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From the definition of convexity:

Set $S$ is convex if $x, y \in S$ and the line segment $\theta x + (1-\theta) y$ is also in the set.

How would you find the convexity of something like the following?

$$S = \{(P_{mn}, Q_{mn}, v_m, l_{mn}):P^2_{mn}+Q^2_{mn}=v_ml_{mn}\}$$

where $v_m > 0$ and $l_{mn} \geq 0$. How would you apply the definition of convexity to this?

Rodrigo de Azevedo
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jian
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1 Answers1

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If $(P_{mn}, Q_{mn}, v_m, l_{mn})$ and $(P'_{mn}, Q_{mn}', v'_m, l'_m)$ are in $S$. Then for an arbitrary $\lambda\in (0,1)$, we need to check whether $$(\lambda P_{mn}+(1-\lambda)P'_{mn},\lambda Q_{mn}+(1-\lambda)Q'_{mn}, \lambda v_{m}+(1-\lambda)v'_{m} , \lambda l_{mn}+(1-\lambda)l'_{mn})\in S,$$ or more explicitly, whether $$[\lambda P_{mn}+(1-\lambda)P'_{mn}]^2+[\lambda Q_{mn}+(1-\lambda)Q'_{mn}]^2=[\lambda v_{m}+(1-\lambda)v'_{m}][ \lambda l_{mn}+(1-\lambda)l'_{mn}],$$ or equivalently, whether \begin{equation}2(P_{mn}P'_{mn}+Q_{mn}Q'_{mn})=v_ml'_{mn}+v'_ml_{mn}.\,\,\,\,\,\,\,\,(*)\end{equation} Then it is easy to see that $S$ is not convex, as for any $v_m,l_{mn}>0$, we have $$(\sqrt{v_ml_{mn}}, 0, v_m, l_{mn}), (0,\sqrt{v_ml_{mn}}, v_m, l_{mn})\in S.$$ But plug these two points back to $(*)$, it is easy to see that the LHS $=0$ but the RHS is strictly positive.

OnoL
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  • Would you mind going over how equation 2 is equivalent to (*). I tried factoring it and the best I could come up with is this. $$2\lambda(1-\lambda)P_{mn}P'{mn} + 2\lambda(1-\lambda)Q{mn}Q'{mn} + \lambda^2 P{mn}^2 + (1-\lambda)^2P_{mn}'^2 + \lambda^2Q_{mn}^2+(1-\lambda)^2Q_{mn}'^2 = \lambda(1-\lambda)l_{mn}v_m' + \lambda(1-\lambda)v_ml_{mn}' + 2\lambda^2v_ml_{mn}+(1-\lambda)v_m'l_{mn}'$$ – jian Mar 11 '18 at 00:39
  • I'm not sure how to get rid of the last few terms so I can just divide everything by $\lambda(1-\lambda)$ – jian Mar 11 '18 at 00:43
  • @dinoblue Well, we first notice that by condition $\lambda^2P^2_{mn}+\lambda^2 Q_{mn}^2=\lambda^2 v_ml_{mn}$ and $(1-\lambda)^2P'^2_{mn}+(1-\lambda)^2 Q'^2_{mn}=(1-\lambda)^2 v_m'l_{mn}'$. Cancel those terms and then you can divide by $\lambda(1-\lambda)$ on both sides. – OnoL Mar 11 '18 at 01:33
  • @dinoblue By the way, there is a typo in your algebra on the right hand side: there should be no $2$ before $\lambda^2 v_ml_{mn}$. – OnoL Mar 11 '18 at 01:35