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I need some help proving that $$ \sqrt{x^T \Sigma x}, $$ with $\Sigma$ being a positive definite matrix, is convex. I already found some questions about this topic, indicating that one should use a conic representation but I did not understand these approaches, I was hoping maybe someone has another idea or could explain it better (maybe with a link to a resp. paper or book).

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When $\Sigma\succeq 0$ then we can define $\sqrt{\mathbf{x}\Sigma\mathbf{x}}=\|\mathbf{x}\|_{\Sigma}$ to get a normed vector space, which has all the properties of the Euclidean norm. Specifically, it is absolutely homogeneous and the triangle inequality holds. Then the solution becomes very easy: \begin{aligned} f(\lambda\mathbf{x}+(1-\lambda)\mathbf{y})&=\|\lambda\mathbf{x}+(1-\lambda)\mathbf{y}\|_{\Sigma}\\&\leq \|\lambda\mathbf{x}\|_{\Sigma}+\|(1-\lambda)\mathbf{y}\|_{\Sigma}\\ &=|\lambda|\|\mathbf{x}\|_{\Sigma}+|(1-\lambda)|\|\mathbf{y}\|_{\Sigma}\\ &=\lambda f(\mathbf{x})+(1-\lambda)f(\mathbf{y}) &\square \end{aligned}

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There are at least two ways to prove this:

  1. $(x,y) \mapsto x^\top \Sigma \, y$ is a scalar product and your function is the corresponding norm.

  2. You have $\sqrt{x^\top \Sigma x} = \| \Sigma^{1/2} x\|$, i.e., the combination of a convex function and a linear function.

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