Negative Entropy.


Notation: Denoting standard simplex in \(\mathbb{R}^n\) as \(\triangle_n \triangleq \left\{ \boldsymbol { \lambda } \in \mathbb { R } _ { + } ^ { n } : \langle \boldsymbol{1}, \boldsymbol { \lambda } \rangle = 1 \right\}\).

Strongly convex property

We consider the negative entropy defined as

\[\eta(\mathbf{x}) = \langle \mathbf{x}, \ln \mathbf{x} \rangle, \mathbf{x} \in \triangle_n.\]

Theorem 1. The negative entropy is a strongly convex on \(\triangle_n\) in the \(\ell_1\)-norm with convexity parameter 1, i.e., for any \(\mathbf{h} \in \mathbb{R}^n\), the following inequality holds

\[\left\langle \nabla^2 \eta(\mathbf{x}) \mathbf{h}, \mathbf{h}\right\rangle \leq \| \mathbf{h} \|_1^2.\]

Proof. First note that \(\nabla \eta(\mathbf{x}) = \mathbf{1} + \ln \mathbf{x}\) and \(\nabla^2 \eta(\mathbf{x}) = \operatorname{diag}(1/\mathbf{x})\), and thus

\[\begin{align} \left\langle \nabla^2 \eta(\mathbf{x}) \mathbf{h}, \mathbf{h}\right\rangle = \left\langle \mathbf{h}, \frac{\mathbf{h}}{\mathbf{x}} \right\rangle = \sum_{i=1}^n \frac{h_i^2}{x_i}. \end{align}\]

Now we shall find

\[\min f(\mathbf{x}) \triangleq \left\langle \mathbf{h}, \frac{\mathbf{h}}{\mathbf{x}} \right\rangle \quad \mathrm{s.t.} \quad \langle \mathbf{1}, \mathbf{x} \rangle = 1.\]

According to the Proposition 1 the optimal \(\mathbf{x}^*\) and optimal dual multiplier \(\lambda^*\) satisfy

\[\nabla f(\mathbf{x}^*) = -\left(\frac{\mathbf{h}}{\mathbf{x}^*}\right)^2 = \mathbf{1} \lambda^* \quad \Rightarrow \quad \left(\frac{h_i}{x^*_i}\right)^2 = \lambda^* \quad \Rightarrow \quad x_i^* = \frac{|h_i|}{\sqrt{\lambda^*}}\]

in the first step, we have absorbed the minus operator into \(\lambda^*\) since it is a free variable; the second step follows the fact that \(h_i \in \mathbb{R}\) and \(x_i^* \in \mathbb{R}_+\).

Taking \(x_i^* = \frac{\vert h_i \vert}{\sqrt{\lambda^*}}\) along with \(\sum_{i=1}^n x_i^* = 1\), we get

\[f(\mathbf{x}^*) = \left( \sum_{i}^n |h_i| \right)^2 = \| \mathbf{h} \|_1^2.\]

Proposition 1. Let \(\mathbf{x}^*\) be a local minimum of a differentiable function \(f\) subject to the linear equality constraints

\[\mathbf{x} \in \mathcal{E} = \{ \mathbf{x} \in \mathbb{R}^n | A \mathbf{x} = \mathbf{b} \} \neq \emptyset,\]

where \(A\) is an \(m \times n\)-matrix with full row rank, and \(m < n\). Then there exists a vector of multipliers \(\boldsymbol{\lambda}^*\) such that

\[\nabla f(\mathbf{x}^*) = A^\top \boldsymbol{\lambda}^*.\]

Proof. Let \(\mathbf{u}_i \in \mathbb{R}^n, i = 1 \ldots k\) be the basis of the null space of \(A\) and \(g(\mathbf{v}) \triangleq \mathbf{x}^* + U \mathbf{v}\) where \(\mathbf{v} \in \mathbb{R}^k\), \(U = [\mathbf{u}_1, \ldots, \mathbf{u}_k]\). We observe that \(g(\mathbf{v}) \in \mathcal{E}\) for any \(\mathbf{v} \in \mathbb{R}^k\), and let \(\phi(\mathbf{v}) = f(g(\mathbf{v}))\). Since \(\phi(0)\) is a local minimum of \(\phi\) then

\[\nabla \phi(\mathbf{v})|_{\mathbf{v} = 0} = U^\top \nabla f(\mathbf{x}^*) = 0,\]

i.e., \(\nabla f(\mathbf{x}^*)\) is orthogonal to the null space of \(A\), hence \(\nabla f(\mathbf{x}^*)\) must stay in the row space of \(A\) which completes the proof.

Remark: \(\mathrm{d} \phi(\mathbf{v}) = \mathrm{d} f(g(\mathbf{v})) = \langle \nabla f, \mathrm{d} g(\mathbf{v}) \rangle = \langle \nabla f, U \mathrm{d} \mathbf{v} \rangle = \langle U^\top \nabla f, \mathrm{d} \mathbf{v} \rangle\).


Differentiable dynamic programming

The \(\max\) operator can be formulated as

\[\max(\mathbf{x}) = \sup_{\mathbf{q} \in \triangle_+} \langle \mathbf{q}, \mathbf{x} \rangle\]

However, it is not a smooth operator w.r.t \(\mathbf{x}\) and we can smooth it with negative entropy to yield

\[\operatorname{max}_{\eta}(\mathbf{x}) = \sup_{\mathbf{q} \in \triangle_+} \langle \mathbf{q}, \mathbf{x} \rangle - \eta(\mathbf{q})\]

where the r.h.s is a strongly concave function w.r.t \(\mathbf{q}\), more details is available here.

Smoothed Wasserstain distance

Recall the Wasserstain distance between two measure \(\mathbf{a}, \mathbf{b}\): \(\mathrm{L}_{C}(\mathbf{a}, \mathbf{b}) \triangleq \min_{P \in U(\mathbf{a}, \mathbf{b})} \langle P, C \rangle\) the r.h.s is a convex function w.r.t \(P\) and we can transform it to a \(\epsilon\)-strongly convex function with negative entropy \(\mathrm{L}_{C}^{\epsilon}(\mathbf{a}, \mathbf{b}) \triangleq \min_{P \in U(\mathbf{a}, \mathbf{b})} \langle P, C \rangle + \epsilon \eta(P).\)