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Is the following set convex?

$$\{(x_1,x_2,x_3): x_3= |x_2|, x_1\leq 4\}$$

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    It contains $(0,1,1)$ and $(0,-1,1)$, but not the point $(0,0,1)$ in between – 2017-02-05
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    So because the point (0,0,1) is on the segment from (0,1,1) and (0,-1,1) you can evaluate it? With respect to lambda how could show this algebraically ? – 2017-02-05
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    A convex set $\Omega$ has the property that if points $A$ and $B$ belong to $\Omega$, then every point on the line segment connecting $A$ and $B$ also belongs to $\Omega$. Can you visualize your set and just see visually that this is not the case? – 2017-02-05
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    Honestly I could visualize the half plane and see the point but I was trying to approach the problem algebraically to look at it differently. and use lambda and 1-lambda to show it. – 2017-02-05
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    I think you should keep it simple and just visualize the set, which makes it obvious that the set is not convex. Once you have visualized the set, it's easy to pick two points such as the ones that @HagenvonEitzen selected, then you can give an algebraic proof of non-convexity. – 2017-02-05
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    As a general rule, nonlinear equality constraints are almost never convex. The exceptions are almost entirely manufactured. – 2017-02-06

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Just writing the Hagen's example. Every point of the form

$$(0,1,1)\cdot k+(1-k)(0,-1,1)=(0,2k-1,1) \text{ for $k \in [0,1]$}$$

It should be in the set. But if you take $k=1/2$ then you will get the point $(0,0,1)$ which is not in the set and so it is not convex.

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    This is exactly what I was wondering and also makes it so much more clear! – 2017-02-05
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    I am happy that you got it. – 2017-02-05
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To say a set $S$ is convex means that if $u \in S$ and $v \in S$ then every point on the line segment joining $u$ and $v$ belongs to $S$.

Visually, it's clear that the line segment joining the points $u = (0,1,1)$ and $v = (0,-1,1)$ goes outside your set. To give an algebraic proof, just select a value of $\lambda$ such that $0 < \lambda < 1$ and $\lambda u + (1 - \lambda) v$ does not belong to your set. In this case, we can just pick $\lambda = 1/2$, so that $\lambda u + (1 - \lambda) v$ is the midpoint of $u$ and $v$. In other words: $$ \lambda u + (1 - \lambda) v = (0,0,1), $$ which is a point that does not belong to your set.