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I need to prove that the following set is convex:

$$ \{(x,y):x^2 +y^2 \lt 2\} $$

Obviously, this an open disk of radius $\sqrt2$.

My intuition is to use triangle inequality for this proof because a similar example was done in class. However, I've never used triangle inequality or proved that a set is convex, so if someone could help me out with that, I'd greatly appreciate it. Thanks!

thisisourconcerndude
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  • Take a look here: http://math.stackexchange.com/questions/251414/proof-that-this-set-is-convex – Frnnd Feb 13 '15 at 22:15

2 Answers2

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You need to show that if $a\in S=\{(x,y):x^2+y^2<2\}$ and $b\in S$ then $\lambda a + (1-\lambda b)\in S$ for any $0\leq \lambda \leq 1$. Use what you know about being in $a$ and $b$ being in $S$ to show that $\lambda a + (1-\lambda)b$ is in $S$.

The triangle inequality is that $\Vert a + b\Vert\leq \Vert a\Vert + \Vert b\Vert$. What you know is that $\Vert a\Vert < 2$ and $\Vert b\Vert < 2$. Apply the triangle inequality to $\lambda a + (1-\lambda)b$.

TravisJ
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Your set is equal to any element with norm less than $\sqrt{2}$, then take $a$ and $b$ with that property:

$$||\lambda a + (1-\lambda)b||\leq \lambda||a||+(1-\lambda)||b||< \lambda \sqrt{2}+(1-\lambda) \sqrt{2}=\sqrt{2} $$

rlartiga
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    Well, there it is. Great! Thank you very much. – thisisourconcerndude Feb 13 '15 at 22:38
  • You mean the set contains any element whose 2-norm (or euclidean norm) is smaller than sqrt(2). The 2-norm squared is also a norm, in that case it is the set of numbers where that norm is smaller than 2. – TravisJ Feb 16 '15 at 13:46