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I'm trying to prove that the following set ($B$) defined by an ellipse is convex, but I am getting stuck.

$$B = \left\{(x_1,x_2):3(x_1-3)^2+2(x_2-3)^2 \leq 1 \right\} $$

My sad attempt:

$$ \textrm{Let }x,y \: \in \: B \:, \: \lambda \: \in \: [0,1]$$

$$ \lambda x+(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\: \lambda x_2 +(1-\lambda)y_2 )$$ $$3(\lambda x_1 + (1-\lambda)y_1 -3)^2+2(\lambda x_2+(1-\lambda)y_2-3)^2=3(\lambda ^2x_1^2 +(1-\lambda)^2y_1^2+9+2\lambda x_1(1-\lambda)y_1-6\lambda x_1 -6(1-\lambda)y_1)+2(\lambda ^2x_2^2 +(1-\lambda)^2y_2^2+9+2\lambda x_2(1-\lambda)y_2-6\lambda x_2 -6(1-\lambda)y_2)$$

and this is where I have lost all confidence in proving it. I was able to factor some of the terms into the form $(\lambda x_1 -3)^2$, $((1-\lambda) y_1 -3)^2$, $(\lambda x_2 -3)^2$, and $((1-\lambda) x_2 -3)^2$, but I don't think that helps the situation any.

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We have to prove that if $(a_1,b_1) \in B$, and $(a_2,b_2) \in B$ then $$(a,b)=(\lambda a_1+(1-\lambda)a_2,\lambda b_1+(1-\lambda)b_2) \in B.$$

The trick is to observe that $a-3 = \lambda(a_1-3)+(1-\lambda)(a_2-3)$ and $b-3 = \lambda(b_1-3)+(1-\lambda)(b_2-3)$.

Consequently, $$3(a-3)^2 + 2(b-3)^2 = 3(\lambda (a_1-3)+(1-\lambda)(a_2-3))^2 + 2(\lambda (b_1-3)+(1-\lambda)(b_2-3))^2.$$ Simplifying the RHS, we get $$3(a-3)^2 + 2(b-3)^2 = \lambda^2(3(a_1-3)^2+2(b_1-3)^2) + (1-\lambda)^2(3(a_2-3)^2+2(b_2-3)^2) \\+ 2\lambda(1-\lambda)(3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3)).$$ Since $3(a_1-3)^2+2(b_1-3)^2 \le 1$ and $3(a_2-3)^2+2(b_2-3)^2 \le 1$, we get $$3(a-3)^2 + 2(b-3)^2 \le \lambda^2 + (1-\lambda)^2+ 2\lambda(1-\lambda)(\color{red}{3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3)}).$$ By Cauchy-Schwarz inequality, the red expression can be bounded as $$3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3) \le \sqrt{(3(a_1-3)^2+2(b_1-3)^2)(3(a_2-3)^2+2(b_2-3)^2)}=1.$$ Therefore, $$3(a-3)^2 + 2(b-3)^2 \le \lambda^2 + (1-\lambda)^2+ 2\lambda(1-\lambda) =1.$$ Q.E.D.

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  • $\begingroup$ Thank you! It was definitely the "trick" part that I was overlooking. $\endgroup$ – Pat Dec 2 '17 at 0:03
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By shifting it is equivalent to prove that $B' = \{3x^2 + 2y^2 \leq 1\}$ is convex.

Let $(x_1,y_1), (x_2, y_2) \in B'$ and $0\leq \lambda \leq 1$. For brevity let $\mu = 1-\lambda$.

Then we need to prove that $(\lambda x_1 + \mu x_2, \lambda y_1 + \mu y_2) \in B'$. In other words, $3(\lambda x_1 + \mu x_2)^2 + 2(\lambda y_1 + \mu y_2)^2 \leq 1$.

Claim: $3(\lambda x_1 + \mu x_2)^2 + 2(\lambda y_1 + \mu y_2)^2 \leq \lambda(3x_1^2 + 2y_1^2) + \mu(3x_2^2 + 2y_2^2)$.

Reason: Subtract the right hand side by the left hand side, we want to prove the output is $\geq 0$. The terms which involves $x_i$'s are $$ 3((\lambda-\lambda^2)x_1^2 + (\mu - \mu^2)x_2^2 - 2\lambda\mu x_1x_2) = 3\lambda\mu(x_1-x_2)^2 \geq 0. $$

And the $y$ terms follows by symmetry. This proves the claim and the convexness. (You need to use the fact $\lambda = 1-\mu$).

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Let's show that $B=\{(x,y):(\frac{x}{a})^2+(\frac{y}{b})^2\leq 1\}$ is convex.

Suppose $P,Q\in B$ with $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. We need to show $tP+(1-t)Q\in B$ for all $t\in[0,1]$. $$tP+(1-t)Q=\Big(tx_1+(1-t)x_2,ty_1+(1-t)y_2\Big)$$ $$\Big(\frac{tx_1+(1-t)x_2}{a}\Big)^2+\Big(\frac{ty_1+(1-t)y_2}{b}\Big)^2=t^2\Big((\frac{x_1}{a})^2+(\frac{y_1}{b})^2\Big)+(1-t)^2\Big((\frac{x_2}{a})^2+(\frac{y_2}{b})^2\Big)+2t(1-t)\Big(\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}\Big)$$ $$\leq t^2+(1-t)^2+2t(1-t)\Big(\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}\Big).$$ Using parametrization $x_1=ar\cos t, y_1=br\sin t$ and $x_1=ar'\cos t', y_1=br'\sin t'$ with $0\leq r,r'\leq 1$ we have $$\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}=\frac{a^2rr'\cos t\cos t'}{a^2}+\frac{b^2rr'\sin t\sin t'}{b^2}=rr'(\cos t\cos t'+\sin t\sin t')=rr'\cos(t-t')\leq rr'\leq1.$$ So $$t^2+(1-t)^2+2t(1-t)\Big(\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}\Big)\leq t^2+(1-t)^2+2t(1-t)=(t+(1-t))^2=1.$$ Note that convexity is invariant under translation so $\{(x,y): (\frac{x-x_0}{a})^2+(\frac{y-y_0}{b})^2\leq 1\}$ is also convex.

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