly yields Theorem 1 by averaging.

1.2. The same polytope

To prove Theorem 3, we use the same polytope we saw before. Recall the class of test functionals ${\mathcal F = \{ \pm \mathrm{Re}\,\chi_\alpha, \pm \mathrm{Im}\,\chi_\alpha : \alpha \in G\} \subseteq L^2(G) }$

We defined ${P(f,\eta) \subseteq L^2(G)}$ by

$\displaystyle P(f,\eta) = \left\{ g \in L^2(G) : \langle g, \varphi \rangle \geq \langle f,\varphi\rangle - \eta\quad\forall \varphi \in \mathcal F\right\}\,.$

Let us consider a slightly different convex optimization:

$\displaystyle \textrm{minimize } \mathrm{Ent}_{\mu}(g)+K \mathbf{\varepsilon} \qquad \textrm{ subject to } g \in P(f,\mathbf{\varepsilon}) \cap Q_{\mu}\,. \ \ \ \ \ (1)$

Here, ${K}$ is a constant that we will set soon. On the other hand, ${\varepsilon \geq 0}$ is now intended as an additional variable over which to optimize. We allow the optimization to trade off the entropy term and the accuracy ${\varepsilon}$. The constant ${K > 0}$ represents how much we value one vs. the other.

Notice that, since ${f \in P(f,0) \cap Q_{\mu}}$, this convex program satisfies Slater’s condition (there is a feasible point in the relative interior), meaning that strong duality holds (see Section 5.2.3).

1.3. The optimal solution

As in our first post on this topic, we can set the gradient of the Lagrangian equal to zero to obtain the form of the optimal solution: For some dual variables ${\{\lambda^*_{\varphi} \geq 0: \varphi \in \mathcal F\} \subseteq \mathbb R}$,

$\displaystyle g^*(x) = \frac{\exp\left(\sum_{\varphi \in \mathcal F} \lambda^*_{\varphi} \varphi(x)\right)}{\mathop{\mathbb E}_{\mu} \left[\exp\left(\sum_{\varphi \in \mathcal F} \lambda^*_{\varphi} \varphi\right)\right]} \ \ \ \ \ (2)$

Furthermore, corresponding to our new variable ${\varepsilon}$, there is a new constraint on the dual variables:

$\displaystyle \sum_{\varphi \in \mathcal F} \lambda^*_{\varphi} \leq K\,.$

Observe now that if we put ${K = 2 \frac{\mathrm{Ent}_{\mu}(f)}{\eta}}$ then we can bound ${\varepsilon^*}$ (the error in the optimal solution): Since ${f}$ is a feasible solution with ${\varepsilon=0}$, we have

$\displaystyle \mathrm{Ent}_{\mu}(g^*) + K \varepsilon^* \leq \mathrm{Ent}_{\mu}(f)\,,$

which implies that ${\varepsilon^* \leq \frac{\mathrm{Ent}_{\mu}(f)}{K} = \frac{\eta}{2}}$ since ${\mathrm{Ent}_{\mu}(g^*) \geq 0}$.

To summarize: By setting ${K}$ appropriately, we obtain ${g^* \in P(f,\eta/2) \cap Q_{\mu}}$ of the form (2) and such that

$\displaystyle \sum_{\varphi \in \mathcal F} \lambda_{\varphi}^* \leq 2\frac{\mathrm{Ent}_{\mu}(f)}{\eta}\,. \ \ \ \ \ (3)$

Note that one can arrive at the same conclusion using the algorithm from our previous post: The version unconcerned with sparsity finds a feasible point after time ${T \leq \frac{\mathrm{Ent}_{\mu}(f)}{\varepsilon}}$. Setting ${\varepsilon = \eta/2}$ yields the same result without using duality.

1.4. A Taylor expansion

Let us slightly rewrite ${g^*}$ by multiplying the numerator and denominator by ${\exp\left({\sum_{\varphi \in \mathcal F} \lambda^*_{\varphi}}\right)}$. This yields:

$\displaystyle g^* = \frac{\exp\left(\sum_{\varphi \in \mathcal F} \lambda^*_{\varphi} (1+\varphi)\right)}{\mathop{\mathbb E}_{\mu} \exp\left(\sum_{\varphi \in \mathcal F} \lambda^*_{\varphi} (1+\varphi)\right)}$

The point of this transformation is that now the exponent is a sum of positive terms (using ${\|\varphi\|_{\infty} \leq 1}$), and furthermore by (3), the exponent is always bounded by

$\displaystyle B \stackrel{\mathrm{def}}{=} 4 \frac{\mathrm{Ent}_{\mu}(f)}{\eta}\,. \ \ \ \ \ (4)$

Let us now Taylor expand ${e^x = \sum_{j \geq 0} \frac{x^j}{j!}}$. Applying this to the numerator, we arrive at an expression

$\displaystyle g^* = \sum_{\vec \alpha} y_{\vec \alpha} T_{\vec \alpha}$

where ${y_{\vec \alpha} \geq 0}$, ${\sum_{\vec \alpha} y_{\vec \alpha} = 1}$, and each ${T_{\vec \alpha} \in Q_{\mu}}$ is a density. Here, ${\vec \alpha}$ ranges over all finite sequences of elements from ${\mathcal F}$ and

$\displaystyle T_{\vec \alpha} = \frac{\prod_{i=1}^{|\vec \alpha|} (1+\varphi_{\vec \alpha_i})}{\mathop{\mathbb E}_{\mu} \prod_{i=1}^{|\vec \alpha|} (1+\varphi_{\vec \alpha_i})}=\prod_{i=1}^{|\vec \alpha|} (1+\varphi_{\vec \alpha_i})\,,$

where we use ${|\vec \alpha|}$ to denote the length of the sequence ${\vec \alpha}$.

1.5. The random subset

We now define a random function ${\mathbf{T} \in L^2(G)}$ by taking ${\mathbf{T}=T_{\vec \alpha}}$ with probability ${y_{\vec \alpha}}$.

Consider some ${\gamma \in \Delta_{\eta}(f)}$. Since ${g^* \in P(f,\eta/2)}$, we know that ${\gamma \in \Delta_{\eta/2}(g^*)}$. Thus

$\displaystyle \frac{\eta}{2} < |\langle g^*, \chi_{\gamma}\rangle| \leq \sum_{\vec \alpha} y_{\vec \alpha} |\langle T_{\vec \alpha}, \chi_{\gamma}\rangle| = \mathop{\mathbb E}\,|\langle \mathbf{T},\chi_{\gamma}\rangle|\,.$

But we also have ${|\langle T_{\vec \alpha}, \chi_{\gamma}\rangle| \leq \|T_{\vec \alpha}\|_1 \cdot \|\chi_{\gamma}\|_{\infty} \leq 1}$. This implies that ${\mathbb P[|\langle \mathbf{T}, \chi_{\gamma}\rangle| > 0] > \eta/2}$.

Equivalently, for any ${\gamma \in \Delta_{\eta}(f)}$, it holds that ${\mathbb P[\gamma \in \Delta_0(\mathbf{T})] > \eta/2}$. We would be done with the proof of Theorem 3 if we also knew that ${\mathbf{T}}$ were supported on functions ${T_{\vec \alpha}}$ for which ${|\vec \alpha| \leq O(B)}$ because ${\dim_{\mathbb F_p}(\mathrm{span}(\Delta_0(T_{\vec \alpha}))) \leq |\vec \alpha|}$. This is not necessarily true, but we can simply truncate the Taylor expansion to ensure it.

1.6. Truncation

Let ${p_k(x) = \sum_{j \leq k} \frac{x^j}{j!}}$ denote the Taylor expansion of ${e^x}$ to degree ${k}$. Since the exponent in ${g^*}$ is always bounded by ${B}$ (recall (4)), we have

$\displaystyle \sum_{|\vec \alpha| > k} y_{\vec \alpha} \leq \sup_{x \in [0,B]} \frac{|e^x - p_k(x)|}{e^x} \leq \frac{B^{k+1}}{(k+1)!}\,.$

By standard estimates, we can choose ${k \leq 3 B + O(\log(1/\eta))}$ to make the latter quantity at most ${\eta/4}$.

Since ${\sum_{|\vec \alpha| > k} y_{\vec \alpha} \leq \eta/4}$, a union bound combined with our previous argument immediately implies that for ${\gamma \in \Delta_{\eta}(f)}$, we have

$\displaystyle \mathbb P\left[|\langle \mathbf{T}, \chi_{\gamma}\rangle| > 0 \textrm{ and } \dim_{\mathbb F_p}(\mathrm{span}(\Delta_0(\mathbf{T}))) \leq k \right] \geq \frac{\eta}{4}\,.$

This completes the proof of Theorem 3.

1.7. Prologue: A structure theorem

Generalizing the preceding argument a bit, one can prove the following.

Let ${G}$ be a finite abelian group and use ${\hat G}$ to denote the dual group. Let ${\mu}$ denote the uniform measure on ${G}$. For every ${\gamma \in \hat G}$, let ${\chi_{\gamma}}$ denote the corresponding character. Let us define a degree-${k}$ Reisz product to be a function of the form

$\displaystyle R(x) = \prod_{i=1}^k (1+\varepsilon_{i} \Lambda_i \chi_{\gamma_i}(x))$

for some ${\gamma_1, \ldots, \gamma_k \in \hat G}$ and ${\varepsilon_1, \ldots, \varepsilon_k \in \{-1,1\}}$ and ${\Lambda_1, \ldots, \Lambda_k \in \{\mathrm{Re},\mathrm{Im}\}}$.

Theorem 4 For every ${\eta > 0}$, the following holds. For every ${f : G \rightarrow \mathbb R_+}$ with ${\mathbb E_{\mu} [f]=1}$, there exists a ${g : G \rightarrow \mathbb R_+}$ with ${\mathbb E_{\mu} [g] = 1}$ such that ${\|\hat f - \hat g\|_{\infty} \leq \eta}$ and ${g}$ is a convex combination of degree-${k}$ Reisz products where

$\displaystyle k \leq O(1) \frac{\mathrm{Ent}_{\mu}(f)}{\eta} + O(\log (1/\eta))\,.$

1.8. A prologue’s prologue

To indicate the lack of algebraic structure required for the preceding statement, we can set things up in somewhat greater generality.

For simplicity, let ${X}$ be a finite set equipped with a probability measure ${\mu}$. Recall that ${L^2(X)}$ is the Hilbert space of real-valued functions on ${X}$ equipped with the inner product ${\langle f,g\rangle = \mathop{\mathbb E}_{\mu}[fg]}$. Let ${\mathcal F \subseteq L^2(X)}$ be a set of functionals with the property that ${\|\varphi\|_{\infty} \leq 1}$ for ${\varphi \in \mathcal F}$.

Define a degree-${k}$ ${\mathcal F}$-Riesz product as a function of the form

$\displaystyle R(x) = \prod_{i=1}^k (1+\varphi_i(x))$

for some functions ${\varphi_1, \ldots, \varphi_k \in \mathcal F}$. Define also the (semi-) norm ${\|f\|_{\mathcal F} = \sup_{\varphi \in \mathcal F} |\langle f,\varphi\rangle_{L^2(X)}|}$.

Theorem 5 For every ${\eta > 0}$, the following holds. For every ${f : X \rightarrow \mathbb R_+}$ with ${\mathbb E_{\mu} [f]=1}$, there exists a ${g : X \rightarrow \mathbb R_+}$ with ${\mathbb E_{\mu} [g] = 1}$ such that ${\|f - g\|_{\mathcal F} \leq \eta}$ and ${g}$ is a convex combination of degree-${k}$ ${\mathcal F}$-Riesz products where

$\displaystyle k \leq O(1) \frac{\mathrm{Ent}_{\mu}(f)}{\eta} + O(\log (1/\eta))\,.$

# Entropy optimality: A potential function

Our goal now is to prove Lemma 3 from the last post, thereby completing the entropy-maximization proof of Chang’s Lemma. We will prove it now in greater generality because it shows how relative entropy provides a powerful potential function for measuring progress.

Let ${X}$ be a finite set equipped with a measure ${\mu}$. We use ${L^2(X,\mu)}$ to denote the space of real-valued functions ${f : X \rightarrow \mathbb R}$ equipped with the inner product ${\langle f,g\rangle = \sum_{x \in X} \mu(x) f(x) g(x)}$. Let ${Q_{\mu} \subseteq L^2(X,\mu)}$ denote the convex polytope of probability densities with respect to ${\mu}$:

$\displaystyle Q_{\mu} = \left\{ g \in L^2(X,\mu) : g \geq 0, \mathop{\mathbb E}_{\mu}[g]=1 \right\}\,.$

1.1. A ground truth and a family of tests

Fix now a density ${f \in Q_{\mu}}$ and a family ${\mathcal T \subseteq L^2(X,\mu)}$ that one can think of as “tests” (or properties of a function we might care about). Given an error parameter ${\varepsilon > 0}$, we define a convex polytope ${P_{\varepsilon}(f, \mathcal T) \subseteq L^2(X,\mu)}$ as follows:

$\displaystyle P_{\varepsilon}(f,\mathcal T) = \left\{ g : \langle \varphi, g \rangle \geq \langle \varphi,f\rangle - \varepsilon \quad \forall \varphi \in \mathcal T\right\}\,.$

This is the set of functions that have “performance” similar to that of ${f}$ on all the tests in ${\mathcal T}$. Note that the tests are one-sided; if we wanted two-sided bounds for some ${\varphi}$, we could just add ${-\varphi}$ to ${\mathcal T}$.

1.2. (Projected) coordinate descent in the dual

We now describe a simple algorithm to find a point ${g \in P_{\varepsilon}(f,\mathcal T) \cap Q_{\mu}}$. It will be precisely analogous to the algorithm we saw in the previous post for the natural polytope coming from Chang’s Lemma. We define a family of functions ${\{g_t : t \geq 0\} \subseteq Q_{\mu}}$ indexed by a continuous time parameter:

$\displaystyle g_t = \frac{\exp\left(\int_0^t \varphi_s\,ds\right)}{\mathop{\mathbb E}_{\mu}\left[\exp\left(\int_0^t \varphi_s\,ds\right)\right]}\,.$

Here ${\varphi_t \in \mathcal T}$ is some test we have yet to specify. Intuitively, it will be a test that ${g_t}$ is not performing well on. The idea is that at time ${t}$, we exponentially average some of ${\varphi_t}$ into ${g_t}$. The exponential ensures that ${g_t}$ is non-negative, and our normalization ensures that ${g_t \in Q_{\mu}}$. Observe that ${g_0 = \mathbf{1}}$ is the constant ${1}$ function.

1.3. The potential function

For two densities ${h,g \in Q_{\mu}}$, we define the relative entropy

$\displaystyle D_{\mu}(h\,\|\,g) = \sum_{x \in X} \mu(x) h(x) \log \frac{h(x)}{g(x)}\,.$

Our goal is to analyze the potential function ${D_{\mu}(f \,\|\,g_t)}$. This functional measures how far the “ground truth” ${f}$ is from our hypothesis ${g_t}$.

Note that ${D_{\mu}(f \,\|\,g_0) = D_{\mu}(f \,\|\, \mathbf{1}) = \mathrm{Ent}_{\mu}(f)}$, using the notation from the previous post. Furthermore, since the relative entropy between two probability measures is always non-negative, ${D_{\mu}(f\,\|\,g_t) \geq 0}$ for all ${t \geq 0}$.

Now a simple calculation gives

$\displaystyle \frac{d}{dt} D_{\mu}(f\,\|\,g_t) = \langle \varphi_t, g_t-f\rangle\,. \ \ \ \ \ (1)$

In other words, as long as the constraint ${\langle \varphi_t, g_t\rangle \geq \langle \varphi_t, f\rangle - \varepsilon}$ is violated by ${g_t}$, the potential is decreasing by at least ${\varepsilon}$!

Since the potential starts at ${\mathrm{Ent}_{\mu}(f)}$ and must always be non-negative, if we choose at every time ${t}$ a violated test ${\varphi_t}$, then after time ${T \leq \frac{\mathrm{Ent}_{\mu}(f)}{\varepsilon}}$, it must be that ${g_t \in P_{\varepsilon}(f,\mathcal T)}$ for some ${t \in [0,T]}$.

1.4. A sparse solution

But our goal in Chang’s Lemma was to find a “sparse” ${g \in P_{\varepsilon}(f,\mathcal T)}$. In other words, we want ${g}$ to be built out of only a few constraints. To accomplish this, we should have ${\varphi_t}$ switch between different tests as little as possible.

So when we find a violated test ${\varphi \in \mathcal T}$, let’s keep ${\varphi_t = \varphi}$ until ${\langle \varphi, g_t\rangle = \langle \varphi,f\rangle}$. How fast can the quantity ${\langle \varphi, g_t-f\rangle}$ drop from larger than ${\varepsilon}$ to zero?

Another simple calculation (using ${\varphi_t = \varphi}$) yields

$\displaystyle \frac{d}{dt} \langle \varphi, g_t\rangle = - \left\langle \vphantom{\bigoplus} \varphi, g_t (\varphi - \langle \varphi,g_t\rangle)\right\rangle = - \langle \varphi^2, g_t \rangle + \langle \varphi, g_t\rangle^2\,.$

This quantity is at least ${- \|g_t\|_1 \cdot \|\varphi\|_{\infty}^2 = -\|\varphi\|_{\infty}^2}$.

In order to calculate how much the potential drops while focused on a single constraint, we can make the pessimistic assumption that ${\frac{d}{dt} \langle \varphi, g_t\rangle = - \|\varphi\|_{\infty}^2}$. (This is formally justified by Grönwall’s inequality.) In this case (recalling (1)), the potential drop is at least

$\displaystyle \frac{1}{\|\varphi\|_{\infty}^2} \int_0^{\varepsilon} x \,dx = \frac12 \frac{\varepsilon^2}{\|\varphi\|_{\infty}^2}\,.$

Since the potential can drop by at most ${\mathrm{Ent}_{\mu}(f)}$ overall, this bounds the number of steps of our algorithm, yielding the following result.

Theorem 1 For every ${\varepsilon > 0}$, there exists a function ${g \in P_{\varepsilon}(f,\mathcal T) \cap Q_{\mu}}$ such that for some ${\{\lambda_i \geq 0\}}$ and ${\{\varphi_i\} \subseteq \mathcal T}$

$\displaystyle g = \frac{\exp\left(\sum_{i=1}^k \lambda_i \varphi_i\right)}{\mathop{\mathbb E}_{\mu}\left[\exp\left(\sum_{i=1}^k \lambda_i \varphi_i\right)\right]}\,,$

for some value

$\displaystyle k \leq 2 \frac{\mathrm{Ent}_{\mu}(f)}{\varepsilon^2} \sup_{\varphi \in \mathcal T} \|\varphi\|_{\infty}^2\,.$

In order to prove Lemma 3 from the preceding post, simply use the fact that our tests were characters (and their negations), and these satisfy ${\|\chi_{\alpha}\|_{\infty} \leq 1}$.

1.5. Bibliographic notes

Mohit Singh has pointed out to me the similarity of this approach to the Frank-Wolfe algorithm. The use of the potential ${D(f \,\|\,g_t)}$ is a common feature of algorithms that go by the names “boosting” or “multiplicative weights update” (see, e.g., this survey).

# Entropy optimality: Chang’s Lemma

In the last post, we saw how minimizing the relative entropy allowed us to obtain a “simple” feasible point of a particular polytope. We continue exploring this theme by giving a related proof of Chang’s Lemma, a Fourier-analytic result that arose in additive combinatorics. Our proof will reveal another aspect of entropy optimality: “Projected sub-gradient descent” and a corresponding convergence analysis.

1.1. Discrete Fourier analysis and Chang’s Lemma

Fix ${n \geq 1}$ and a prime ${p}$. Let ${\mathbb F_p}$ denote the finite field with ${p}$ elements. We will work over ${\mathbb F_p^n}$. For convenience, denote ${G=\mathbb F_p^n}$. Let ${\mu}$ denote the uniform measure on ${G}$, and for a non-negative function ${f : G \rightarrow [0,\infty)}$, define the relative entropy

$\displaystyle \mathrm{Ent}_{\mu}(f) = \mathbb E_{\mu} \left[f \log \frac{f}{\mathbb E_{\mu} f}\right]\,,$

where we use the shorthand ${\mathop{\mathbb E}_{\mu} [f] = \sum_{x \in G} \mu(x) f(x)}$. If ${\mathop{\mathbb E}_{\mu} [f] = 1}$, we say that ${f}$ is a density (with respect to ${\mu}$). In that case, ${\mathrm{Ent}_{\mu}(f) = D(f \mu \,\|\,\mu)}$ where ${D}$ denotes the relative entropy (introduced in the last post).

We briefly review some notions from discrete Fourier analysis. One could skip this without much loss; our proof will only use boundedness of the Fourier characters and no other features. For functions ${f,g : G \rightarrow \mathbb C}$, we will use the inner product ${\langle f,g\rangle = \mathop{\mathbb E}_{\mu}[f\bar g]}$ and the norms ${\|f\|_p = (\mathop{\mathbb E}_{\mu}[|f|^p])^{1/p}}$ for ${p \geq 1}$.

For an element ${\alpha \in G}$, we use

$\displaystyle \chi_{\alpha}(x) = \exp\left(\frac{-2\pi}{p} \sum_{i=1}^n \alpha_i x_i\right)$

to denote the corresponding Fourier character. Now every ${f : G \rightarrow \mathbb C}$ can be written in the Fourier basis as

$\displaystyle f = \sum_{\alpha \in G} \hat f(\alpha) \chi_{\alpha}\,,$

where ${\hat f(\alpha) = \langle f,\chi_{\alpha}\rangle}$. (For simplicity of notation, we have implicitly identified ${G}$ and ${\hat G}$.)

We will be interested in the large Fourier coefficients of a function. To that end, for ${\eta > 0}$, we define

$\displaystyle \Delta_{\eta}(f) = \left\{ \alpha \in G : |\hat f(\alpha)| > \eta \|f\|_1 \right\}\,,$

Now we are ready to state Chang’s Lemma; it asserts that if ${\mathrm{Ent}_{\mu}(f)}$ is small, then all of ${f}$‘s large Fourier coefficients are contained in a low-dimensional subspace.

Lemma 1 If ${f : G \rightarrow [0,\infty)}$ is a density, then for every ${\eta > 0}$ the set ${\Delta_{\eta}(f)}$ is contained in an ${\mathbb F_p}$-subspace of dimension at most ${d}$, where

$\displaystyle d \leq 2 \sqrt{2} \frac{\mathrm{Ent}_{\mu}(f)}{\eta^2}\,.$

In the special case ${p=2}$, the proof will actually give constant ${2}$ instead of ${2 \sqrt{2}}$.

1.2. A slightly stronger statement

In fact, we will prove the following slightly stronger statement.

Theorem 2 If ${f : G \rightarrow [0,\infty)}$ is a density, then for every ${\eta > 0}$ there exists a density ${g : G \rightarrow [0,\infty)}$ such that ${\Delta_{\eta}(f) \subseteq \Delta_0(g)}$ and moreover ${g(x) = \Psi(\chi_{\alpha_1}(x), \ldots, \chi_{\alpha_d}(x))}$ for some ${d \leq 1+2\sqrt{2} \frac{\mathrm{Ent}_{\mu}(f)}{\eta^2}}$ and ${\{\alpha_i\} \subseteq G}$ and some function $\Psi : \mathbb C^d \to \mathbb R$. In other words, ${g}$ is a function of at most ${d}$ characters.

It is an exercise in linear algebra to see that for such a density ${g}$, the set of non-zero Fourier coefficients ${\Delta_0(g)}$ is contained in the span of ${\alpha_1, \ldots, \alpha_d}$. Since ${\Delta_{\eta}(f) \subseteq \Delta_0(g)}$, this yields Lemma 1.

1.3. Entropy optimality

Let ${L^2(G)}$ denote the Hilbert space of real-valued functions on ${G}$ equipped with the inner product ${\langle \cdot, \cdot\rangle}$. Assume now that ${f : G \rightarrow [0,\infty)}$ is a density and fix ${\eta > 0}$.

Dtnote by ${\mathcal F}$ the family of functions of the form ${\pm \mathrm{Re}\,\chi_\alpha}$ or ${\pm \mathrm{Im}\, \chi_\alpha}$ for some ${\alpha \in G}$.

Let us associate to ${f}$ the convex polytope ${P(f, \eta) \subseteq L^2(G)}$ of functions ${g : G \rightarrow [0,\infty)}$ satisfying

$\displaystyle \langle g,\varphi\rangle \geq \langle f,\varphi\rangle - \eta$

for all ${\varphi \in \mathcal F}$.

Just as in the matrix scaling setting, ${P(f,\eta)}$ encapsulates the family of “tests” that we care about (which Fourier coefficient magnitudes are large). Note that if ${g \in P(f,\eta/\sqrt{2})}$, then ${\Delta_{\eta}(f) \subseteq \Delta_0(g)}$. We also let ${Q}$ denote the probability (density) simplex: The set of all ${g : G \rightarrow [0,\infty)}$ such that ${\mathop{\mathbb E}_{\mu} g = 1}$.

Of course, ${P(f,\eta) \cap Q}$ is non-empty because ${f \in P(f,\eta) \cap Q}$. As before, we will try to find a “simple” representative from ${P(f,\eta) \cap Q}$ by doing entropy maximization (which is the same thing as relative entropy minimization):

$\displaystyle \textrm{minimize } \mathrm{Ent}_{\mu}(g) = \sum_{x \in G} \mu(x) g(x) \log g(x) \qquad \textrm{ subject to } g \in P(f,\eta) \cap Q\,.$

Soon we will explore this optimization using convex duality. But in this post, let us give an algorithm that tries to locate a feasible point of ${P(f,\eta) \cap Q}$ by a rather naive form of (sub-)gradient descent.

We start with the initial point ${g_0 = \mathbf{1}}$ which is the function that is identically ${1}$ on all of ${G}$. Note that by strict convexity, this is the unique maximizer of ${\mathrm{Ent}_{\mu}(g)}$ among all ${g \in Q}$. But clearly it may be that ${g_0 \notin P(f,\eta)}$.

So suppose we have a function ${g_t \in Q}$ for some ${t \geq 0}$. If ${g_t \in P(f,\eta)}$, we are done. Otherwise, there exists a violated constraint: Some ${\varphi_t \in \mathcal F}$ such that ${\langle g_t, \varphi_t\rangle < \langle f,\varphi_t\rangle - \eta}$. Our goal is now to update ${g_t}$ to ${g_{t+1} \in Q}$ that doesn't violate this constraint quite as much.

Notice the dichotomy being drawn here between the “soft'' constraints of ${P(f,\eta)}$ (which we try to satisfy slowly) and the “hard'' constraints of ${Q}$ that we are enforcing at every step. (This procedure and the analysis that follows are a special case of the mirror descent framework; see, for example, Section 4 of Bubeck’s monograph. In particular, there is a more principled way to design the algorithm that follows.)

We might try to write ${g_{t+1} = g_t + \varepsilon \varphi_t}$ for some small ${\varepsilon > 0}$, but then ${g_{t+1}}$ could take negative values! Instead, we will follow the approximation ${e^x \approx 1+x}$ (for ${x}$ small) and make the update

$\displaystyle g_{t+1} = \frac{g_t \exp\left(\varepsilon \varphi_t\right)}{\mathbb E_{\mu}[g_t \exp\left(\varepsilon \varphi_t\right)]}\,.$

Observe that the exponential weight update keeps ${g_{t+1}}$ positive, and our explicit renormalization ensures that ${\mathop{\mathbb E}_{\mu} [g_{t+1}]=1}$. Thus ${g_{t+1} \in Q}$.

Of course, one might feel fishy about this whole procedure. Sure, we have improved the constraint corresponding to ${\varphi_t}$, but maybe we messed up some earlier constraints that we had satisfied. While this may happen, it cannot go on for too long. In the next post, we will prove the following.

Lemma 3 For an appropriate choice of step size ${\varepsilon > 0}$, it holds that ${g_T \in P(f,\eta)}$ for some ${T \leq 2 \frac{\mathrm{Ent}_{\mu}(f)}{\eta^2}}$.

Notice that this lemma proves Theorem 2 and thus finishes our proof of Chang’s Lemma. That’s because by virtue of ${g_T \in P(f,\eta/\sqrt{2})}$, we know that ${\Delta_{\eta}(f) \subseteq \Delta_0(g_T)}$. Moreover, ${g_T}$ is built out of at most ${T+1}$ characters ${\alpha_0, \alpha_1, \ldots, \alpha_{T}}$, and thus it satisfies the conclusion of Theorem 2.

For the case ${p=2}$, we do not have to approximate the real and complex parts separately, so one saves a factor of ${\sqrt{2}}$.

1.5. Bibliographic notes

This argument bears some similarity to the proof of Impagliazzo, Moore, and Russell. It was discovered in a joint work with Chan, Raghavendra, and Steurer on extended formulations. In both those arguments, the “descent” is implicit and is done by hand—the characters are carefully chosen not to interact. It turns out that this level of care is unnecessary. The analysis in the next post will allow the descent to choose any violated constraint at any time, and moreover will only use the fact that the characters are bounded.

# Entropy optimality: Matrix scaling

This is the first in a series of posts on the surprising power of “entropy optimality.” I have recently encountered many incarnations of this principle in a number of d\ifferent settings (functional analysis, stochastic processes, additive number theory, communication complexity, etc.). It is also connected to many well-studied phenomena in machine learning, optimization, and statistical physics. In the spirit of blogs, my goal is for each post to highlight a single interesting aspect of the theory.

For the first post, we’ll look at a very simple but compelling example: matrix scaling. This is a problem that arises in statistics, numerical analysis, and a number of other areas (see Section 1.2 here for references).

Suppose we are given an ${n \times n}$ target matrix ${T=(t_{ij})}$ with nonnegative entries. We also have an ${n \times n}$ input matrix ${X=(x_{ij})}$ and our goal is to multiply the rows and columns of ${X}$ by positive weights so that the resulting matrix has the same row and column sums as ${T}$.

Stated differently, we look for diagonal matrices ${D_1}$ and ${D_2}$ such that ${D_1 X D_2}$ has the same row and column sums as ${T}$. A typical choice of target might be ${t_{ij} = 1/n}$ for all ${1 \leq i,j \leq n}$, meaning that we want to rescale ${X}$ to be doubly stochastic.

Theorem 1 If ${x_{ij} = 0 \iff t_{ij} = 0}$ for all ${1 \leq i,j \leq n}$, then such a weighting exists.

To prove the theorem, we will search for a matrix ${Y = (y_{ij})}$ with the same row and column sums as ${T}$: For all ${i,j \in [n]}$:

$\displaystyle \sum_{i} y_{ij}=\sum_i t_{ij} \quad \textrm{ and } \quad \sum_j y_{ij} = \sum_j t_{ij}\,. (1)$

Clearly such a matrix exists; we could just take ${Y=T}$! But the idea is that we want ${Y}$ to be a “simple perturbation” of ${X}$. We will follow a philosophy analogous to Jaynes’ principle of maximum entropy. What results is precisely the argument of Franklin and Lorenz.

1.1. The relative entropy

Given two probability distributions ${\mu}$ and ${\nu}$ on a finite set ${S}$, we recall that the relative entropy between two measures ${\mu}$ and ${\nu}$ on ${X}$ is given by

$\displaystyle D(\mu\,\|\,\nu) = \sum_{x \in S} \mu(x) \log \frac{\mu(x)}{\nu(x)}\,.$

We take this quantity to be infinite unless ${\nu(x) = 0 \implies \mu(x)=0}$. A fundamental property of ${D}$ is that it is jointly convex in its arguments. For the following, we will only need that ${D}$ is convex in ${\mu}$ which is a simple consequence of the fact that ${y \mapsto y \log y}$ is a convex function on ${\mathbb R_+}$.

In proving Theorem 1, we may clearly assume that ${\sum_{ij} x_{ij} = \sum_{ij} t_{ij} = 1}$. Now let ${P(T) \subseteq \mathbb R^{n^2}}$ denote the affine subspace of all ${Y}$ satisfying the constraints (1). We will consider the optimization problem:

$\displaystyle \textrm{minimize } D(Y \,\|\, X) = \sum_{ij} y_{ij} \log \frac{y_{ij}}{x_{ij}} \quad \textrm{ subject to } Y \in P(T)\quad (2)$

The idea is that the objective function prefers ${Y}$ to be as close to ${X}$ as possible in terms of relative entropy while still satisfying the constraints. Philosophically, this is supposed to encourage ${Y}$ to be “simple” with respect to ${X}$.

Observe that, by our assumption on ${X}$, it follows that ${T \in P(T)}$ is a point satisfying ${D(T \,\|\,X) < \infty}$. Note that since ${P(T)}$ is a convex, compact set, we know that there is an optimizer, and since ${Y \mapsto D(Y \,\|\,X)}$ is strictly convex, the optimal solution ${Y^*}$ is unique. It will turn out that ${Y^*}$ is of the form ${D_1 X D_2}$, and this fact will prove our theorem. The last step is to understand the structure of ${Y^*}$ in terms of the Lagrangian.

1.2. The Lagrangian

Let us introduce ${2n}$ dual variables ${{ \alpha_i, \beta_j \in \mathbb R : i,j \in [n] }}$ and consider the Lagrangian (see the Boyd-Vandenberghe book) corresponding to the optimization (2):

$\displaystyle \sum_{ij} y_{ij} \log \frac{y_{ij}}{x_{ij}} + \sum_i \alpha_i \sum_j (t_{ij}-y_{ij}) + \sum_j \beta_j \sum_i (t_{ij}-y_{ij})\,.$

At optimality, the gradient should be equal to zero. If we differentiate with respect to some ${y_{ij}}$, we see that

$\displaystyle 1+\log y_{ij} + \alpha_i + \beta_j - \log x_{ij} = 0\,,$

which implies that, at optimality,

$\displaystyle y^*_{ij} = x_{ij}, e^{-\alpha^*_i-\beta^*_j-1}\,.$

In particular, we see that ${Y^*}$ is obtained from ${X}$ by multiplying the rows by ${{e^{-\alpha_i^*-1}}}$ and multiplying the columns by ${{e^{-\beta_j^*}}}$, completing the proof of Theorem 1.

An astute reader (or, in this case, commenter) might point out that we did not use the condition ${t_{ij} = 0 \implies x_{ij}=0}$. But clearly it is necessary, as one can see from examining the pair

$\displaystyle T = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \quad X = \left(\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right)\,.$

See these lecture notes for a detailed account.

# What does Kadison-Singer have to do with Quantum Mechanics?

Most accounts of the Kadison-Singer problem mention that it arose from considerations in the foundations of quantum mechanics. Specifically, a question about “extensions of pure states.” See, for instance, Gil’s blog. But trying to reconcile this question with my (limited) knowledge of QM proved frustrating. There is also an intuitive explanation posted at soulphysics, along with mathematical notes. But I couldn’t figure that one out either, because it only deals with “normal” quantum states (those corresponding to countably additive measures; see below). I think the KS problem for normal states is trivial. For the meat of KS, one has to think not only about infinite-dimensional spaces, but about “pathological” (non-constructible) measures. In particular, one has to think about “quantum states” that don’t have density matrices.

What follows are my notes on the matter, all of which are either well-known or wrong. Skip to the end for the very simple formulation with no mention of quantum states. Even better advice is to just Read Nick Harvey’s survey.

How I understand quantum measurement

A (mixed) quantum state ${\rho}$ on the space ${\mathbb C^n}$ is an ${n \times n}$ Hermatian matrix satisfying ${\rho \succeq 0}$ (${\rho}$ is positive semi-definite) and ${\mathrm{Tr}(\rho)=1}$. Suppose we also have a resolution of the identity: A collection ${\mathcal M = \{M_1, M_2, \ldots, M_k\}}$ of PSD matrices with ${\sum_{j=1}^k M_j = I}$. Then when we perform the measurement ${\mathcal M}$, the probability of obtaining outcome ${j}$ is ${\mathrm{Tr}(\rho M_j)}$ for ${j=1,2,\ldots,k}$.

A pure state is given by a density matrix of the form ${\rho = \psi \psi^T}$ for some ${\psi \in \mathbb C^n}$. Equivalently, a state is pure if and only if it cannot be written as a non-trivial convex combination of other states, i.e. if ${\rho = \lambda \rho_1 + (1-\lambda) \rho_2}$ for ${\lambda \in (0,1)}$, then ${\rho=\rho_1=\rho_2}$. The pure states are the extreme points of the convex set of all states.

Quantum logic

There is another way to think about quantum measurement. Let ${\mathcal H}$ be a (separable) Hilbert space and consider a projection ${P : \mathcal H \rightarrow S}$ onto some closed subspace. A measure ${\mu}$ on projections assigns to every such projection a value in ${[0,1]}$. We say that ${\mu}$ is ${\sigma}$-additive if whenever ${Q = \sum_{j=1}^{\infty} P_j}$ and the projections ${\{P_j\}}$ are mutually orthogonal, we have ${\mu(Q) = \sum_{j=1}^{\infty} \mu(P_j)}$. If this only holds for finite sums of projections, we say that ${\mu}$ is finitely additive. We should also have ${\mu(I)=1}$.

Note that we have allowed here the possibility that ${\mathcal H}$ is infinite-dimensional. In the case where the measure ${\mu}$ is ${\sigma}$-additive, it is completely specified by its value at one-dimensional projections, i.e. the values ${\mu(P_x)}$ where ${P_x}$ denotes projection onto ${x \in \mathcal H}$. Now we can state Gleason’s theorem: Our two notions of quantum measurement coincide.

Theorem 1 Let ${\mathcal H}$ be a Hilbert space of dimension at least 3 and let ${\mu}$ be a ${\sigma}$-additive measure on projections. Then there exists a unique linear operator ${\rho}$ such that for any projection ${P}$, we have ${\mu(P) = \mathrm{Tr}(\rho P)}$.

In other words, every ${\sigma}$-additive measure on projections arises from a mixed state.

Restricted purity

Note that a measure on projections corresponds to a pure state if and only if it cannot be written as a non-trivial convex combination of measures on projections. In this way, we can also define purity with respect to a restricted family of projections.

Consider, for example, the Hilbert space ${\ell^2(\mathbb N)}$ of complex sequences, and let ${\{e_j : j \in \mathbb N\}}$ denote the standard basis. Suppose we consider only the projections ${\mathcal P = \{P_j\}}$ onto each of the basis elements. Say that a ${\mathcal P}$-measure is a measure on all projections of the form ${\sum_{i \in A} P_i}$ for ${A \subseteq \mathbb N}$. Such a measure only gives us probabilities for measurements performed in the standard basis. (The algebra generated by ${\mathcal P}$ is a canonical example of a “maximal abelian subalgebra,” i.e. MASA.) We can say that a ${\mathcal P}$-measure is pure if it cannot be written as a non-trivial convex combination of other ${\mathcal P}$-measures. What are the pure ${\mathcal P}$-measures? Well, the only ${\sigma}$-additive pure ${\mathcal P}$-measures are the point measures ${\mu(P_j)=1}$ for some ${j}$. If ${\mu(P_j), \mu(P_{j'}) > 0}$ for ${j \neq j'}$, then by ${\sigma}$-additivity, we can split ${\mu}$ into a convex combination of two measures (one supported only on ${P_j}$ and the other supported off ${P_j}$).

But when ${\sigma}$-additivity is dropped, some tricky things can happen (see the next section). It turns out that, in this case, there are more pure states than just ${e_j e_j^T}$ (in fact, the cardinality of the finitely-additive pure states is gigantic, ${2^{2^{\aleph_0}}}$). Now we are in position to state the Kadison-Singer problem.

Kadison-Singer Problem: Is it the case that every pure, finitely-additive ${\mathcal P}$-measure has a unique extension to a finitely-additive measure on all the projections of ${\mathcal H}$?

The answer is clearly positive for ${\sigma}$-additive measures (in which case the unique extension corresponds to ${e_j e_j^T}$). The answer is also positive for finitely-additive ${\mathcal P}$-measures, as recently proved by Marcus, Spielman, and Srivistava.

In their original work, Kadison and Singer showed that the answer is false if one constructs the MASA using the continuous Hilbert space ${L^2([0,1])}$ instead of ${\ell^2(\mathbb N)}$. One can take the subalgebra ${\mathcal C}$ of multiplication operators ${T_g (f)=gf}$ for ${\|g\|_{\infty} < \infty}$. Here the corresponding family of projections are the maps ${P_S f = \mathbf{1}_S f}$ for ${S \subseteq [0,1]}$. Unlike in the case above, this subalgebra is not “discrete.” It turns out that every maximal abelian subalgebra of bounded operators on a separable Hilbert space is either finite-dimensional, the discrete one, the continuous one, or a direct sum of algebras isomorphic to these.

Note that states corresponding to finitely additive (but not ${\sigma}$-additive measures) do not have density matrices and are often considered to be “unphysical” because they would take an infinite amount of energy to prepare.

A ${\beta}$-world teeming with pure states

In the above setting, we said there are finitely-additive pure ${\mathcal P}$-measures ${\mu}$ other than point measures. To see what one of these must look like, consider the following construction. Let ${P_{\mathrm{even}}}$ denote the projection onto ${\{e_j : j \textrm{ even}\}}$ and define ${P_{\mathrm{odd}}}$ similarly. Define the measures ${\mu_{\mathrm{even}}(P) = \mu(P P_{\mathrm{even}})}$ and ${\mu_{\mathrm{odd}}(P) = \mu(P P_{\mathrm{odd}})}$. Then we can write ${\mu = \mu(P_{\mathrm{even}}) \mu_{\mathrm{even}} + \mu(P_{\mathrm{odd}}) \mu_{\mathrm{odd}}}$. This is a non-trivial convex combination unless either ${\mu(P_{\mathrm{even}})=0}$ or ${\mu(P_{\mathrm{odd}})=0}$.

Indeed, for every partition ${\mathbb N = S \cup \bar S}$, a pure ${\mathcal P}$-measure ${\mu}$ can only put non-zero weight on projections involving coordinates of exactly one of ${S}$ or ${\bar S}$. Furthermore, if ${\mu(\sum_{i \in {\bar S}} P_i) = 0}$ then ${\mu(\sum_{i \in S} P_i)=1}$. In other words, for every set of projections, the measure ${\mu}$ takes only values 0 or 1. Such finitely-additive measures are exactly given by ultrafilters on ${\mathbb N}$. The finitely-additive pure ${\mathcal P}$-measures are in one-to-one correspondence with such ultrafilters, the set of which can be identified with ${\beta \mathbb N}$, the Stone-Cech compactification of the naturals. See Terry Tao’s notes on ultafilters and Stone-Cech compactification.

A simple formulation

Take the Hilbert space ${\ell^2 = \ell^2(\mathbb N)}$ with standard basis ${\{e_j : j \in \mathbb N\}}$, and let ${\mathcal S}$ denote the set of all its closed subspaces. Consider a map ${\mu : \mathcal S \rightarrow [0,1]}$. We say that ${\mu}$ is a finitely-additive measure if ${\mu(\ell^2)=1}$ and whenever ${S_1, S_2, \ldots, S_k \in \mathcal S}$ is a finite collection of mutually orthogonal subspaces, we have ${\mu(\mathrm{span}(S_1, \ldots, S_k)) = \sum_{i=1}^k \mu(S_i)}$. Say that such a measure ${\mu}$ is pure if it cannot be written as a non-trivial convex combination of two finitely-additive measures. A canonical example of a pure measure involves fixing a unit vector ${x \in \ell^2}$ and setting ${\mu(S) = \langle x, P_S x\rangle}$ where $P_S$ denotes the orthogonal projection onto $S$.

Now let ${\mathcal S_{\mathrm{diag}}}$ be the set of all the subspaces of ${\ell^2}$ of the form ${\mathrm{span}(e_i : i \in A)}$ for some subset ${A \subseteq \mathbb N}$, i.e. all the coordinate subspaces. We can define “finitely-additive” and “pure” for measures on ${\mathcal S_{\mathrm{diag}}}$ in the analogous way.

Kadison-Singer problem: Can every finitely-additive pure measure on ${\mathcal S_{\mathrm{diag}}}$ be extended uniquely to a finitely-additive measure on ${\mathcal S}$?

Note that the real question is about uniqueness of the extension. There always exists a pure extension; see the comments.

# Separated sets in unions of frames

In celebration of the recent resolution of the Kadison-Singer problem by Marcus, Spielman, and Srivastava, here is a question on isotropic point sets on which Kadison-Singer does not (seem to) shed any light. A positive resolution would likely have strong implications for the Sparsest Cut problem and SDP hierarchies. The question arose in discussions with Shayan Oveis Gharan, Prasad Raghavendra, and David Steurer.

Open Question: Do there exist constants $\varepsilon, \delta > 0$ such that for any $k \in \mathbb N$, the following holds? Let $\mathcal B_1, \mathcal B_2, \ldots, \mathcal B_k \subseteq \mathbb R^k$ be a collection of $k$ orthonormal bases and define $W = \mathcal B_1 \cup \mathcal B_2 \cup \cdots \cup \mathcal B_k$. Then there are subsets $A,B \subseteq W$ with $|A|,|B| \geq \varepsilon |W|$ and $\min_{x \in A, y \in B} \|x-y\|_2 \geq \delta$.

[Some additional notes: One piece of intuition for why the question should have a positive resolution is that these $k$ orthonormal bases which together comprise at most $k^2$ vectors cannot possibly “fill” $k$-dimensional space in a way that achieves $k$-dimensional isoperimetry. One would seem to need $\exp(O(k))$ points for this.

One can state an equivalent question in terms of vertex expansion. Say that a graph $G=(V,E)$ on $k^2$ vertices is a vertex expander if $|N(S)| \geq 1.1 |S|$ for all subsets $S \subseteq V$ with $|S| \leq |V|/2$. Here, $N(S)$ denotes all the nodes that are in $S$ or are adjacent to $S$. Then one can ask whether there exists a 1-1 mapping from $V$ to $\mathcal B_1 \cup \cdots \cup \mathcal B_k$ for some orthonormal bases $\{\mathcal B_i\}$ such that the endpoints of every edge are mapped at most $o(1)$ apart (as $k \to \infty$).]

# Open question recap

In light of the nearly immediate failure of the last open question, here is a recap of the progress on the few somewhat more legitimate questions appearing here over the past few years. (Note that most of these questions are not original and appeared other places as well.)

# Open question: Separated sets in isotropic measures

[Note: This question is trivially impossible, though I leave the original form below. If $\mu$ is the uniform measure on $S^{k-1}$, then any two sets of $\Omega(1)$ measure are at distance $O(1/\sqrt{k})$ by concentration of measure. Thus it is impossible to find $k$ sets satisfying the desired bound. I will try to think of the right question and repost. Unfortunately, I now recall having gone down this path a few times before, and I fear there may not be a relevant elementary open question after all.]

Here is an interesting and elementary (to state) open question whose resolution would give an optimal higher-order Cheeger esimate. The goal is to replace the factor ${k^{O(1)}}$ in Theorem 1 of the previous post with a ${\mathrm{polylog}(k)}$ factor, or even an optimal bound of ${\sqrt{\log k}}$.

Fix ${k \in \mathbb N}$ and let ${\mu}$ denote a probability measure on the ${(k-1)}$-dimensional sphere ${S^{k-1}}$. For the purpose of this question, one can assume that ${\mu}$ is supported on a finite number of points. Suppose also that ${\mu}$ is isotropic in the sense that, for every ${\theta \in S^{k-1}}$,

$\displaystyle \int_{S^{k-1}} \langle x, \theta\rangle^2 \,d\mu(x)= \frac{1}{k}\,.$

Now our goal is to find, for some ${\varepsilon, \delta > 0}$, a collection of (measurable) subsets ${U_1, U_2, \ldots, U_k \subseteq S^{k-1}}$ such that ${\mu(U_i) \geq \varepsilon/k}$ for each ${i=1,2,\ldots,k}$, and such that for ${i \neq j}$, we have

$\displaystyle \min_{x \in U_i, y \in U_j} \|x-y\|_2 \geq \delta\,.$

In Lemma 5 of the previous post, we proved that this is possible for ${\varepsilon \gtrsim 1}$ and ${\delta \gtrsim k^{-5/2}}$. In this paper, we improve the estimate on ${\delta}$ somewhat, though the best-known is still ${\delta \gtrsim k^{-c}}$ for some ${c > 0}$.

Question:
Is it possible to achieve ${\varepsilon, \delta \gtrsim 1/\mathrm{poly}(\log k)}$? One could even hope for ${\varepsilon \gtrsim 1}$ and ${\delta \gtrsim 1/(\log k)}$.

Two extreme cases are when (1) ${\mu}$ is uniform on ${S^{k-1}}$, or (2) ${\mu}$ is uniform on the standard basis ${\{e_1,e_2,\ldots,e_k\}}$. In the first case, one can easily achieve ${\varepsilon,\delta \gtrsim 1}$ (in fact, one could even get ${100k}$ sets satisfying these bounds). In the second case, taking each set to contain a single basis vector achieves ${\varepsilon = 1}$ and ${\delta = \sqrt{2}}$.

If one only wants to find, say, ${\lceil 3k/4\rceil}$ sets instead of ${k}$ sets, then it is indeed possible to achieve ${\varepsilon \gtrsim 1}$ and ${\delta \gtrsim 1/O(\log k)}$. [Update: This is actually impossible for the reasons stated above. To get the corresponding bounds, we do a random projection into $O(\log k)$ dimensions. This only preserves the original Euclidean distances on average.] Furthermore, for ${\varepsilon \gtrsim 1}$, this bound on ${\delta}$ is asymptotically the best possible.

# No frills proof of higher-order Cheeger inequality

Following some recent applications by Mamoru Tanaka and Laurent Miclo, I was asked where there is a short, no-frills, self-contained, and (possibly) quantitatively non-optimal proof of the higher-order Cheeger inequalities from our paper with Shayan Oveis-Gharan and Luca Trevisan. I thought I would post it here. (If you’re hungering for something new, see this recently posted preprint of my coauthors relating higher eigenvalues to graph expansion.)

[Update: The application of Miclo can also be done using the higher-order Cheeger inequalities of Louis, Raghavendra, Tetali, and Vempala.]

The main simplification comes from doing the random partitioning non-optimally with axis-parallel cubes. For ease of notation, we will deal only with regular graphs, but there will be no quantitative dependence on the degree and this assumption can be removed (see the full paper).

Suppose that ${G=(V,E)}$ is a connected, ${n}$-vertex, ${d}$-regular graph. Define the Laplacian by ${\mathcal L = I - (1/d)A}$, where ${A}$ is the adjacency matrix of ${G}$. We will think of ${\mathcal L}$ as acting on ${\ell^2(V)}$, the space of functions ${f : V \rightarrow \mathbb R}$ equipped with the ${\ell^2}$ norm. ${\mathcal L}$ is positive semi-definite with spectrum

$\displaystyle 0 = \lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_{|V|} \leq 2\,.$

We we define the Rayleigh quotient of a function ${f \in \ell^2(V)}$ by

$\displaystyle \mathcal R(f) = \frac{\sum_{u \sim v} (f(u)-f(v))^2}{d \sum_{u \in V} f(u)^2}\,,$

where the numerator is summed over edges of ${G}$. By the variational principle for eigenvalues, we have

$\displaystyle \lambda_k = \min_{\stackrel{W \subseteq \ell^2(V)}{\dim(W)=k}} \max_{0 \neq f \in W} \mathcal R(f)\,. \ \ \ \ \ (1)$

For a subset ${S \subseteq V}$, we define the expansion of ${S}$ by ${\phi(S) = \mathcal R(\mathbf{1}_S)}$, where ${\mathbf{1}_S}$ is the indicator function of ${S}$.

Finally, for ${k \in \{1,2,\ldots,n\}}$, we define the ${k}$-way expansion constant of ${G}$ by

$\displaystyle \rho_G(k) = \min \left\{ \max_i \phi(S_i) : S_1, S_2, \ldots, S_k \subseteq V \right\}\,,$

where the minimum is over collections of ${k}$ disjoint, non-empty subsets of ${V}$.

The classical discrete Cheeger inequality asserts that

$\displaystyle \frac{\lambda_2}{2} \leq \rho_G(2) \leq \sqrt{2\lambda_2}\,.$

We will now prove the following generalization. See the full paper for a discussion of the surrounding issues and better quantitative bounds.

Theorem 1 For every ${k \in \{1,2,\ldots,n\}}$,

$\displaystyle \frac{\lambda_k}{2} \leq \rho_G(k) \leq 30 k^{3.5} \sqrt{\lambda_k}\,. \ \ \ \ \ (2)$

First, let’s prove the (easy) LHS of (2). Suppose we have ${S_1, S_2, \ldots, S_k \subseteq V}$ which are disjoint and non-empty and which satisfy ${\phi(S_i) = \mathcal R(\mathbf{1}_{S_i}) \leq \rho}$. Then certainly ${W = \mathrm{span}(\mathbf{1}_{S_1}, \ldots, \mathbf{1}_{S_k})}$ is a ${k}$-dimensional subspace of ${\ell^2(V)}$. On the other hand, every ${f \in W}$ satisfies ${\mathcal R(f) \leq 2 \rho}$ because if ${f = \alpha_1 \mathbf{1}_{S_1} + \cdots + \alpha_k \mathbf{1}_{S_k}}$, then

$\displaystyle \sum_{u \sim v} (f(u)-f(v))^2 \leq 2 \sum_{i=1}^k \alpha_i^2 \sum_{u \sim v} |\mathbf{1}_{S_i}(u)-\mathbf{1}_{S_i}(v)|^2\,,$

where we have used the fact that if ${u \in S_i}$ and ${v \in S_j}$, then

$\begin{array}{rcl} |\alpha_i \mathbf{1}_{S_i}(u) - \alpha_j \mathbf{1}_{S_j}(u)|^2 &\leq & 2(\alpha_i^2 + \alpha_j^2) \\ &=& 2\alpha_i^2 |\mathbf{1}_{S_i}(u) - \mathbf{1}_{S_i}(v)|^2 + 2\alpha_j^2 |\mathbf{1}_{S_j}(u)-\mathbf{1}_{S_j}(v)|^2\,.\end{array}$

But now using (1), the subspace ${W}$ witnesses the fact that ${\lambda_k \leq 2\rho}$.

To prove the more difficult RHS of (2), we will use the following discrete Cheeger inequality with boundary conditions.

Lemma 2 For any ${f : V\rightarrow \mathbb R}$, there is a subset ${U \subseteq \{ v \in V : f(v) \neq 0\}}$ such that

$\displaystyle \phi(U) \leq \sqrt{2 \mathcal R(f)}\,.$

Proof: Let ${U_t = \{ v \in V : f(v)^2 \geq t \}}$. Observe that for each ${t > 0}$, one has ${U_t \subseteq \{ v \in V : f(v) \neq 0\}}$. For ${S \subseteq V}$, let ${E(S,\bar S)}$ denote the edges of ${G}$ with exactly one endpoint in ${S}$. Then we have

$\displaystyle \begin{array}{rcl} \int_0^{\infty} |E(U_t, \bar U_t)|\,dt &=& \sum_{\{u,v\} \in E} \left|f(u)^2 - f(v)^2\right| \\ &= & \sum_{\{u,v\} \in E} |f(u) + f(v)| |f(u)-f(v)| \\ &\leq & \sqrt{\sum_{\{u,v\} \in E} (|f(u)|+|f(v)|)^2} \sqrt{\sum_{\{u,v\} \in E} |f(u)-f(v)|^2} \\ &\leq & \sqrt{2 d \sum_{u \in V} f(u)^2}\sqrt{\sum_{\{u,v\} \in E} |f(u)-f(v)|^2}. \end{array}$

On the other hand, ${\int_0^{\infty} d|U_t|\,dt = d \sum_{u \in V} |f(u)|^2,}$ thus

$\displaystyle \int_0^{\infty} |E(U_t, \bar U_t)|\,dt \leq \sqrt{2 \mathcal R(f)} \int_0^{\infty} d|U_t|\,dt\,,$

implying there exists a ${t > 0}$ such that ${\phi(U_t) = \frac{|E(U_t,\bar U_t)|}{d |U_t|} \leq \sqrt{2 \mathcal R(f)}}$. $\Box$

In light of the preceding lemma, to prove the RHS of (2), it suffices to find ${k}$ disjointly supported functions ${\psi_1, \psi_2, \ldots, \psi_k : V \rightarrow \mathbb R}$ such that ${\mathcal R(\psi_i) \leq (30)^2 k^6 \lambda_k}$ for each ${i=1,2,\ldots,k}$. Then Lemma 2 guarantees the existence of disjoint subsets of vertices satisfying our desired conclusion.

Toward this end, let ${f_1, f_2, \ldots, f_k : V \rightarrow \mathbb R}$ denote ${k}$ orthonormal functions satisfying ${\mathcal L f_i = \lambda_i f_i}$ for each ${i=1,2,\ldots,k}$, i.e. consider the first ${k}$ eigenfunctions of the Laplacian. Define the map ${F : V \rightarrow \mathbb R^k}$ via

$\displaystyle F(v) = \left(f_1(v), f_2(v), \ldots, f_k(v)\right)\,.$

We also put ${\hat F(v) = F(v)/\|F(v)\|}$. (Since ${\lambda_1=0}$, the function ${f_1}$ takes value ${1/\sqrt{n}}$ everywhere, hence ${F}$ is never zero and this is well-defined.)

The next lemma shows that, in order to find disjointly supported functions, it suffices to find “separated sets.”

Lemma 3 Suppose that for some ${\varepsilon > 0}$ and ${0 < \delta \leq 1}$, there exist ${k}$ subsets ${U_1, U_2, \ldots, U_k \subseteq V}$ satisfying the conditions:

1. For every ${i=1,2,\ldots,k}$, we have ${\sum_{v \in U_i} \|F(v)\|^2 \geq \varepsilon}$, and
2. For every ${i \neq j}$, if ${u \in U_i}$ and ${v \in U_j}$, then

$\displaystyle \|\hat F(u)-\hat F(v)\| \geq \delta$

Then there are disjointly supported functions ${\psi_1, \psi_2, \ldots, \psi_k : V \rightarrow \mathbb R}$ such that ${\mathcal R(\psi_i) \leq 25k\lambda_k/(\varepsilon\delta^2)}$.

Proof: For each ${i=1,2,\ldots,k}$, we define maps ${\theta_i : V \rightarrow \mathbb R}$ by

$\displaystyle \theta_i(v) = \max\left(0, 1 - \frac{2}{\delta} \min_{u \in U_i} \|\hat F(u)-\hat F(v)\|\right)\,.$

Observe that, by the triangle inequality, for ${u,v \in V}$, we have

$\displaystyle |\theta_i(u)-\theta_i(v)| \leq \frac{2}{\delta} \|\hat F(u)-\hat F(v)\|\,. \ \ \ \ \ (3)$

Next, we define

$\displaystyle \psi_i(v) = \theta_i(v) \|F(v)\|\,.$

Observe that since ${\theta_i}$ is identically ${1}$ on ${U_i}$, we have

$\displaystyle \sum_{v \in V} \psi_i(v)^2 \geq \sum_{v \in U_i} \|F(v)\|^2 \geq \varepsilon \ \ \ \ \ (4)$

by condition (i).

Next, we argue that the functions ${\{\theta_i\}}$ are disjointly supported. This immediately implies that the ${\{\psi_i\}}$ are disjointly supported. If ${u \in \mathrm{supp}(\theta_i) \cap \mathrm{supp}(\theta_j)}$ for some ${i \neq j}$, then there are points ${v \in U_i}$ and ${v' \in U_j}$ such that ${\|\hat F(u)-\hat F(v)\| < \delta/2}$ and ${\|\hat F(u)-\hat F(v')\| < \delta/2}$. But then by the triangle inequality, ${\|\hat F(v)-\hat F(v')\| < \delta}$, violating condition (ii).

Finally, we bound ${\mathcal R(\psi_i)}$. For any ${u,v \in V}$ and ${i=1,2,\ldots,k}$, we have

$\displaystyle |\psi_i(u)-\psi_i(v)| \leq \theta_i(u) \left|\vphantom{\bigoplus}\|F(u)\|-\|F(v)\|\right| + \|F(v)\| \cdot |\theta_i(u)-\theta_i(v)|\,. \ \ \ \ \ (5)$

Now using (3),

$\displaystyle \|F(v)\| \cdot |\theta_i(u)-\theta_i(v)| \leq \frac{2}{\delta} \|F(v)\| \cdot \|\hat F(u)-\hat F(v)\| \leq \frac{4}{\delta} \|F(u)-F(v)\|\,, \ \ \ \ \ (6)$

where we have used the fact that for any non-zero vectors ${x,y \in \mathbb R^k}$, we have

$\displaystyle \|x\| \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| = \left\|x - \frac{\|x\|}{\|y\|} y\right\| \leq \|x-y\| + \left\|y - \frac{\|x\|}{\|y\|} y\right\| \leq 2 \,\|x-y\|\,.$

Using (6) and the fact that ${\theta_i(u) \leq 1}$ in (5) yields

$\displaystyle |\psi_i(u)-\psi_i(v)| \leq \left(1+\frac{4}{\delta}\right) \|F(u)-F(v)\|\,. \ \ \ \ \ (7)$

Finally, observe that

$\displaystyle \sum_{u \sim v} \|F(u)-F(v)\|^2 = \sum_{i=1}^k \sum_{u \sim v} |f_i(u)-f_i(v)|^2 = d (\lambda_1 + \cdots + \lambda_k) \leq dk\lambda_k\,.$

Combining this with (7) and (4) shows that ${\mathcal R(\psi_i) \leq \lambda_k \frac{k}{\varepsilon}(1+\frac{4}{\delta})^2 \leq 25k\lambda_k/(\varepsilon \delta^2).}$ $\Box$

We will first make the task of finding separated sets slightly easier.

Lemma 4 Suppose that for some ${0 < \delta \leq 1}$ and ${m \geq 1}$, there are subsets ${T_1, T_2, \ldots, T_m \subseteq V}$ which satisfy:

1. ${\displaystyle \sum_{i=1}^m \sum_{v \in T_i} \|F(v)\|^2 \geq k - \frac14}$.
2. For every ${i \neq j}$, if ${u \in T_i}$ and ${v \in T_j}$, then ${ \|\hat F(u)-\hat F(v)\| \geq \delta }$
3. For every ${i=1,2,\ldots, m}$,

$\displaystyle \sum_{v \in T_i} \|F(v)\|^2 \leq 1 + \frac{1}{2k}\,.$

Then there are ${k}$ sets ${U_1, U_2, \ldots, U_k \subseteq V}$ satisfying the assumption of Lemma 3 with ${\varepsilon = \frac{1}{4}}$.

Proof: We can form the desired sets ${\{U_i\}}$ by taking disjoint unions of the sets ${\{T_i\}}$. Begin with the family ${\{T_1, T_2, \ldots, T_m\}}$ and repeatedly replace two sets ${T}$ and ${T'}$ by their union if they each satisfy ${\sum_{v \in T} \|F(v)\|^2 < \frac12}$.

When we finish, we are left with a collection of sets each member of which satisfies ${\sum_{v \in T} \|F(v)\|^2 \in [\frac12, 1+\frac{1}{2k}]}$, and possibly one set for which ${\sum_{v \in T} \|F(v)\|^2 < 1/2}$. By (i), we must end with at least ${k}$ sets which each satisfy ${\sum_{v \in T} \|F(v)\|^2 \geq 1/4}$. $\Box$

Our final lemma, which finishes the proof of Theorem 1, simply asserts that such sets exist. We will no longer need the fact that ${F}$ comes from eigenfunctions.

Lemma 5 Suppose that ${g_1, g_2, \ldots, g_k : V \rightarrow \mathbb R}$ are orthornormal in ${\ell^2(V)}$. Let

$\displaystyle G(v) = (g_1(v), g_2(v), \ldots, g_k(v))$

and ${\hat G(v) = G(v)/\|G(v)\|}$. Then there is an ${m \geq 1}$ and subsets ${T_1, T_2, \ldots, T_m \subseteq V}$ such that

1. ${\displaystyle\sum_{i=1}^m \sum_{v \in T_i} \|G(v)\|^2 \geq k - \frac14}$.
2. For every ${i \neq j}$, if ${u \in T_i}$ and ${v \in T_j}$, then

$\displaystyle \|\hat G(u)-\hat G(v)\| \geq \frac{1}{2\sqrt{2} k^{3}}\,.$

3. For every ${i = 1,2,\ldots,m}$,

$\displaystyle \sum_{v \in T_i} \|G(v)\|^2 \leq 1 + \frac{1}{2k}\,.$

Proof: Consider the ${n \times k}$ matrix ${A}$ which has columns ${g_1, g_2, \ldots, g_k}$. For any ${x \in \mathbb R^k}$, we have

$\displaystyle \sum_{v \in V} \langle x, G(v)\rangle^2 = \|Ax\|^2 = x^T A^T A x = x^T x = \|x\|^2\,, \ \ \ \ \ (8)$

since the columns of ${A}$ are orthornormal

For a subset ${X \subseteq S^{k-1}}$ of the unit sphere in ${\mathbb R^k}$, we put ${V(X) = \{ v \in V : \hat G(v) \in X \}}$. Fix some ${x \in X}$ and let ${\mathsf{diam}(X)}$ denote the diameter of ${X}$, then by (8), we have

$\displaystyle 1 = \sum_{v \in V(X)} \langle x, G(v)\rangle^2 = \sum_{v \in V(X)} \|G(v)\|^2 \left(1-\frac{\|\hat G(v)-x\|^2}{2}\right)^2 \geq \sum_{v \in V(X)} \|G(v)\|^2 \left(1-\frac{\mathrm{diam}(X)^2}{2}\right)^2\,.$

We conclude that if ${\mathrm{diam}(X) \leq 1/\sqrt{2k}}$, then

$\displaystyle \sum_{v \in V(X)} \|G(v)\|^2 \leq 1 + \frac{1}{2k}\,. \ \ \ \ \ (9)$

Now, let ${\mathcal{P}}$ be a partition of ${\mathbb R^k}$ into axis-parallel squares of side length ${L = \frac{1}{k \sqrt{2}}}$. For any such square ${Q \in \mathcal P}$, we let ${\tilde Q \subseteq Q}$ denote the set of points which are at Euclidean distance at least ${\frac{L}{4k^2}}$ from every side of ${Q}$. Observe that

$\displaystyle \mathrm{vol}(\tilde Q) \geq \left(1-\frac{1}{4k^2}\right)^k \mathrm{vol}(Q) \geq \left(1-\frac{1}{4k}\right) \mathrm{vol}(Q)\,. \ \ \ \ \ (10)$

Consider now the collection of sets ${\{V(\tilde Q) : Q \in \mathcal P\}}$. Since ${\mathrm{diam}(\tilde Q) \leq L \sqrt{k} = \frac{1}{\sqrt{2k}}}$, (9) implies that ${\sum_{v \in V(\tilde Q)} \|G(v)\|^2 \leq 1 + \frac{1}{2k}}$. Furthermore, by construction, for any ${u \in V(\tilde Q)}$ and ${v \in V(\tilde Q')}$ with ${Q \neq Q'}$, we have

$\displaystyle \|\hat G(u)-\hat G(v)\| \geq 2 \frac{L}{4k^2} \geq \frac{1}{2\sqrt{2} k^{3}}\,.$

Thus the collection of sets ${\{V(\tilde Q) : Q \in \mathcal P\}}$ satisfy both conditions (ii) and (iii) of the lemma. We are left to verify (i).

Note that ${\sum_{v \in V} \|G(v)\|^2 = \sum_{i=1}^k \sum_{v \in V} g_i(v)^2 = k}$. If we choose a uniformly random axis-parallel translation ${\mathcal P'}$ of the partition ${\mathcal P}$, then (10) implies

$\displaystyle \mathbb{E} \sum_{Q \in \mathcal P'} \sum_{v \in V(\tilde Q)} \|G(v)\|^2 \geq \left(1-\frac{1}{4k}\right) \sum_{v \in V(Q)} \|G(v)\|^2 \geq k - \frac14\,.$

In particular, there exists some fixed partition that achieves this bound. For this partition ${\mathcal P}$, the sets ${\{V(\tilde Q) : Q \in \mathcal P\}}$ satisfy all three conditions of the lemma, completing the proof.

# Talagrand’s Bernoulli Conjecture, resolved.

Apparently, Bednorz and Latała have solved Talagrand’s \$5,000 Bernoulli Conjecture. The conjecture concerns the supremum of a Bernoulli process.

Consider a finite subset ${T \subseteq \ell^2}$ and define the value

$\displaystyle b(T) = \mathbb{E} \max_{t \in T} \sum_{i \geq 1} \varepsilon_i t_i\,,$

where ${\varepsilon_1, \varepsilon_2, \ldots}$ are i.i.d. random ${\pm 1}$. This looks somewhat similar to the corresponding value

$\displaystyle g(T) = \mathbb{E} \max_{t \in T} \sum_{i \geq 1} g_i t_i\,,$

where ${g_1, g_2, \ldots}$ are i.i.d. standard normal random variables. But while ${g(T)}$ can be characterized (up to universal constant factors) by the Fernique-Talagrand majorizing measures theory, no similar control was known for ${b(T)}$. One stark difference between the two cases is that ${g(T)}$ depends only on the distance geometry of ${T}$, i.e. ${g(A(T))=g(T)}$ whenever ${A}$ is an affine isometry. On the other hand, ${b(T)}$ can depend heavily on the coordinate structure of ${T}$.

There are two basic ways to prove an upper bound on ${b(T)}$. One is via the trivial bound

$\displaystyle b(T) \leq \max_{t \in T} \|t\|_1\,. \ \ \ \ \ (1)$

The other uses the fact that the tails of Gaussians are “fatter” than those of Bernoullis.

$\displaystyle b(T) \leq \sqrt{\frac{\pi}{2}} g(T)\,. \ \ \ \ \ (2)$

This can be proved easily using Jensen’s inequality.

Talagrand’s Bernoulli conjecture is that ${b(T)}$ can be characterized by these two upper bounds.

Bernoulli conjecture: There exists a constant ${C > 0}$ such that for every ${T \subseteq \ell^2}$, there are two subsets ${T_1, T_2 \subseteq \ell^2}$ such that

$\displaystyle T \subseteq T_1 + T_2 = \{ t_1 + t_2 : t_1 \in T_1, t_2 \in T_2 \}\,,$

and

$\displaystyle g(T_1) + \sup_{t \in T_2} \|t\|_1 \leq C b(T)\,.$

Note that this is a “characterization” because given such sets ${T_1}$ and ${T_2}$, equations (1) and (2) imply

$\displaystyle b(T) \leq \sqrt{\frac{\pi}{2}} g(T_1) + \sup_{t \in T_2} \|t\|_1\,.$

The set ${T_1}$ controls the “Gaussian” part of the Bernoulli process, while the set ${T_2}$ controls the part that is heavily dependent on the coordinate structure.

This beautiful problem finally appears to have met a solution. While I don’t know of any applications yet in TCS, it does feel like something powerful and relevant.