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$$S_{1}=\{ (x_{1},x_{2}):x_{1}^2+x_{2}^2\leq 1 \}$$

I cant solve trying with an alpha value $\in[0,1]$ by two dummy vectors. Please can anyone tell me how to prove it is not convex set ?

what I tried is

Defined two vectors $$a=(a_{1},a_{2})\in S_{1}$$ so $$a_{1}^2+a_{2}^2\leq 1$$, $$b=(b_{1},b_{2})\in S_{1}$$ so $$b_{1}^2+b_{2}^2\leq1$$,

and tried to prove $$\alpha\begin{bmatrix} a_{1}\\a_{2} \end{bmatrix}+(1-\alpha)\begin{bmatrix} b_{1}\\b_{2} \end{bmatrix}\in S_{1} ? ?$$

it was so hard to me continuing this

Rodrigo de Azevedo
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Samet Sökel
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2 Answers2

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Let $\overline{x},\overline{y}\in S$. We'll show that for any $\lambda\in[0,1]$, the vector $\lambda\overline{x}+(1-\lambda)\overline{y}\in S$. That is to say we need to show that $|\lambda\overline{x}+(1-\lambda)\overline{y}|\leq 1$. This follows more or less immediatley from the triangle inequality:

\begin{align*} |\lambda\overline{x} + (1-\lambda)\overline{y}| &\leq |\lambda\overline{x}| + |(1-\lambda)\overline{y}|\\ & = |\lambda||\overline{x}| + |1-\lambda||\overline{y}|\\ & \leq |\lambda| + |1-\lambda|\\ & = 1 \end{align*}

Nathan Lowry
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  • Thanks for solution, where did $\overline{x}$ and $\overline{y}$ go at the 3rd step ? I don't understand – Samet Sökel Oct 13 '20 at 21:06
  • That is where we use the assumption that $\overline{x}$ and $\overline{y}$ belong to $S$. By definition, they have norm less than or equal to 1. – Nathan Lowry Oct 13 '20 at 21:07
  • Thank you, I got it. Just one more question; does "equals 1" mean that it is impossible to be greater than 1 ? How can I be sure ? – Samet Sökel Oct 13 '20 at 21:11
  • We proved the desired inequality by looking at a string of inequalities. Some of these happen to be on-the-nose equalities, but that doesn't really matter. Since "less than or equal to" is a transitive property, we proved that $|\lambda\overline{x} + (1-\lambda)\overline{y}|\leq 1$. This means, by definition of "less than or equal to", that this quantity is no greater than 1. – Nathan Lowry Oct 13 '20 at 22:13
  • It should be noted that the norm squared does not satisfy the triangle inequality, and this proof is true because $a_1^2+a_2^2\leq 1 \Rightarrow |(a_1,a_2)|=\sqrt{a_1^2+a_2^2}\leq \sqrt{1}=1$. – Zim Oct 14 '20 at 14:29
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The Hessian of the function $\left(x_1,x_2\right) \mapsto x^2_1 + x^2_2$ is $2I$, so this function is convex. Any set of the form $g\left(x\right) \leq c$ where $g$ is a convex function is a convex set.

sven svenson
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