# Primer: Lifts of polytopes and non-negative matrix factorization

In preparation for the next post on entropy optimality, we need a little background on lifts of polytopes and non-negative matrix factorization.

1.1. Polytopes and inequalities

A ${d}$-dimensional convex polytope ${P \subseteq \mathbb R^d}$ is the convex hull of a finite set of points in ${\mathbb R^d}$. Equivalently, it is a compact set defined by a family of linear inequalities

$\displaystyle P = \{ x \in \mathbb R^d : Ax \leq b \}$

for some matrix ${A \in \mathbb R^{m \times d}}$.

Let us give a measure of complexity for ${P}$: Define ${\gamma(P)}$ to be the smallest number ${m}$ such that for some ${C \in \mathbb R^{s \times d}, y \in \mathbb R^s, A \in \mathbb R^{m \times d}, b \in \mathbb R^m}$, we have

$\displaystyle P = \{ x \in \mathbb R^d : Cx=y \textrm{ and } Ax \leq b\}\,.$

In other words, this is the minimum number of inequalities needed to describe ${P}$. If ${P}$ is full-dimensional, then this is precisely the number of facets of ${P}$ (a facet is a maximal proper face of ${P}$).

Thinking of ${\gamma(P)}$ as a measure of complexity makes sense from the point of view of optimization: Interior point methods can efficiently optimize linear functions over ${P}$ (to arbitrary accuracy) in time that is polynomial in ${\gamma(P)}$.

1.2. Lifts of polytopes

Many simple polytopes require a large number of inequalities to describe. For instance, the cross-polytope

$\displaystyle C_d = \{ x \in \mathbb R^d : \|x\|_1 \leq 1 \} = \{ x \in \mathbb R^d : \pm x_1 \pm x_2 \cdots \pm x_d \leq 1 \}$

has ${\gamma(C_d)=2^d}$. On the other hand, ${C_d}$ is the projection of the polytope

$\displaystyle Q_d = \left\{ (x,y) \in \mathbb R^{2d} : \sum_{i=1}^n y_i = 1, \,\, - y_i \leq x_i \leq y_i\,\forall i\right\}$

onto the ${x}$ coordinates, and manifestly, ${\gamma(Q_d) \leq 2d}$. Thus ${C_d}$ is the (linear) shadow of a much simpler polytope in a higher dimension.

[image credit: Fiorini, Rothvoss, and Tiwary]

A polytope ${Q}$ is called a lift of the polytope ${P}$ if ${P}$ is the image of ${Q}$ under a linear projection. Again, from an optimization stand point, lifts are important: If we can optimize linear functionals over ${Q}$, then we can optimize linear functionals over ${P}$. Define now ${\bar \gamma(P)}$ to be the minimal value of ${\gamma(Q)}$ over all lifts ${Q}$ of ${P}$. (The value ${\bar \gamma(P)}$ is sometimes called the (linear) extension complexity of ${P}$.)

1.3. The permutahedron

Here is a somewhat more interesting family of examples where lifts reduce complexity. The permutahedron ${\Pi_n \subseteq \mathbb R^n}$ is the convex hull of the vectors ${(i_1, i_2, \ldots, i_n)}$ where ${\{i_1, \ldots, i_n\} = \{1,\ldots,n \}}$. It is known that ${\gamma(\Pi_n)=2^n-2}$.

Let ${B_n \subseteq \mathbb R^{n^2}}$ denote the convex hull of the ${n \times n}$ permutation matrices. The Birkhoff-von Neumann theorem tells us that ${B_n}$ is precisely the set of doubly stochastic matrices, thus ${\gamma(B_n)=n^2}$ (corresponding to the non-negativity constraints on each entry).

Observe that ${\Pi_n}$ is the linear image of ${B_n}$ under the map ${A \mapsto A (1, 2, \ldots, n)^T}$, i.e. we multiply a matrix ${A \in B_n}$ on the right by the column vector ${(1, 2, \ldots, n)}$. Thus ${B_n}$ is a lift of ${\Pi_n}$, and we conclude that ${\bar \gamma(\Pi_n) \leq n^2 \ll \gamma(\Pi_n)}$.

1.4. The cut polytope

If ${P \neq NP}$, there are certain combinatorial polytopes we should not be able to optimize over efficiently. A central example is the cut polytope: ${\mathrm{CUT}_n \subseteq \mathbb R^{n^2}}$ is the convex hull of all matrices of the form ${(A_S)_{ij} = |\mathbf{1}_S(i)-\mathbf{1}_S(j)|}$ for some subset ${S \subseteq \{1,\ldots,n\}}$. Here, ${\mathbf{1}_S}$ denotes the characteristic function of ${S}$.

Note that the MAX-CUT problem on a graph ${G=(V,E)}$ can be encoded in the following way: Let ${W_{ij} = 1}$ if ${\{i,j\} \in E}$ and ${W_{ij}=0}$ otherwise. Then the value of the maximum cut in ${G}$ is precisely the maximum of ${\langle W, A\rangle}$ for ${A \in \mathrm{CUT}_n}$. Accordingly, we should expect that ${\bar \gamma(\mathrm{CUT}_n)}$ cannot be bounded by any polynomial in ${n}$ (lest we violate a basic tenet of complexity theory).

1.5. Non-negative matrix factorization

The key to understanding ${\bar \gamma(\mathrm{CUT}_n)}$ comes from Yannakakis’ factorization theorem.

Consider a polytope ${P \subseteq \mathbb R^d}$ and let us write in two ways: As a convex hull of vertices

$\displaystyle P = \mathrm{conv}\left(x_1, x_2, \ldots, x_n\right)\,,$

and as an intersection of half-spaces: For some ${A \in \mathbb R^{m \times d}}$ and ${b \in \mathbb R^m}$,

$\displaystyle P = \left\{ x \in \mathbb R^d : Ax \leq b \right\}\,.$

Given this pair of representations, we can define the corresponding slack matrix of ${P}$ by

$\displaystyle S_{ij} = b_i - \langle A_i, x_j \rangle \qquad i \in \{1,2,\ldots,m\}, j \in \{1,2,\ldots,n\}\,.$

Here, ${A_1, \ldots, A_m}$ denote the rows of ${A}$.

We need one more definition. In what follows, we will use ${\mathbb R_+ = [0,\infty)}$. If we have a non-negative matrix ${M \in \mathbb R_+^{m \times n}}$, then a rank-${r}$ non-negative factorization of ${M}$ is a factorization ${M = AB}$ where ${A \in \mathbb R_+^{m \times r}}$ and ${B \in \mathbb R_+^{r \times n}}$. We then define the non-negative rank of ${M}$, written ${\mathrm{rank}_+(M)}$, to be the smallest ${r}$ such that ${M}$ admits a rank-${r}$ non-negative factorization.

Theorem 1 (Yannakakis) For every polytope ${P}$, it holds that ${\bar \gamma(P) = \mathrm{rank}_+(S)}$ for any slack matrix ${S}$ of ${P}$.

The key fact underlying this theorem is Farkas’ Lemma. It asserts that if ${P = \{ x \in \mathbb R^d : Ax \leq b \}}$, then every valid linear inequality over ${P}$ can be written as a non-negative combination of the defining inequalities ${\langle A_i, x \rangle \leq b_i}$.

There is an interesting connection here to proof systems. The theorem says that we can interpret ${\bar \gamma(P)}$ as the minimum number of axioms so that every valid linear inequality for ${P}$ can be proved using a conic (i.e., non-negative) combination of the axioms.

1.6. Slack matrices and the correlation polytope

Thus to prove a lower bound on ${\bar \gamma(\mathrm{CUT}_n)}$, it suffices to find a valid set of linear inequalities for ${\mathrm{CUT}_n}$ and prove a lower bound on the non-negative rank of the corresponding slack matrix.

Toward this end, consider the correlation polytope ${\mathrm{CORR}_n \subseteq \mathbb R^{n^2}}$ given by

$\displaystyle \mathrm{CORR}_n = \mathrm{conv}\left(\left\{x x^T : x \in \{0,1\}^n \right\}\right)\,.$

It is an exercise to see that ${\mathrm{CUT}_{n+1}}$ and ${\mathrm{CORR}_n}$ are linearly isomorphic.

Now we identify a particularly interesting family of valid linear inequalities for ${\mathrm{CORR}_n}$. (In fact, it turns out that this will also be an exhaustive list.) A quadratic multi-linear function on ${\mathbb R^n}$ is a function ${f : \mathbb R^n \rightarrow \mathbb R}$ of the form

$\displaystyle f(x) = a_0 + \sum_i a_{ii} x_i + \sum_{i < j} a_{ij} x_i x_j\,,$

for some real numbers ${a_0}$ and ${\{a_{ij}\}}$.

Suppose ${f}$ is a quadratic multi-linear function that is also non-negative on ${\{0,1\}^n}$. Then “${f(x) \geq 0 \,\,\forall x \in \{0,1\}^n}$” can be encoded as a valid linear inequality on ${\mathrm{CORR}_n}$. To see this, write ${f(x)=\langle A,xx^T\rangle + a_0}$ where ${A=(a_{ij})}$. (Note that ${\langle \cdot,\cdot\rangle}$ is intended to be the standard inner product on ${\mathbb R^{n^2}}$.)

Now let ${\textrm{QML}^+_n}$ be the set of all quadratic multi-linear functions that are non-negative on ${\{0,1\}^n}$, and consider the matrix (represented here as a function) ${\mathcal M_n : \textrm{QML}^+_n \times \{0,1\}^n \rightarrow \mathbb R_+}$ given by

$\displaystyle \mathcal M_n(f,x) = f(x)\,.$

Then from the above discussion, ${\mathcal M_n}$ is a valid sub-matrix of some slack matrix of ${\mathrm{CORR}_n}$. To summarize, we have the following theorem.

Theorem 2 For all ${n \geq 1}$, it holds that ${\bar \gamma(\mathrm{CUT}_{n+1}) \geq \mathrm{rank}_+(\mathcal M_n)}$.

It is actually the case that ${\bar \gamma(\mathrm{CUT}_{n+1}) = \mathrm{rank}_+(\mathcal M_n)}$. The next post will focus on providing a lower bound on ${\mathrm{rank}_+(\mathcal M_n)}$.

# Entropy optimality: Forster’s isotropy

In this post, we will see an example of entropy optimality applied to a determinantal measure (see, for instance, Terry Tao’s post on determinantal processes and Russ Lyons’ ICM survey). I think this is an especially fertile setting for entropy maximization, but this will be the only post in this vein for now; I hope to return to the topic later.

Our goal is to prove the following theorem of Forster.

Theorem 1 (Forster) Suppose that ${x_1, x_2, \ldots, x_n \in \mathbb R^k}$ are unit vectors such that every subset of ${k}$ vectors is linearly independent. Then there exists a linear mapping ${A : \mathbb R^k \rightarrow \mathbb R^k}$ such that

$\displaystyle \sum_{i=1}^n \frac{(A x_i) (A x_i)^T}{\|A x_i\|^2} = \frac{n}{k} I\,. \ \ \ \ \ (1)$

This result is surprising at first glance. If we simply wanted to map the vectors ${\{x_1, \dots x_n\}}$ to isotropic position, we could use the matrix ${A = (\sum_{i=1}^n x_i x_i^T)^{-1/2}}$. But Forster’s theorem asks that the unit vectors

$\displaystyle \left\{ \frac{Ax_1}{\|A x_1\|}, \ldots, \frac{A x_n}{\|A x_n\|} \right\}$

are in isotropic position. This seems to be a much trickier task.

Forster used this as a step in proving lower bounds on the sign rank of certain matrices. Forster’s proof is based on an iterative argument combined with a compactness assertion.

There is another approach based on convex programming arising in the work of Barthe on a reverse Brascamp-Lieb inequality. The relation to Forster’s theorem was observed in work of Hardt and Moitra; it is essentially the dual program to the one we construct below.

We first recall a few preliminary facts about determinants. For any ${x_1, \ldots, x_n \in \mathbb R^k}$, we have the Cauchy-Binet formula

$\displaystyle \det\left(\sum_{i=1}^n x_i x_i^T\right) = \sum_{|S|=k} \det\left(\sum_{i \in S} x_i x_i^T\right)\,.$

We also have a rank-one update formula for the determinant: If a matrix ${A}$ is invertible, then

$\displaystyle \det\left(A+u u^T\right) = \det(A) \left(1+u^T A^{-1} u\right)\,.$

Finally, for ${k}$ vectors ${x_1, x_2, \ldots, x_k \in \mathbb R^k}$ and nonnegative coefficients ${c_1, c_2, \ldots, c_k \geq 0}$, we have

$\displaystyle \det\left(\sum_{i=1}^k c_i \,x_i x_i^T\right) = \left(\prod_{i=1}^k c_i\right) \det\left(\sum_{i=1}^k x_i x_i^T\right)\,.$

This follows because replacing ${x_i}$ by ${c_i x_i}$ corresponds to multiplying the ${i}$th row and column of ${XX^T}$ by ${\sqrt{c_i}}$, where ${X}$ is the matrix that has the vectors ${x_1, \ldots, x_k}$ as columns.

1.2. A determinantal measure

To prove Theorem 1, we will first define a probability measure on ${{[n] \choose k}}$, i.e., on the ${k}$-subsets of ${\{1,2,\ldots,n\}}$ by setting:

$\displaystyle D_S = \frac{\det\left(\sum_{i \in S} x_i x_i^T\right)}{\det\left(\sum_{i=1}^n x_i x_i^T\right)}\,.$

The Cauchy-Binet formula is precisely the statement that ${\sum_{|S|=k} D_S=1}$, i.e. the collection ${\{D_S\}}$ forms a probability distribution on ${{[n] \choose k}}$. How can we capture the fact that some vectors ${x_1, \ldots, x_n}$ satisfy ${\sum_{i=1}^n x_i x_i^T = \frac{n}{k} I}$ using only the values ${D_S}$?

Using the rank-one update formula, for an invertible ${k \times k}$ matrix ${B}$, we have ${\frac{\partial}{\partial t}\big|_{t=0} \log \det(B+t u u^T) = \langle u, B^{-1} u\rangle}$. Thus ${B}$ is the ${k \times k}$ identity matrix if and only if for every ${u \in \mathbb R^k}$,

$\displaystyle \frac{\partial}{\partial t}\left|_{t=0} \log \det(B+t uu^T)=\|u\|^2\,.\right.$

Note also that using Cauchy-Binet,

$\displaystyle \frac{\partial}{\partial t}\left|_{t=0}\vphantom{\bigoplus}\right. \log \det\left(\sum_{j=1}^n x_j x_j^T + t x_i x_i^T\right)\qquad\qquad\qquad$

$\displaystyle = \frac{\sum_{S : i \in S} \frac{\partial}{\partial t}\big|_{t=0} \det\left(\sum_{j \in S} x_j x_j^T + t x_i x_i^T\right)} {\det\left(\sum_{i=1}^n x_i x_i^T\right)}$

$\displaystyle = \frac{\sum_{S : i \in S} \frac{\partial}{\partial t}\big|_{t=0} (1+t)\det\left(\sum_{j \in S} x_j x_j^T\right)} {\det\left(\sum_{i=1}^n x_i x_i^T\right)} = \sum_{S : i \in S} D_S\,.$

In particular, if ${\sum_{i=1}^n x_i x_i^T = \frac{n}{k} I}$, then for every ${i=1,2,\ldots,n}$, we have

$\displaystyle \frac{k}{n} = \frac{\partial}{\partial t}\big|_{t=0} \log \det\left(\sum_{i=1}^n x_i x_i^T +t x_i x_i^T\right) = \sum_{S : i \in S} D_S\,. \ \ \ \ \ (2)$

Of course, our vectors ${x_1, \ldots, x_n}$ likely don’t satisfy this condition (otherwise we would be done). So we will use the max-entropy philosophy to find the “simplest” perturbation of the ${D_S}$ values that does satisfy it. The optimal solution will yield a matrix ${A}$ satisfying (1).

1.3. Entropy maximization

Consider the following convex program with variables ${\{p_S : S \subseteq [n], |S|=k\}}$.

$\displaystyle \textrm{minimize} \qquad \sum_{|S|=k} p_S \log \frac{p_S}{D_S}$

$\displaystyle \textrm{subject to} \qquad \sum_{S : i \in S} p_S = \frac{k}{n} \qquad \forall i=1,2,\ldots,n$

$\displaystyle \sum_{|S|=k} p_S = 1$

$\displaystyle p_S \geq 0 \qquad \forall |S|=k\,.$

In other words, we look for a distributon on ${[n] \choose k}$ that has minimum entropy relative to ${D_S}$, and such that all the “one-dimensional marginals” are equal (recall (2)). Remarkably, the optimum ${p^*_S}$ will be a determinantal measure as well.

Note that the uniform distribution on subsets of size ${k}$ is a feasible point and the objective is finite precisely because ${D_S \neq 0}$ for every ${S}$. The latter fact follows from our assumption that every subset of ${k}$ vectors is linearly independent.

1.4. Analyzing the optimizer

By setting the gradient of the Lagrangian to zero, we see that the optimal solution has the form

$\displaystyle p_S(\lambda_1, \ldots, \lambda_n) = \frac{\exp\left(\sum_{i \in S} \lambda_i \right) D_S}{\sum_{|S|=k} D_S \exp\left(\sum_{i \in S} \lambda_i\right)} = \frac{\exp\left(\sum_{i \in S} \lambda_i \right) D_S}{\det\left(\sum_{i=1}^n e^{\lambda_i} x_i x_i^T\right)} = \frac{\det\left(\sum_{i \in S} e^{\lambda_i} x_i x_i^T\right)}{\det\left(\sum_{i=1}^n e^{\lambda_i} x_i x_i^T\right)} \ \ \ \ \ (3)$

for some dual variables ${(\lambda_1, \ldots, \lambda_n)}$. Note that the ${\{\lambda_i\}}$ dual variables are unconstrained because they come from equality constraints.

Let us write ${U = \sum_{i=1}^n e^{\lambda_i} x_i x_i^T}$. We use ${p_S^*}$, ${U_*}$, and ${\{\lambda_i^*\}}$ to denote the values at the optimal solution. Using again the rank-one update formula for the determinant,

$\displaystyle \frac{\partial}{\partial \lambda_i} \log \det(U) = e^{\lambda_i} \langle x_i, U^{-1} x_i\rangle\,.$

But just as in (2), we can also use Cauchy-Binet to calculate the derivative (from the second expression in (3)):

$\displaystyle \frac{\partial}{\partial \lambda_i} \log \det(U) = \sum_{S : i \in S} p_S\,,$

where we have used the fact that ${\frac{\partial}{\partial \lambda_i} p_S = p_S}$ if ${i \in S}$ (and otherwise equals ${0}$). We conclude that

$\displaystyle \langle x_i, U_*^{-1} x_i\rangle = e^{-\lambda^*_i} \sum_{S : i \in S} p^*_S = e^{-\lambda^*_i} \frac{k}{n}\,.$

Now we can finish the proof: Let ${A = U_*^{-1/2}}$. Then:

$\displaystyle \sum_{i=1}^n \frac{(A x_i) (A x_i)^T}{\|A x_i\|^2} = U_*^{-1/2} \sum_{i=1}^n \frac{x_i x_i^T}{\langle x_i, U_*^{-1} x_i\rangle} U_*^{-1/2}$

$\displaystyle = U_*^{-1/2} \sum_{i=1}^n \frac{n}{k} e^{\lambda_i^*} x_i x_i^T U_*^{-1/2} = \frac{n}{k} I\,.$

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