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I am having trouble determining if the following set is convex

$$\ \{x \in R^{n}: \sum_{i=0}^n i\sin(x_i) \le 1 \}$$

My intuition tells me that it is not convex, because of the periodicity of trig functions, but on the other hand, a sin(x) is less than 1 on all periods due to that being the largest value it ever takes on. How should I go about determining the convexity of a set that is defined by trig functions?

Steven White
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  • I think it shouldn't be too hard to find two points in this set, such that their average is not in the set. Try something like (0,0,...,0,0) and (0,0,...,0,pi). – TMM Apr 16 '17 at 01:16
  • I am just starting out learning convexity analysis and this is an example of an answer that is frustratingly obvious after it is pointed out, so I must be going about this in the wrong way. In future problems, how should I know when to just find an example that shows the set is not convex, as opposed to trying to prove it with the definition of convexity? – Steven White Apr 16 '17 at 01:25
  • What I did was I started with $n = 2$ and I used the function $\sin x + 2 \sin y$ because the term with $0$ can be ignored. If you plot this (through Desmos for example) you will see a whole bunch of holes. So the set isn't convex. Now think about how to prove this. First, I saw that $(0,\dots,0)$ was in the set because $\sin(0) = 0$. From there I thought about how can I have $\sin(x) = 0$ and $\sin(x/2) = 1$ so that $(x,\dots,x)$ is in the set but $(x/2,\dots,x/2)$ is not. There was an obvious choice here. – Trevor Gunn Apr 18 '17 at 03:57

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Note that $(0,0,\dots,0)$ and $(\pi,\pi,\dots,\pi)$ are both in the set but their midpoint is not.

Trevor Gunn
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