Concavity

dm1.tex

@@ -19,6 +19,7 @@
 \newcommand{\N}{\mathbb{N}}
 \newcommand{\trans}{\mathsf{T}}
 \newcommand{\dist}{\text{\textbf{dist}}}
+\newcommand{\pmat}[1]{\begin{pmatrix} #1 \end{pmatrix}}
 
 \author{Yoann Coudert-\,-Osmont}
 
@@ -350,9 +351,21 @@
 	Now we use the fact that $u \geqslant - \| y \|_2$:
 	$$ g(x, t, y, u) = - \| y \|_2 + \log \left( 2 \lambda \| y \|_2 + 1 \right) \xrightarrow[\lambda \rightarrow +\infty]{} +\infty $$
 	So when $\| y \|_2 \geqslant -u$, we have $f^*(y, u) = +\infty$. \\[2mm]
-	Now we suppose $\| y \|_2 < -u$. We compute the partial derivatives of $g$ with respect to $x$ and $t$:
+	Now we suppose $\| y \|_2 < -u$. The function $(x, t) \mapsto y^\trans x + tu$ is linear then it is concave. Now we will show that $h : (x, t) \mapsto log \left( t^2 - x^\trans x \right)$ is concave to show that $g(\cdot, \cdot, y, u)$ is concave. We let:
+	$$ X = \pmat{x \\ t} \quad J = \pmat{I_n & 0 \\ 0 & -1} \qquad \text{such that } h(X) = \log \left( - X^\trans J X \right) $$
+	Now we compute the gradient of $h$ and its Hessian:
+	$$ \nabla h(X) = \dfrac{2 J X}{X^\trans J X} \qquad \nabla^2 h(X) = 2 \dfrac{\left( X^\trans J X \right) J - 2 X J^\trans J X^\trans}{\left( X^\trans J X \right)^2} = 2 \dfrac{\left( X^\trans J X \right) J - 2 X X^\trans}{\left( X^\trans J X \right)^2} $$
+	Now we consider a vector $Z = \pmat{ z^\trans v }^\trans$ of $\R^{n+1}$. We look at the sign of $Z^\trans \nabla^2 h(X) Z$ that is to say the sign of $Z^\trans \left( X^\trans J X J - 2 X X^\trans \right) Z$:
+	$$ \begin{array}{lll}
+	Z^\trans \left( X^\trans J X J - 2 X X^\trans \right) Z & = & (x^\trans x - t^2) (z^\trans z - v^2) - 2 \left[ \left( z^\trans x \right)^2 + (tv)^2 \right] \\
+	& = & \| x \|_2^2 \| z \|_2^2 - t^2 \| z \|_2^2 - v^2 \| x \|_2^2 - 2 \left( z^\trans x \right)^2 - (tv)^2
+	\end{array} $$
+	We also now that $t > \| x \|_2$ which gives:
+	$$ Z^\trans \left( X^\trans J X J - 2 X X^\trans \right) Z \leqslant - 2 v^2 \| x \|_2^2 - 2 \left( z^\trans x \right)^2 \leqslant 0 $$
+	This means that $\nabla^2 h \preceq 0$. So $h$ is concave. Then $g(\cdot, \cdot, y, u)$ is concave on $K$.
+	We compute the partial derivatives of $g$ with respect to $x$ and $t$:
 	$$ \dfrac{\partial g}{\partial x}(x, t, y, u) = y - \dfrac{2x}{t^2 - x^\trans x} \qquad \dfrac{\partial g}{\partial t}(x, t, y, u) = u + \dfrac{2t}{t^2 - x^\trans x} $$
-	Let $\alpha = \frac{1}{2} \left( t^2 - x^\trans x \right)$. The maxima satisfies:
+	Let $\alpha = \frac{1}{2} \left( t^2 - x^\trans x \right)$. Because $g(\cdot, \cdot, y, u)$ is concave, he maxima satisfies $\nabla g = 0$. That is to say:
 	$$ x = \alpha y \qquad t = - \alpha u $$
 	We inject this two new expressions in the derivative of $g$ with respect to $x$ and we obtain:
 	$$ y = \dfrac{2 \alpha y}{\alpha^2 u^2 - \alpha^2 y^\trans y} = \frac{1}{\alpha} \dfrac{2y}{u^2 - y^\trans y} \quad \Rightarrow \quad \alpha = \dfrac{2}{u^2 - y^\trans y} \geqslant 0 $$