Jtdylktuy

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:

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

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

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.

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
$$3(a-3)^2 + 2(b-3)^2 = lambda^2(3(a_1-3)^2+2(b_1-3)^2) + (1-lambda)^2(3(a_2-3)^2+2(b_2-3)^2) \+ 2lambda(1-lambda)(3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3)).$$
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
$$3(a-3)^2 + 2(b-3)^2 le lambda^2 + (1-lambda)^2+ 2lambda(1-lambda)(color{red}{3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3)}).$$
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,
$$3(a-3)^2 + 2(b-3)^2 le lambda^2 + (1-lambda)^2+ 2lambda(1-lambda) =1.$$
Q.E.D.

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 $0leq lambda leq 1$. 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
$$
3((lambda-lambda^2)x_1^2 + (mu - mu^2)x_2^2 - 2lambdamu x_1x_2) = 3lambdamu(x_1-x_2)^2 geq 0.
$$

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

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

Suppose $P,Qin B$ with $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. We need to show $tP+(1-t)Qin B$ for all $tin[0,1]$.
$$tP+(1-t)Q=Big(tx_1+(1-t)x_2,ty_1+(1-t)y_2Big)$$
$$Big(frac{tx_1+(1-t)x_2}{a}Big)^2+Big(frac{ty_1+(1-t)y_2}{b}Big)^2=t^2Big((frac{x_1}{a})^2+(frac{y_1}{b})^2Big)+(1-t)^2Big((frac{x_2}{a})^2+(frac{y_2}{b})^2Big)+2t(1-t)Big(frac{x_1x_2}{a^2}+frac{y_1y_2}{b^2}Big)$$
$$leq t^2+(1-t)^2+2t(1-t)Big(frac{x_1x_2}{a^2}+frac{y_1y_2}{b^2}Big).$$
Using parametrization $x_1=arcos t, y_1=brsin t$ and $x_1=ar'cos t', y_1=br'sin t'$ with $0leq r,r'leq 1$ we have
$$frac{x_1x_2}{a^2}+frac{y_1y_2}{b^2}=frac{a^2rr'cos tcos t'}{a^2}+frac{b^2rr'sin tsin t'}{b^2}=rr'(cos tcos t'+sin tsin t')=rr'cos(t-t')leq rr'leq1.$$
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 ${(x,y): (frac{x-x_0}{a})^2+(frac{y-y_0}{b})^2leq 1}$ is also convex.