$S=\left\{ x\in \mathbb{R} ^n:x^{\mathsf{T}}Ax+b^{\mathsf{T}}x+c\le 0 \right\} $, and $T=\left\{ x\in \mathbb{R} ^n:g^{\mathsf{T}}x+h=0 \right\} $.

Question

Let set $S=\left\{ x\in \mathbb{R} ^n:x^{\mathsf{T}}Ax+b^{\mathsf{T}}x+c\le 0 \right\} $, and $T=\left\{ x\in \mathbb{R} ^n:g^{\mathsf{T}}x+h=0 \right\} $.

(1) if $A$ is positive semi-definite matrix, prove the set $S$ is convex.

(2) if $A+gg^{\mathsf{T}}$ is positive semi-definite matrix, prove the set $S\bigcap{T}$ is convex.

What I know is that if $\forall x^{(1)},x^{(2)} \in S$ and $\forall \lambda\in\left[0,1\right]$, we all have $\lambda x^{(1)}+(1-\lambda)x^{(2)} \in S$, then the set $S$ is named as convex set, however, I cannot employ this convex set definition to prove the above conclusion.

Answer 1

Let $\;S=\left\{x\in\mathbb{R}^n:x^{\mathsf{T}}Ax+b^{\mathsf{T}}x+c\leqslant0\right\}\;.$

Now, we will suppose that $\,A\,$ is a positive semi-definite matrix and prove the set $\,S\,$ is convex.

$\forall\,x_1,x_2\in S\;$ and $\;\forall\lambda\in\left[0,1\right]\;$, it results that

$\lambda\left(x_1^{\mathsf{T}}Ax_1+b^{\mathsf{T}}x_1+c\right)+(1-\lambda)\left(x_2^{\mathsf{T}}Ax_2+b^{\mathsf{T}}x_2+c\right)\leqslant0\;\;,$

$\lambda x_1^{\mathsf{T}}\!Ax_1+(1\!-\!\lambda)x_2^{\mathsf{T}}\!Ax_2 +b^{\mathsf{T}}\big[\lambda x_1+(1\!-\!\lambda)x_2\big]+c\leqslant0\,.\;\;\color{blue}{(1)}$

Since $\,A\,$ is a positive semi-definite matrix, it follows that

$-\lambda(1-\lambda)(x_1-x_2)^{\mathsf{T}}A(x_1-x_2)\leqslant0\;\;,$

$\left(\lambda^2\!-\!\lambda\right)\! x_1^{\mathsf{T}}\!Ax_1\!+\!\lambda(1\!-\!\lambda)\!\left(x_1^{\mathsf{T}}\!Ax_2\!+\!x_2^{\mathsf{T}}\!Ax_1\!-\!x_2^{\mathsf{T}}\!Ax_2\right)\!\leqslant0\,.\;\;\color{blue}{(2)}$

By adding $(1)$ and $(2)$, side-by-side, we get that

$\lambda x_1^{\mathsf{T}}\!A\lambda x_1\!+\!\lambda x_1^{\mathsf{T}}\!A(1\!-\!\lambda)x_2\!+\!(1\!-\!\lambda)x_2^{\mathsf{T}}\!A\lambda x_1\!+\!(1\!-\!\lambda)x_2^{\mathsf{T}}\!A(1\!-\!\lambda)x_2+b^{\mathsf{T}}\big[\lambda x_1\!+\!(1\!-\!\lambda)x_2\big]+c\leqslant0\;\;,$

$\lambda x_1^{\mathsf{T}}A\big[\lambda x_1+(1-\lambda)x_2\big]+(1-\lambda)x_2^{\mathsf{T}}A\big[\lambda x_1+(1-\lambda)x_2\big]+b^{\mathsf{T}}\big[\lambda x_1+(1-\lambda)x_2\big]+c\leqslant0\;\;,$

$\big[\lambda x_1\!+\!(1\!-\!\lambda)x_2\big]^{\!\mathsf{T}}\!\!A\big[\lambda x_1\!+\!(1\!-\!\lambda)x_2\big]\!\!+\!b^{\mathsf{T}}\big[\lambda x_1\!+\!(1\!-\!\lambda)x_2\big]\!\!+\!c\leqslant0$

consequently ,

$\lambda x_1+(1-\lambda)x_2 \in S\;.$

In this way, we have proved the set $\,S\,$ is convex.