Global Optimal of Convex Optimization

$$\text{ (P) }\underset{x\in \R^n}{\min}f(x) \text{ s.t. } x\in C$$

• $f:\R^n\to\R$ convex and continuously differentiable
• $C\subseteq \R^n$ convex

If $\bar{x}$ is a local minimizer, then $\bar{x}$ is a global min

• level sets of convex functions

• first and second order optimization: $\nabla f(x)=0$; $\nabla^2f(x)\ge0$
• Optimality for constrained problems

➡️ Fact: If $f:\R^n\to\R$ convex, then the level sets $[f\le \tau]:=\{x\in \R^n \mid f(x)\in \tau\}$ are convex sets

Suppose $\bar{x}\in C$ solves (P)

$\underbrace{f(\bar{x})}_{\equiv \tau}\le f(x), \forall x\in C$

$[f\le \tau]= \{x\mid f(x)\le \tau \}$

$[f\le \tau]\cap C$ ⇒ solution set

Take any $x,y\in [f\le \tau]$

Show that $\lambda x+(1-\lambda)y\in [f\le \tau]$ for $\lambda \in [0,1]$, $f(x)\le \tau$, $f(y)\le \tau$.

Because $f$ is conex,

$\begin{aligned}f(\lambda x+(1-\lambda)y)&\le\lambda f(x)+(1-\lambda)f(y) \\ & \le \lambda\tau+(1-\lambda)\tau \\ &=\tau\end{aligned}$$\checkmark$

First-Order Characterization

✅ Theorem. Let $f:C\to\R$ be continuously differentiable over $C\subseteq \R^n$, where $C$ means convex set, then $f$ is convex iff

$$f(x)+\nabla f(x)^T(z-x)\le f(z), \forall x,z\in C$$

Proof Sketch: (⇒ only if direction)

By convexity of $f$:

$f(\lambda z+(1-\lambda)x)\le \lambda f(z)+(1-\lambda)f(x)$

$f(\lambda z+(1-\lambda)x)-f(x)\le \lambda[f(z)-f(x)]$

Dividing $\lambda$ by both sides, $\frac{f(\lambda z+(1-\lambda)x)-f(x)}{\lambda}\le [f(z)-f(x)]$

$\begin{aligned}&\underset{\lambda\to0}{\lim}\frac{f(\lambda z+(1-\lambda)x)-f(x)}{\lambda}\le [f(z)-f(x)] \\ =&\underset{\lambda\to0}{\lim}\frac{f(x+\lambda(z-x))-f(x)}{\lambda}\le [f(z)-f(x)]\end{aligned}$

$f’(x;z-x)\le f(z)-f(x)$

Because $f$ continuously differentiable, $f’(x;z-x)=\nabla f(x)^T(z-x)$

⇒ $\nabla f(x)^T(z-x)\le f(z)-f(x)$

⇒ $f(x)+\nabla f(x)^T(z-x)\le f(z), \forall x,z$

Consider the unconstrained differentiable problem

$$\underset{x\in \R^n}{\min}f(x) \text{ (f convex)}$$

By convexity,

$$f(x)+\nabla f(x)^T(z-x)\le f(z),\forall x,z\in \R^n$$

If $x^*$ is a local min, then

$$\nabla f(x^*)=0$$

$x=x^*$

$$f(x^*)\le f(z), \forall x\in\R^n$$

For convex unconstrained problems,

$$\text{Stationarity} \Leftrightarrow \text{Global Optimality}$$

For $f:C\to\R$ ($C\subseteq \R^n$ convex) twice continuous differentiable then $f$ is convex iff

$$\nabla^2 f(x)\ge 0, \forall x\in C$$

Example. $f(x)=\frac{1}{2}\lVert x\rVert^2$

$\nabla f(x)=x$

$\nabla^2 f(x)=I\gt 0$

Unconstrained Case ($C=\R^n$)

$$\begin{aligned}x^* \in \underset{x\in\R^n}{\argmin}f(x)&\Leftrightarrow f’(x^*;x-x^*)=\nabla f(x^*)^T(x-x^*)\ge 0\\&\Leftrightarrow f(x^*)=0\end{aligned}$$

Constrained Case ($C \subset \R^n$)

$$x^* \in \underset{x\in C}{\argmin}f(x)\Leftrightarrow f’(x^*;x-x^*)=\nabla f(x^*)^T(x-x^*)\ge 0,\forall x\in C$$

⇒ All feasible directions are non-decreasing on $f$.

⇒ $-\nabla f(x^*)^T(x-x^*)\le0,\forall x\in C$