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Problem 2.10 of Boyd & Vandenberghe's Convex Optimization:

Let $C \subseteq \Re^n$ be the solution set of a quadratic inequality, $$C = \left\{ x \in \Re^n \mid x^TAx +b^Tx + c \leq 0 \right\}$$ where $A \in \Re^{n \times n}$, b $\in \Re^n$ and c $\in \Re$. We want to show that $C$ is convex if $A \succeq 0$.

The solution is provided. But I want to prove it through another way in which we assume that if $x_1$ and $x_2$ are in the set then for $0\leq \lambda\leq1$ the point $\lambda x_1+(1-\lambda)x_2$ is also in the set.

I tried it and I end up with following equation $$\lambda^2x_1^TAx_1+\lambda x_1^TAx_2-\lambda^2 x_1^TAx_2+\lambda x_2^TAx_1-\lambda^2 x_2^TAx_1+(1-\lambda)^2x_2^TAx_2+b^T\lambda x_1+(1-\lambda)b^Tx_2+c$$ I do not know how to show that the above quantity is less than or equal to zero. Any help in this regard will be much appreciated. Thanks in advance.

Rodrigo de Azevedo
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Frank Moses
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  • Group the terms related to the affine part and those related to the quadratic part. Then use the fact that $A \geq 0$. – LinAlg May 04 '18 at 11:32

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If $f(x)=x^TAx+b^Tx+c$ and $\lambda_1+\lambda_2=1$ then I think

$$ f(\lambda_1x_1+\lambda_2x_2) = \lambda_1f(x_1)+\lambda_2f(x_2)-\lambda_1\lambda_2(x_1-x_2)^TA(x_1-x_2) $$

Michal Adamaszek
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  • how can you not have a term with $\lambda_1^2$? – LinAlg May 04 '18 at 11:29
  • @LinAlg On the right-hand side $x_1^TAx_1$ appears with coefficient $\lambda_1-\lambda_1\lambda_2=\lambda_1^2$. And I implicitly assumed $A$ is symmetric. – Michal Adamaszek May 04 '18 at 11:38
  • Thank you for your answer. I think even if we do not assume $A$ to be symmetric the right side of your equation matches perfectly with my equation. But I do not know which steps you took to obtain the right side of your equation from my equation. – Frank Moses May 08 '18 at 02:38
  • @FrankMoses I just tied to group the terms the best I could. – Michal Adamaszek May 08 '18 at 06:48