[笔记] Convex Optimization 2015.11.25

$\lVert y \rVert _* = \sup \{ x^T y : \lVert x \rVert \le 1 \} \implies x^T y \le \lVert x \rVert \cdot \lVert y \rVert _*$
(because $\frac{x^T}{\lVert x \rVert} y \le \lVert y \rVert _*$)
Want inequality of type: $x^T y \le f(x) + "f^*(y)"$ for “general” $f$ (Fenchel’s Inequality)

• Definition: For $f : \mathbb{R}^n \to \mathbb{R}$, the conjugate $f^*$ of $f$ is defined by $f^*(y) = \underset{x}{\sup} (x^T y - f(x))$
  with $dom \, f^* =$ set of $y$’s for which $\sup$ is $\lt \infty$.

  • Example:

    1. $f(x) = a^T x + b (x \in \mathbb{R}^n)$
       $f^*(y) = \underset{x}{\sup} x^T y - a^T x - b = \begin{cases} \infty & \text{if } y \neq a \\ -b & \text{if } y = a \end{cases}$
    2. $f(x) = -\log x (x \gt 0)$
       $(xy + \log x)' = y + \frac{1}{x} = 0 \implies x = -\frac{1}{y}$
       $f^*(y) = \underset{x \gt 0}{\sup} x^T y + \log x = \begin{cases} \infty & \text{if } y \ge 0 \\ -\log(-y) - 1 & \text{if } y \lt 0 \end{cases}$
    3. $f(x) = e^x (x \in \mathbb{R})$
       $(xy - e^x)' = y - e^x = 0 \implies x = \log y$
       $f^*(y) = \underset{x}{\sup} x^T y - e^x = \begin{cases} \infty & \text{if } y \lt 0 \\ y\log y - y & \text{if } y \ge 0 \end{cases}$
    4. $f(x) = x \log x (x \ge 0)$
       $(xy - x \log x)' = y - \log x - 1 = 0 \implies x = e^{y - 1}$
       $f^*(y) = \underset{x \ge 0}{\sup} x^T y - x \log x = y e^{y - 1} - (y - 1) e^{y - 1} = e^{y - 1}$
    5. $f(x) = \frac{1}{2} x^T Qx$ with $Q \in S_{++}^n$
       $f^*(y) = \underset{x}{\sup} x^T y - \frac{1}{2} x^T Qx = y^T Q^{-1} y - \frac{1}{2} y^T Q^{-1} y = \frac{1}{2} y^T Q^{-1} y$
       ($\underset{x}{\inf} x^T Ax + x^T b \implies \text{best} x = -\frac{1}{2} A^{-1} b$)
       So $x = Q^{-1} y$
       $\implies x^T y \le \frac{1}{2} x^T Qx + \frac{1}{2} y^T Q^{-1} y$, for all $Q \succ 0$
    6. $f(x) = \log \left( \sum _{i = 1}^n e^{x_i} \right)$
       $f*(y) = \underset{x}{\sup} x^T y - \log \left( \sum _{i = 1}^n e^{x_i} \right)$
       $\left(xy - \log \left( \sum _{i = 1}^n e^{x_i} \right) \right)' = y - \frac{e^{x_i}}{\sum _{i = 1}^n e^{x_i}} = 0$
       $\implies y_i = \frac{e^{x_i}}{\sum _{i = 1}^n e^{x_i}}, y \succeq 0, 1^T y = 1$
       assume for simplicity, $y \succ 0$
       put $x_i = \log (y_i)$, then $\sum e^{x_i} = 1^T y = 1$ and optimality conditions hold
       then $f^*(y) = \sum _{i = 1}^n y_i \log (y_i) - \log (1^T y) = \sum _{i = 1}^n y_i \log (y_i)$
    7. $f(x) = \lVert x \rVert$
       $f^*(y) = \underset{x}{\sup} x^T y - \lVert x \rVert = \begin{cases} 0 & \text{if } \lVert y \rVert _* \le 1 \\ \infty & \text{if } \lVert y \rVert _* \gt 1 \end{cases}$
       $x^T y - \lVert x \rVert \le \lVert x \rVert \cdot \lVert y \rVert _* - \lVert x \rVert = \lVert x \rVert (\lVert y \rVert _* - 1) \le 0$ if $\lVert y \rVert _* - 1 \le 0$
    8. $f(x) = \frac{1}{2} \lVert x \rVert ^2$
       $f^*(y) = \underset{x}{\sup} x^T y - \frac{1}{2} \lVert x \rVert ^2 = \frac{1}{2} \lVert y \rVert _*^2$
       $x^T y - \frac{1}{2} \lVert x \rVert ^2 \le \lVert x \rVert \cdot \lVert y \rVert _* - \frac{1}{2} \lVert x \rVert ^2 \le \frac{1}{2} \lVert y \rVert _*^2$ ($\lVert x \rVert = \lVert y \rVert_*$)
       $\implies x^T y \le \frac{1}{2} \lVert x \rVert ^2 + \frac{1}{2} \lVert y \rVert _*^2$

  • Proof of general hyperplane seperation:
    Let $C \subseteq \mathbb{R}^n$ be a convex set, $H \subseteq \mathbb{R}$ be the affine subspace of smallest dimention containing $C$, we write $C_{\varepsilon} = \{ x : B_{\varepsilon} (x) \bigcap H \subseteq C \}$
    then $C_{\varepsilon} \subseteq "\text{relint} (C)" = \underset{\varepsilon \gt 0}{\bigcup} C_{\varepsilon}$. (relint: relative interior)
    ($C \subseteq \overline{\text{relint}(C)}$, $C$ is a subset of closure of $\text{relint}(C)$)
    Let $C, D$ be disjoint convex sets. Then for every $\varepsilon \gt 0$ the sets $A_{\varepsilon} = \overline{C_{\varepsilon}} \bigcap B_{\frac{1}{\varepsilon}}(0)$, $\overline{D}$ are closed disjoint convex sets with $\overline{C_{\varepsilon}} \bigcap B_{\frac{1}{\varepsilon}}(0)$ bounded, and $\text{dist}(A_{\varepsilon}, \overline{D}) \ge \varepsilon \gt 0$.
    So $\exists A_{\varepsilon} \in \mathbb{R}^n$, $a_{\varepsilon} \neq 0$, $b_{\varepsilon} \in \mathbb{R}$ s.t. $(a_{\varepsilon}, b_{\varepsilon})$ define a seperating hyperplane for $A_{\varepsilon}, \overline{D}$.
    $a_{\varepsilon}^T x \le b_{\varepsilon} \; \forall x \in A_{\varepsilon}$, $a_{\varepsilon}^T x \ge b_{\varepsilon} \; \forall x \in \overline{D}$
    WLOG $\lVert a_{\varepsilon} \rVert = 1$
    The sequence $(\vec{a}_\frac{1}{n})_{n = 1}^{\infty}$ is a sequence of unit vectors and so has a convergent subsequence, say WLOG convergent to $a_0 \in \mathbb{R}^n$.
    can assume sequence $b_{\frac{1}{n}}$ is bonded (or else one of the sets $C, D$ is empty)
    and so also convergent to some value $b_0 \in \mathbb{R}$.
    Want to show $(a_0, b_0)$ is SH for $C, D$, i.e., that
    $a_0^T x \le b_0 \; \forall x \in C, \, a_0^T x \ge b_0 \; \forall x \in D$
    (Assume $C$ is not a point, proof like above; then assume D is not a point, switch $C, D$.
    If $C, D$ are points, obious true.)

  Log-convexity and log-concavity
  - Definition: $f : \mathbb{R}^n \to \mathbb{R}_{\gt 0}$ is log-convex (log-concave) if $\log (f)$ is convex (concave).
  - Convexity:
  $\log (f(\theta x + (1 - \theta) y)) \le \theta \log (f(x)) + (1 - \theta) \log (f(y)) = \log (f(x)^{\theta} f(y)^{1 - \theta})$
  $\iff f(\theta x + (1 - \theta)y) \le f(x)^{\theta} f(y)^{1 - \theta}$
  - Remark 2: log-convex $\implies$ convex, $f(x) = e^{\log f(x)}$, (composition function, QED)
  concave $\implies$ log-concave