# Prove a set is convex algebraically

• Jan 30th 2011, 11:32 AM
MSUMathStdnt
Prove a set is convex algebraically
How can I determine if a set is convex without graphing it?

I have that a set, $S$, is convex if, given any two points, $x_1, x_2 \in S$, the line connecting them is entirely within $S$, or:

$\lambda x_1 + (1-\lambda) x_2 \in S \qquad \forall \lambda \in [0,1]$

Some of the sets in question:
b. $\{(x,y,z) \colon z=\left| y \right|, x \le 3 \}$
c. $\{(x,y): y-3x^2=0\}$
d. $\{(x,y,z): y \ge x^2, x+2y+z \le 4\}$
e. $\{(x,y):x=3, \left| y \right|, \le 4 \}$
f. $\{(x,y,z): z= \left| y \right|, x \le 3 \}$

I have to hand in f. I'm not good enough to draw it, but I can envision it. It looks like a "V" going from $x=3$ down to $-\infty$. Clearly, it is not convex. But how would I show that mathematically?

Thanks.
• Jan 30th 2011, 12:20 PM
FernandoRevilla
Quote:

Originally Posted by MSUMathStdnt
Some of the sets in question (I can't make the braces in math mode?):

If you write \{a\}, you'll obtain $\{a\}$ .

Quote:

b. { $(x,y,z) \colon z=\left| y \right|, x \le 3$}

It is not convex. Choose:

$x_1=(0,-1,1),\;x_2=(0,1,1),\;\lambda=1/2$.

Fernando Revilla