Shen Shen
Feb 23, 2021
Def: A set \(\Omega\) is a convex set if
\(\lambda x+(1-\lambda) y \in \Omega, \) \(\forall x,y \in \Omega, \forall \lambda\in[0,1]\)
convex combination of \(x\) and \(y\)
convex combination coefficient
Convex Sets
Non-convex Sets
because...
- Hyperplanes: \(\left\{x \mid a^{T} x=b\right\}\left(a \in \mathbb{R}^{n}, b \in \mathbb{R}, a \neq 0\right)\)
- Halfspaces: \(\left\{x \mid a^{T} x \leq b\right\} \left(a \in \mathbb{R}^{n}, b \in \mathbb{R}, a \neq 0\right)\)
\(n=3: 3x_1+4x_2+5x_3=1\)
\(n=2: 3x_1+4x_2+\leq 2\)
- Euclidean balls: \(\left\{x \mid \left||x-x_{c}\right|| \leq r\right\} \) \( (x_{c} \in \mathbb{R}^{n}, r \in \mathbb{R},\|.\|\) 2-norm)
- Ellipsoids: \(\left\{x \mid\left(x-x_{c}\right)^{T} P\left(x-x_{c}\right) \leq r\right\}\left(x_{c} \in \mathbb{R}^{n}, r \in \mathbb{R}, P \succ 0\right)\)
\(n=2\)
\(n=3\)
- The set of PSD matrices
- The set of co-positive matrices (HW)
A common recipe
1. Pick two arbitrary members, say \(x\) and \(y\), from \(\Omega\)
2. Write out the convex combination \(c\)
3. Make use of that \(x\) and \(y\) are from \(\Omega\), and \(\lambda \in [0,1]\)
4. Claim the \(c\) also belongs in \(\Omega\)
Proof of the PSD claim:
Pick \(A\succeq 0, B\succeq 0\)
N.t.s \(\lambda A+(1-\lambda) B \succeq 0\)
i.e., \(x^T(\lambda A+(1-\lambda) B)x \geq 0 \forall x\)
\(=\lambda x^{T} A x+(1-\lambda) x^{T} B x\)
\(x^T(\lambda A+(1-\lambda) B)x\)
\(\geq 0\)
\(\geq 0\)
\(\geq 0\)
Useful fact: \(\Omega_{1}\) convex, \(\Omega_{2}\) convex \(\Rightarrow \Omega_{1} \cap \Omega_{2}\) convex.
- Proof in last lecture.
- The union of convex sets may not be convex
- Polyhedrons: \(\{x \mid A x \leq b\}\) are convex sets
e.g. m =4 (4 halfspaces), n=2 (2d)
\(A=\left[\begin{array}{rr}-1 & 0 \\ 0 & -1 \\ 0 & 1 \\ 1 & 1\end{array}\right], b=\left[\begin{array}{l}0 \\ 0 \\ 1 \\ 3\end{array}\right]\)
Def: A function \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is convex if its domain is a convex set and
\(f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)\) \(\forall x,y \in \text{domain}(f), \forall \lambda\in[0,1]\)
Simple examples
Convex functions
Non-convex functions
\(f\) is called a concave function if \(-f\) is convex
- All affine functions \(f(x)=a^{T} x+b \quad\) (for any \(\left.a \in \mathbb{R}^{n}, b \in \mathbb{R}\right)\)
- Some quadratic functions (details later)
- All norms
Recall norm is a function \(f\) that satisfies:
a. \(f(\alpha x)=|\alpha| f(x), \forall \alpha \in \mathbb{R}\)
b. \(f(x+y) \leq f(x)+f(y)\)
c. \(f(x) \geq 0, \forall x, f(x)=0 \Rightarrow x=0\)
\(f(\lambda x+(1-\lambda) y) \)
\(= \lambda f(x)+(1-\lambda) f(y)\)
b.c. (b)
b.c. (a) and \(\lambda \geq 0\)
Proof (convexity of any norm):
\(\leqslant f(\lambda x)+f((1-\lambda) y)\)
Theorem: If a function \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is convex, then all its sublevelsets are convex sets.
Remarks:
- Recall the sub-levelsets definition:
The \(\alpha\) -sublevel set of a function \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is the set \(S_{\alpha}=\{x \in \operatorname{domain}(f) \mid f(x) \leq \alpha\}\)
Theorem: If a function \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is convex, then all its sublevelsets are convex sets.
Remarks:
- Recall the sub-levelsets definition:
The \(\alpha\) -sublevel set of a function \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is the set \(S_{\alpha}=\{x \in \operatorname{domain}(f) \mid f(x) \leq \alpha\}\)
- Instead, a function whose sub-levelsets are convex sets is called quasiconvex.
- Converse is not true:
Theorem: Suppose\(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is twice differentiable over its domain. Then, the following are equivalent:
(i) \(f\) is convex.
(ii) \(f(y) \geq f(x)+\nabla f^{T}(x)(y-x), \quad \forall x, y \in \operatorname{dom}(f)\)
(iii) \(\nabla^{2} f(x) \succeq 0, \quad \forall x \in \operatorname{dom}(f)\) (i.e., the Hessian is psd \(\left.\forall x \in \operatorname{dom}(f)\right)\).
Intuition for (ii) \(f(y) \geq f(x)+\nabla f^{T}(x)(y-x), \quad \forall x, y \in \operatorname{dom}(f)\)
Intuition for (iii) \(\nabla^{2} f(x) \succeq 0, \quad \forall x \in \operatorname{dom}(f)\) (i.e., the Hessian is psd \(\left.\forall x \in \operatorname{dom}(f)\right)\).
Function has nonnegative curvature everywhere: "it curves up"
Recall we said, some quadratic functions are convex, this second order characterization makes identifying the "some" really easy
In one dimension: \(f^{\prime \prime}(x) \geq 0, \forall x \in \operatorname{dom}(f)\)
Recall theorem says \(f\) convex \(\Leftrightarrow \nabla^{2} f(x) \succeq 0, \quad \forall x \in \operatorname{dom}(f)\)
Consider a quadratic function \(f(x)=x^{T} A x+b x+c \quad\) (A symmetric)
when is \(f\) convex?
we know \(\nabla^{2} f(x)=2 A\)
Text
\(f\) convex \(\Leftrightarrow A \succeq 0\) (independent of \(b, c)\)
(for convex functions)
If \(f_{1}, \ldots, f_{n}\) are convex functions and \(\omega_{1}, \ldots, \omega_{n} \geq 0,\) then
\(f(x)=\omega_{1} f_{1}(x)+\cdots+\omega_{n} f_{n}(x)\) is also convex.
If \(f_{1}, \ldots, f_{m}\) are convex functions then their pointwise maximum
\(f(x)=\max \left\{f_{1}(x), f_{2}(x), \ldots f_{m}(x)\right\}\)
with \(\operatorname{dom}(f)=\operatorname{dom}\left(f_{1}\right) \cap \operatorname{dom}\left(f_{2}\right) \cap \cdots \cap \operatorname{dom}\left(f_{m}\right),\) is also convex.
Suppose \(f: \mathbb{R}^{\mathrm{n}} \rightarrow \mathbb{R}, \mathrm{A} \in \mathbb{R}^{\mathrm{n} \times \mathrm{m}}\) and \(b \in \mathbb{R}^{\mathrm{n}}\). Define \(g: \mathbb{R}^{\mathrm{m}} \rightarrow \mathbb{R}\) by
\(g(x)=f(A x+b)\)
with \(\operatorname{dom}(g)=\{x \mid A x+b \in \operatorname{dom}(f)\}\). Then, if \(f\) is convex, so is \(g\); if \(f\) is concave so is \(g\).
Let \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) be a convex function and fix some \(a, b, \in \mathbb{R}^{n} .\) Then, the function \(g: \mathbb{R} \rightarrow \mathbb{R}\) given by \(g(x)=f(ax+b)\) is convex.