# How to find the explicit support function of convex set?

Let $S=A\cup B$ where $A=\{(x_1,x_2):x_1<0,x_1^2+x_2^2\le4\}$ and $B=\{(x_1,x_2):x_1\ge0,-2\le x_2 \le2\}.$ Find the support function.

The support function is defined as $f(y)=\text{sup} \{y^tx:x\in S\},$ where $S$ is a convex and bounded set in $\mathbb R^n$ and $f:\mathbb R^n\to\mathbb R.$

From here, $$f(y)=\text{sup} \{y^t(x_1,x_2):(x_1,x_2)\in A\cup B\}=\text{sup} \{(y_1x_1+x_2y_2):(x_1,x_2)\in A\cup B\}$$

What do I do from here?

Should I consider cases for $(x_1,x_2)?$ i.e. if $(x_1,x_2)\in A$ or in B or in both?

• "S" is not convex in your question ! does that matter at all? – Red shoes Dec 31 '17 at 2:15
• @Redshoes Sorry I forgot it was a circle. – user441848 Dec 31 '17 at 2:16
• By $\sup$ do you mean least upper bound? I'm confused because of the word "support". – Shine On You Crazy Diamond Dec 31 '17 at 3:47
• @EnjoysMath yes, sup means the least upper bound – user441848 Dec 31 '17 at 3:55
• Note that even if $S$ is not convex, the support function is convex, being the supremum of an (potentially infinite) set of affine functions. – Michael Grant Jan 2 '18 at 17:27

The dot product $y^t x$ equals $\cos \theta |y||x|$ where $\theta$ is the smaller angle between the two vectors. So to maximize $y^t x$ you choose a vector $x$ that is parallel to $y$ since then $\cos (0) = 1 =$ the maximum value $\cos$ can take.
Thus the $x$ maximizing the orange $y$ vector pictured is always going to be on the boundary of $S$. Thus $f(y)$ is simply the magnitude $|y| \equiv \sqrt{y^t y}$ times the distance from the origin to where the boundary of $S$ intersects the ray along $y$.
If you have to compute the explict description of it, you can find where the ray $r(t) = y t$ intersects each line in the square and similarly where it intersects the circle.
• That said, thanks to your diagram, we can determine the expression by inspection: $$f(y) = \begin{cases} +\infty & y_1 > 0 \\ 2 \|y\|_2 & y_1 \leq 0 \end{cases}$$ This follows directly from the fact that, on the right side of the plot, we can choose $x_1$ arbitrarily large, so $x^T y$ is unbounded above. And on the left side, $\|x\|_2=2$ on the boundary, and $x^T y \leq \|y\|_2 \|x\|_2 \leq 2\|y\|_2$. – Michael Grant Jan 2 '18 at 17:37
• Why $x^T y \leq \|y\|_2 \|x\|_2 \leq 2\|y\|_2$? – user441848 Jan 16 '18 at 7:05
• The first inequality is simply the Cauchy Schwartz inequality. (If you don't know it, look it up; you need it to do just about anything useful in convex analysis). The second inequality follows from the fact that $\|x\|_2\leq 2$, given that the left-half side of the region is a circle of radius 2. – Michael Grant Jan 17 '18 at 5:09