0
$\begingroup$

So if I have to show the Set $S_1$ where $S_1=\{(x_1,x_2), x_2 \geq x_1^2\}$ is Convex how do I explain the soution using the Cauchy-Schwarz inequality?

\begin{align*} &\lambda x_2+(1-\lambda)x_4-[\lambda x_1+(1-\lambda)x_3]^2\\ &=... \\\ &=\lambda(1-\lambda)(x_1-x_3)^2\\ &\geq0 \end{align*}

I understand the algebra but need help explaining why this works in the form of a discriminant?

  • 0
    Do you want to know how to use Cauchy-Schwartz to get the inequalities, that you have written? – 2017-02-06

1 Answers 1

0

Indeed you can solve the problem by applying the CS inequality, but be mindful of just where you apply it...

Let $a=(x_1, y_1), b=(x_2, y_2) \in S:=\{(x,y) \in \mathbb R^2 | y \ge x^2\}$ and $ 0 \le t \le 1$. By the CS inequality, it holds that

$$2|x_1x_2| \le x_1^2 + x_2^2. $$ Now, a direct computation shows that $$ \begin{split} (tx_1 + (1-t)x_2)^2 &= t^2x_1^2 + 2t(1-t)x_1x_2 + (1-t)^2x_2^2\\ &\stackrel{\text{CS}}{\le} t^2x_1^2 + t(1-t)(x_1^2+x_2^2) + (1-t)^2x_2^2\\ &= (t^2 + t(1-t))x_1^2 + ((1-t)^2 + t(1-t))x_2^2\\ &= t(t + 1 - t)x_1^2 + (1-t)(1-t + t)x_2^2 = tx_1^2 + (1-t)x_2^2\\ &\stackrel{\text{def. of } S}{\le} ty_1 + (1-t)y_2. \end{split} $$ Thus $ta + (1-t)b \in S$, proving that $S$ is convex.