ECE1505H Convex Optimization. Lecture 4: Sets and convexity. Taught by Prof. Stark Draper

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, covering [1] content.

### Today

• more on various sets: hyperplanes, half-spaces, polyhedra, balls, ellipses, norm balls, cone of PSD
• generalize inequalities
• operations that preserve convexity
• separating and supporting hyperplanes.

## Hyperplanes

Find some $$\Bx_0 \in \mathbb{R}^n$$ such that $$\Ba^\T \Bx_0 = \Bb$$, so

\label{eqn:convexOptimizationLecture4:20}
\begin{aligned}
\setlr{ \Bx | \Ba^\T \Bx = \Bb }
&=
\setlr{ \Bx | \Ba^\T \Bx = \Ba^\T \Bx_0 } \\
&=
\setlr{ \Bx | \Ba^\T (\Bx – \Bx_0) } \\
&=
\Bx_0 + \Ba^\perp,
\end{aligned}

where

\label{eqn:convexOptimizationLecture4:40}
\Ba^\perp = \setlr{ \Bv | \Ba^\T \Bv = 0 }.

fig. 1. Parallel hyperplanes.

Recall

\label{eqn:convexOptimizationLecture4:60}
\Norm{\Bz}_\conj = \sup_\Bx \setlr{ \Bz^\T \Bx | \Norm{\Bx} \le 1 }

Denote the optimizer of above as $$\Bx^\conj$$. By definition

\label{eqn:convexOptimizationLecture4:80}
\Bz^\T \Bx^\conj \ge \Bz^\T \Bx \quad \forall \Bx, \Norm{\Bx} \le 1

This defines a half space in which the unit ball

\label{eqn:convexOptimizationLecture4:100}
\setlr{ \Bx | \Bz^\T (\Bx – \Bx^\conj \le 0 }

Start with the $$l_1$$ norm, duals of $$l_1$$ is $$l_\infty$$

fig. 2. Half space containing unit ball.

Similar pic for $$l_\infty$$, for which the dual is the $$l_1$$ norm, as sketched in fig. 3.  Here the optimizer point is at $$(1,1)$$

fig. 3. Half space containing the unit ball for l_infinity

and a similar pic for $$l_2$$, which is sketched in fig. 4.

fig. 4. Half space containing for l_2 unit ball.

## Polyhedra

\label{eqn:convexOptimizationLecture4:120}
\begin{aligned}
\mathcal{P}
&= \setlr{ \Bx |
\Ba_j^\T \Bx \le \Bb_j, j \in [1,m],
\Bc_i^\T \Bx = \Bd_i, i \in [1,p]
} \\
&=
\setlr{ \Bx | A \Bx \le \Bb, C \Bx = d },
\end{aligned}

where the final inequality and equality are component wise.

Proving $$\mathcal{P}$$ is convex:

• Pick $$\Bx_1 \in \mathcal{P}$$, $$\Bx_2 \in \mathcal{P}$$
• Pick any $$\theta \in [0,1]$$
• Test $$\theta \Bx_1 + (1-\theta) \Bx_2$$. Is it in $$\mathcal{P}$$?

\label{eqn:convexOptimizationLecture4:140}
\begin{aligned}
A \lr{ \theta \Bx_1 + (1-\theta) \Bx_2 }
&=
\theta A \Bx_1 + (1-\theta) A \Bx_2 \\
&\le
\theta \Bb + (1-\theta) \Bb \\
&=
\Bb.
\end{aligned}

## Balls

Euclidean ball for $$\Bx_c \in \mathbb{R}^n, r \in \mathbb{R}$$

\label{eqn:convexOptimizationLecture4:160}
\mathcal{B}(\Bx_c, r)
= \setlr{ \Bx | \Norm{\Bx – \Bx_c}_2 \le r },

or
\label{eqn:convexOptimizationLecture4:180}
\mathcal{B}(\Bx_c, r)
= \setlr{ \Bx | \lr{\Bx – \Bx_c}^\T \lr{\Bx – \Bx_c} \le r^2 }.

Let $$\Bx_1, \Bx_2$$, $$\theta \in [0,1]$$

\label{eqn:convexOptimizationLecture4:200}
\begin{aligned}
\Norm{ \theta \Bx_1 + (1-\theta) \Bx_2 – \Bx_c }_2
&=
\Norm{ \theta (\Bx_1 – \Bx_c) + (1-\theta) (\Bx_2 – \Bx_c) }_2 \\
&\le
\Norm{ \theta (\Bx_1 – \Bx_c)}_2 + \Norm{(1-\theta) (\Bx_2 – \Bx_c) }_2 \\
&=
\Abs{\theta} \Norm{ \Bx_1 – \Bx_c}_2 + \Abs{1 -\theta} \Norm{ \Bx_2 – \Bx_c }_2 \\
&=
\theta \Norm{ \Bx_1 – \Bx_c}_2 + \lr{1 -\theta} \Norm{ \Bx_2 – \Bx_c }_2 \\
&\le
\theta r + (1 – \theta) r \\
&= r
\end{aligned}

## Ellipse

\label{eqn:convexOptimizationLecture4:220}
\mathcal{E}(\Bx_c, P)
=
\setlr{ \Bx | (\Bx – \Bx_c)^\T P^{-1} (\Bx – \Bx_c) \le 1 },

where $$P \in S^n_{++}$$.

• Euclidean ball is an ellipse with $$P = I r^2$$
• Ellipse is image of Euclidean ball $$\mathcal{B}(0,1)$$ under affine mapping.

fig. 5. Circle and ellipse.

Given

\label{eqn:convexOptimizationLecture4:240}
F(\Bu) = P^{1/2} \Bu + \Bx_c

\label{eqn:convexOptimizationLecture4:260}
\begin{aligned}
\setlr{ F(\Bu) | \Norm{\Bu}_2 \le r }
&=
\setlr{ P^{1/2} \Bu + \Bx_c | \Bu^\T \Bu \le r^2 } \\
&=
\setlr{ \Bx | \Bx = P^{1/2} \Bu + \Bx_c, \Bu^\T \Bu \le r^2 } \\
&=
\setlr{ \Bx | \Bu = P^{-1/2} (\Bx – \Bx_c), \Bu^\T \Bu \le r^2 } \\
&=
\setlr{ \Bx | (\Bx – \Bx_c)^\T P^{-1} (\Bx – \Bx_c) \le r^2 }
\end{aligned}

## Geometry of an ellipse

Decomposition of positive definite matrix $$P \in S^n_{++} \subset S^n$$ is:

\label{eqn:convexOptimizationLecture4:280}
\begin{aligned}
P &= Q \textrm{diag}(\lambda_i) Q^\T \\
Q^\T Q &= 1
\end{aligned},

where $$\lambda_i \in \mathbb{R}$$, and $$\lambda_i > 0$$. The ellipse is defined by

\label{eqn:convexOptimizationLecture4:300}
(\Bx – \Bx_c)^\T Q \textrm{diag}(1/\lambda_i) (\Bx – \Bx_c) Q \le r^2

The term $$(\Bx – \Bx_c)^\T Q$$ projects $$\Bx – \Bx_c$$ onto the columns of $$Q$$. Those columns are perpendicular since $$Q$$ is an orthogonal matrix. Let

\label{eqn:convexOptimizationLecture4:320}
\tilde{\Bx} = Q^\T (\Bx – \Bx_c),

this shifts the origin around $$\Bx_c$$ and $$Q$$ rotates into a new coordinate system. The ellipse is therefore

\label{eqn:convexOptimizationLecture4:340}
\tilde{\Bx}^\T
\begin{bmatrix}
\inv{\lambda_1} & & & \\
&\inv{\lambda_2} & & \\
& \ddots & \\
& & & \inv{\lambda_n}
\end{bmatrix}
\tilde{\Bx}
=
\sum_{i = 1}^n \frac{\tilde{x}_i^2}{\lambda_i} \le 1.

An example is sketched for $$\lambda_1 > \lambda_2$$ below.

Ellipse with $$\lambda_1 > \lambda_2$$.

• $$\lambda_i$$ tells us length of the semi-major axis.
• Larger $$\lambda_i$$ means $$\tilde{x}_i^2$$ can be bigger and still satisfy constraint $$\le 1$$.
• Volume of ellipse if proportional to $$\sqrt{ \det P } = \sqrt{ \prod_{i = 1}^n \lambda_i }$$.
• When any $$\lambda_i \rightarrow 0$$ a dimension is lost and the volume goes to zero. That removes the invertibility required.

Ellipses will be seen a lot in this course, since we are interested in “bowl” like geometries (and the ellipse is the image of a Euclidean ball).

## Norm ball.

The norm ball

\label{eqn:convexOptimizationLecture4:360}
\mathcal{B} = \setlr{ \Bx | \Norm{\Bx} \le 1 },

is a convex set for all norms. Proof:

Take any $$\Bx, \By \in \mathcal{B}$$

\label{eqn:convexOptimizationLecture4:380}
\Norm{ \theta \Bx + (1 – \theta) \By }
\le
\Abs{\theta} \Norm{ \Bx } + \Abs{1 – \theta} \Norm{ \By }
=
\theta \Norm{ \Bx } + \lr{1 – \theta} \Norm{ \By }
\lr
\theta + \lr{1 – \theta}
=
1.

This is true for any p-norm $$1 \le p$$, $$\Norm{\Bx}_p = \lr{ \sum_{i = 1}^n \Abs{x_i}^p }^{1/p}$$.

Norm ball.

The shape of a $$p < 1$$ norm unit ball is sketched below (lines connecting points in such a region can exit the region).

## Cones

Recall that $$C$$ is a cone if $$\forall \Bx \in C, \theta \ge 0, \theta \Bx \in C$$.

Impt cone of PSD matrices

\label{eqn:convexOptimizationLecture4:400}
\begin{aligned}
S^n &= \setlr{ X \in \mathbb{R}^{n \times n} | X = X^\T } \\
S^n_{+} &= \setlr{ X \in S^n | \Bv^\T X \Bv \ge 0, \quad \forall v \in \mathbb{R}^n } \\
S^n_{++} &= \setlr{ X \in S^n_{+} | \Bv^\T X \Bv > 0, \quad \forall v \in \mathbb{R}^n } \\
\end{aligned}

These have respectively

• $$\lambda_i \in \mathbb{R}$$
• $$\lambda_i \in \mathbb{R}_{+}$$
• $$\lambda_i \in \mathbb{R}_{++}$$

$$S^n_{+}$$ is a cone if:

$$X \in S^n_{+}$$, then $$\theta X \in S^n_{+}, \quad \forall \theta \ge 0$$

\label{eqn:convexOptimizationLecture4:420}
\Bv^\T (\theta X) \Bv
= \theta \Bv^\T \Bv
\ge 0,

since $$\theta \ge 0$$ and because $$X \in S^n_{+}$$.

Shorthand:

\label{eqn:convexOptimizationLecture4:440}
\begin{aligned}
X &\in S^n_{+} \Rightarrow X \succeq 0
X &\in S^n_{++} \Rightarrow X \succ 0.
\end{aligned}

Further $$S^n_{+}$$ is a convex cone.

Let $$A \in S^n_{+}$$, $$B \in S^n_{+}$$, $$\theta_1, \theta_2 \ge 0, \theta_1 + \theta_2 = 1$$, or $$\theta_2 = 1 – \theta_1$$.

Show that $$\theta_1 A + \theta_2 B \in S^n_{+}$$ :

\label{eqn:convexOptimizationLecture4:460}
\Bv^\T \lr{ \theta_1 A + \theta_2 B } \Bv
=
\theta_1 \Bv^\T A \Bv
+\theta_2 \Bv^\T B \Bv
\ge 0,

since $$\theta_1 \ge 0, \theta_2 \ge 0, \Bv^\T A \Bv \ge 0, \Bv^\T B \Bv \ge 0$$.

fig. 8. Cone.

Inequalities:

Start with a proper cone $$K \subseteq \mathbb{R}^n$$

• closed, convex
• non-empty interior (“solid”)
• “pointed” (contains no lines)

The $$K$$ defines a generalized inequality in \R{n} defined as “$$\le_K$$”

Interpreting

\label{eqn:convexOptimizationLecture4:480}
\begin{aligned}
\Bx \le_K \By &\leftrightarrow \By – \Bx \in K
\Bx \end{aligned}

Why pointed? Want if $$\Bx \le_K \By$$ and $$\By \le_K \Bx$$ with this $$K$$ is a half space.

Example:1: $$K = \mathbb{R}^n_{+}, \Bx \in \mathbb{R}^n, \By \in \mathbb{R}^n$$

fig. 12. K is non-negative “orthant”

\label{eqn:convexOptimizationLecture4:500}
\Bx \le_K \By \Rightarrow \By – \Bx \in K

say:

\label{eqn:convexOptimizationLecture4:520}
\begin{bmatrix}
y_1 – x_1
y_2 – x_2
\end{bmatrix}
\in R^2_{+}

Also:

\label{eqn:convexOptimizationLecture4:540}
K = R^1_{+}

(pointed, since it contains no rays)

\label{eqn:convexOptimizationLecture4:560}
\Bx \le_K \By ,

with respect to $$K = \mathbb{R}^n_{+}$$ means that $$x_i \le y_i$$ for all $$i \in [1,n]$$.

Example:2: For $$K = PSD \subseteq S^n$$,

\label{eqn:convexOptimizationLecture4:580}
\Bx \le_K \By ,

means that

\label{eqn:convexOptimizationLecture4:600}
\By – \Bx \in K = S^n_{+}.

• Difference $$\By – \Bx$$ is always in $$S$$
• check if in $$K$$ by checking if all eigenvalues $$\ge 0$$.
• $$S^n_{++}$$ is the interior of $$S^n_{+}$$.

Interpretation:

\label{eqn:convexOptimizationLecture4:620}
\begin{aligned}
\Bx \le_K \By &\leftrightarrow \By – \Bx \in K \\
\Bx \end{aligned}

We’ll use these with vectors and matrices so often the $$K$$ subscript will often be dropped, writing instead (for vectors)

\label{eqn:convexOptimizationLecture4:640}
\begin{aligned}
\Bx \le \By &\leftrightarrow \By – \Bx \in \mathbb{R}^n_{+} \\
\Bx < \By &\leftrightarrow \By – \Bx \in \textrm{int} \mathbb{R}^n_{++}
\end{aligned}

and for matrices

\label{eqn:convexOptimizationLecture4:660}
\begin{aligned}
\Bx \le \By &\leftrightarrow \By – \Bx \in S^n_{+} \\
\Bx < \By &\leftrightarrow \By – \Bx \in \textrm{int} S^n_{++}.
\end{aligned}

## Intersection

Take the intersection of (perhaps infinitely many) sets $$S_\alpha$$:

If $$S_\alpha$$ is (affine,convex, conic) for all $$\alpha \in A$$ then

\label{eqn:convexOptimizationLecture4:680}
\cap_\alpha S_\alpha

is (affine,convex, conic). To prove in homework:

\label{eqn:convexOptimizationLecture4:700}
\mathcal{P} = \setlr{ \Bx | \Ba_i^\T \Bx \le \Bb_i, \Bc_j^\T \Bx = \Bd_j, \quad \forall i \cdots j }

This is convex since the intersection of a bunch of hyperplane and half space constraints.

1. If $$S \subseteq \mathbb{R}^n$$ is convex then\label{eqn:convexOptimizationLecture4:720}
F(S) = \setlr{ F(\Bx) | \Bx \in S }
is convex.
2. If $$S \subseteq \mathbb{R}^m$$ then\label{eqn:convexOptimizationLecture4:740}
F^{-1}(S) = \setlr{ \Bx | F(\Bx) \in S }
is convex. Such a mapping is sketched in fig. 14.

fig. 14. Mapping functions of sets.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.