[笔记] Convex Optimization 2015.10.21

$int(K)$: interior of $K$

• Definition: Cone $K$ is proper if it is closed, convex, pointed (contains no line), solid (nonempty interior)
  If $K$ is proper, then $K^*$ is proper, $K^* = \{ \lambda : \lambda ^T x \ge 0 \; \forall x \in K \}$
  • Fact1: $x \succeq _k y \iff \lambda ^T x \ge \lambda ^T y \; \forall \lambda \; s.t. \; \lambda \succeq _{K^*} 0$
  • Proof: $x \succeq _k y \iff x - y \in K \overset{because\, K^{**} = K}{\iff} \lambda ^T (x - y) \ge 0 \; \forall \lambda \in K^*$
  • Lemma: Let $K$ be proper, then $x \in int(K)$ if and only if $\lambda ^T x \gt 0 \; \forall \lambda \in K^* \backslash \{ 0 \}$
  • Proof: If $x \in int(K)$ then obviously $\lambda ^T x \ge 0$ for all $\lambda \in K^* \backslash \{ 0 \}$.
    If $\lambda \in K^* \backslash \{ 0 \}$ and $\lambda ^T x = 0$ then there exists $u \in \mathbb{R}^n$ such that $\lambda ^T u \lt 0$, and $x + tu \in K$ for $t$ sufficiently small, a contradiction.
    Next, assume that $\lambda ^T x \gt 0$ for all $\lambda \in K^* \backslash \{ 0 \}$.
    Then $x \in K^{**} = K$, so $x \in K$.
    If $x \notin int(K)$ then $\exists (x_i)_{i = 1}^{\infty} \notin K \; s.t. \; \underset{i \to \infty}{\lim} x_i = x$.
    Since $x_i \notin K$, $\exists \lambda _i \in K^* \; s.t. \; \lambda _i^T x_i \lt 0$ (for all $i$).
    WLOG $\lambda'_i$s are unit vectors.
    Then $\{ \lambda _i : i \in \mathbb{N} \}$ lives inside a compact set (the unit ball of $\mathbb{R}^n$).
    So we can assume WLOG that $\lambda _1, \lambda _2, \cdots$ is convergent: $\underset{i \to \infty}{\lim} \lambda _i = \lambda$.
    Then $\lambda \in K^*$ because $K^*$ is closed and $\underset{i \to \infty}{\lim} (\lambda ^T x - \lambda _i^T x_i) = \underset{i \to \infty}{\lim} (\lambda ^T x - \lambda _i^T x) + \underset{i \to \infty}{\lim} (\lambda _i^T x - \lambda _i^T x_i) = 0 + 0 = 0$.
    … contradiction!
    QED.
  • Fact2: $x \succ _k y \iff \lambda ^T x \gt \lambda ^T y \; \forall \lambda \in K^* \backslash \{ 0 \}$
  • Proof: $x \succ _k y \iff x - y \in int(K) \underset{\text{by lemma}}{\iff} \lambda ^T (x - y) \gt 0 \; \forall \lambda \in K^* \backslash \{ 0 \}$

  Convex Functions:
  - A function $f : \mathbb{R}^n \to \mathbb{R}$ is convex if and only if $dom \, f$ is convex and $f(\theta x + (1 - \theta) y) \le \theta f(x) + (1 - \theta) f(y)$ for all $x, y \in dom \, f$ and all $0 \lt \theta \lt 1$.
  - $f : \mathbb{R}^n \to \mathbb{R}$ is convex if and only if $epi(f) \subseteq \mathbb{R}^{n + 1} = \{(x, t) : x \in dom \, f, t \ge f(x)\}$ is convex.

  • Example: $f(x) = \max(x_1, \cdots, x_n)$ is convex.

  • Proof: $\max(\theta x + (1 - \theta) y) = \underset{i}{\max} [\theta x_i + (1 - \theta) y_i] \le \theta \underset{i}{\max} x_i + (1 - \theta) \underset{j}{\max} y_j = \theta \max (x) + (1 - \theta) \max (y)$

  • Observation: $f$ is convex iff the restriction of $f$ to every line is convex.
    Namely, $f$ is convex if and only if the function $g(t) = f(x + tu)$ from $\mathbb{R}$ to $\mathbb{R}$ is convex for every $x \in dom \, f$ and $u \in \mathbb{R}^n$.

  • Derivatives: A function $f : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at a point $x \in int(dom \, f)$ if and only if there exists a matrix $"Df(x)" \in \mathbb{R}^{m \times n}$ such that $f(x + z) = f(x) + Df(x)z + err(z)$ where $\underset{z \to 0}{\lim} \frac{err(z)}{\lVert z \rVert _2} = 0$
    (Namely, $\underset{z \to 0 }{\lim} \frac{f(x + z) - f(z) - Df(x)z}{\lVert z \rVert _2} = 0$)

  • Proposition: The derivative, if it exists, is unique.

  • Proof: Assume that $A, B \in \mathbb{R}^{m \times n}$ both satisfy the conditions of the derivative and that $A \neq B$
    Snce $A \neq B$ there exists some vector $u \in \mathbb{R}^n \; s.t. \; Au \neq Bu$, WLOG $u$ is a unit vector.
    Then $0 = \underset{t \to 0}{\lim} \frac{f(x + tu) - f(x + tu)}{\lVert tu \rVert _2} = \underset{t \to 0}{\lim} \frac{f(x) + Atu + err_A(tu) - f(x) - Btu - err_B(tu)}{\lVert tu \rVert _2} = (A - B)_u + \underset{t \to 0}{\lim} \frac{err_A(tu) - err_B(tu)}{\lVert tu \rVert _ 2} = (A - B)_u \neq 0$

  • Example: Let $f(x) = q^T x + c$ for $q \in \mathbb{R}^n, c \in \mathbb{R}$
    Then $Df(x) = q^T$ is the derivative.

  • Example 2: Let $f(x) = x^T P x$ for $P \in S^n$
    Have $f(x + z) - f(x) = (x^T + z^T) P(x + z) - x^T P x = z^T P x + x^T P z + z^T P z = 2 x^T P z + z^T P z$
    so will have $Df(x) = 2x^T P$ if we can show $\underset{z \to 0}{\lim} \frac{\lVert z^T P z \rVert _2}{\lVert z \rVert _2} = 0$
    $\lVert z^T P z \rVert _2 \le \lVert z \rVert _2 \lVert Pz \rVert _2 \le \lVert z \rVert _2 \lVert P \rVert _2 \lVert z \rVert _2$

  • Example: Let $f : S_{++}^n \to \mathbb{R}$ be defined by $f(X) = \log (\det X)$, $Z \in S^n$
    

    $$\begin{align*} f(X + Z) - f(x) = & \log \det(X + Z) - \log \det X \\ =& \log \det (X(I + X^{-1} Z)) - \log \det X \\ =& \log \det X + \log \det (I + X^{-1} Z) - \log \det X \\ & \text{(Let $I = QQ^T, X^{-1} Z = Q \text{diag}(\lambda_1, cdots , \lambda _n) Q^T$)} \\ =& \log \prod_{i = 1}^n (1 + \lambda _i) \\ =& \sum_{i = 1}^n \log (1 + \lambda _i) \\ =& \sum_{i = 1}^n \lambda _i + o(\lambda _i) \\ =& tr(X^{-1} Z) \\ =& \left \langle X^{-1}, Z \right \rangle \\ \end{align*}$$

  $tr(B^T A) = \left \langle B, A \right \rangle$

  Second derivative (for functions of range $\mathbb{R}$, m = 1)
  $f : \mathbb{R}^n \to \mathbb{R}, Df(x) = \nabla f(x)^T, D(Df(x)^T) = D(\nabla f(x)) \in \mathbb{R}^{n \times n}$
  $\nabla f(x) = [\frac{\partial f}{\partial x_1}, \cdots , \frac{\partial f}{\partial x_n}]^T$
  $D(\nabla f(x)) = \begin{bmatrix} \frac{\partial ^2 f}{\partial x_1^2} & \cdots & \frac{\partial ^2 f}{\partial x_1 x_n} \\ \vdots & & \vdots \\ \frac{\partial^2 f}{\partial x_n x_1} & \cdots & \frac{\partial^2 f}{\partial x_n^2} \end{bmatrix} = \nabla ^2 f''(x)$
  means the Hessian of $f'$.
  $f = \begin{bmatrix} f_1 \\ \vdots \\ f_m \end{bmatrix}$
  $Df = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n} \end{bmatrix}$

  First order condition for convexities
  - Fact: Assume $f : \mathbb{R}^n \to \mathbb{R}$ is differentiable (This means $dom \, f$ is open.)
  Then $f$ is convex iff $dom \, f$ is convex and $f(y) \ge f(x) + Df(x)(y - x)$ for all $x, y \in dom \, f$
  - Proof: $\impliedby$ Let $x, y \in dom \, f$, $0 \lt \theta \lt 1$, $z = \theta x + (1 - \theta) y$
  We have, by $f(y) \ge f(x) + Df(x) (y - x)$,
  $f(x) \ge f(z) + Df(z)(x - z)$, $f(y) \ge f(z) + Df(z)(y - z)$
  so $\theta f(x) + (1 - \theta)f(y) \ge f(z) + \theta Df(z) x + (1 - \theta) Df(z) y - Df(z) z = f(z) \implies \text{convex}$
  $\implies$ Fix $x, y \in dom \, f$ and $x \neq y$, Let $g(t) = f(x + t(y - x)) = f(ty + (1 - t)x)$
  Then $g(t) \le tf(y) + (1 - t)f(x)$ by convexity (for $t \in [0, 1]$)
  so $\frac{f(x + t(y - x)) + (1 -t)f(x)}{t} \le f(y) \implies \frac{f(x + t(y - x)) -f(x)}{t} + f(x) \le f(y)$
  $\lim _{t \to 0} \frac{Df(x)(t(y - x)) +err_x(t(y - x))}{t} + f(x) \le f(y)$
  $Df(x)(y - x) + \lim_ {t \to 0} \frac{err_x(t(y - x))}{t} + f(x) \le f(y)$
  $f(y) \ge Df(x)(y - x) + f(x)$