There is an example for proof to convexity of a unit circle in Professor Arora's "Introduction to optimum design".
Show the convexity of the set
$$
S = \left\{\mathbf{x} \ \middle|\ x_1^2 + x_2^2 -1.0 \leq 0 \right\}
$$
I know a proof to use triangular inequality to prove the convexity of the above set.
But, the book suggests the following:
$$\mathbf{x} = \alpha \mathbf{x}^{(2)} + (1-\alpha)\mathbf{x}^{(1)} ; \quad 0 \leq \alpha \leq 1 \tag{4.71}$$
We can also use Eq. (4.71) to show convexity of $S$. To do this, take any two points $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$ in the set $S$. Use of Eq. (4.71) to calculate $\mathbf{x}$ and the condition that the distance between $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$ is non-negative (that is, $\lVert \mathbf{x}^{(1)}- \mathbf{x}^{(2)} \rVert \geq 0$) will show $\mathbf{x} \in S$.
I am trying the manner suggested in the book rather than using triangular inequality. But I can not find a solution.
How can $\lVert \mathbf{x}^{(1)}- \mathbf{x}^{(2)} \rVert \geq 0$ be applied to this problem?
Best Answer
Consider two points $(x_1,y_1)$ and $(x_2,y_2)$ that belong to the set $S$. Then we must prove: $$(tx_1+(1-t)x_2)^2+(ty_1+(1-t)y_2)^2\le 1.$$ We expand the brackets and regroup: $$t^2(x_1^2+y_1^2)+2t(1-t)(x_1x_2+y_1y_2)+(1-t)^2(x_2^2+y_2^2)\le t^2+2t(1-t)+(1-t)^2=1,$$ because: $$x_1^2+y_1^2\le 1; x_2^2+y_2^2\le 1;$$ $$x_1x_2+y_1y_2\le \frac{x_1^2+x_2^2}{2}+\frac{y_1^2+y_2^2}{2}=\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)}{2}\le \frac{1+1}{2}=1.$$