# Convexity of root

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).

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}
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.