Show that this function is convex given that Q is positive semi-definite

Question

For this problem it's given that $Q\in \mathbb{R}^{n\times n}$ is positive semi-definite, and now I'm trying to show that $$f(x) = \sqrt{x^TQx+1}$$ is convex over $\mathbb{R}^n_{++}$

I've tried to solve this problem by computing the Hessian of f, but I kind of got nowhere as I wasn't able to show that $$v^T\nabla^2f(x)v \geq 0$$

Has anyone done this type of problem before? I'm just not sure if there's a better/easier way to do this than by computing the Hessian. . .I tried using the definition of convex functions but again I'm kind of getting no where. Any pointers in the right direction would be helpful. Thank you!

Answer 1

This question is already answered here Prove the convexity of $f(x) = \sqrt{x^T Qx + 1}$ over $\mathbb R^n$, with $Q \succcurlyeq 0$. Note in particular the second answer that uses Hessian and relies on Cauchy schwartz inequality on the inner product $<x,y>_Q=x^{T}Qy$ towards the end.

Answer 2

Guide:

Let $Q=UDU^T$

\begin{align} f(x) &= \sqrt{x^TQx+1}\\ &= \sqrt{x^TUDU^Tx+1}\\ &= \left\|\begin{bmatrix} D^\frac12 U^Tx \\ 1 \end{bmatrix}\right\| \end{align}

Now, you can use triangle inequality of norm to prove convexity.

Answer 3

Let $\mathbf{x} \colon= \left[ x_1, \ldots, x_n \right]^T$ and $\mathbf{y} \colon= \left[ y_1, \ldots, y_n \right]^T$ be any points in $\mathbb{R}^n$ and let $\lambda \in \mathbb{R}$ be such that $0 < \lambda < 1$. Now let us put $$ \mathbf{z} \colon= (1 - \lambda) \mathbf{x} + \lambda \mathbf{y} = \left[ (1-\lambda)x_1 + \lambda y_1, \ldots, (1-\lambda)x_n + \lambda y_n \right]^T. $$

We need to show that $$ f( \mathbf{z} ) \leq ( 1-\lambda) f(\mathbf{x}) + \lambda f( \mathbf{y} ). $$

Now we note that $$ \begin{align} f( \mathbf{z} ) &= \sqrt{ \mathbf{z}^T Q z + 1} \\ &= \sqrt{ \left( (1-\lambda) \mathbf{x}^T + \lambda \mathbf{y}^T \right)Q \left( (1-\lambda) \mathbf{x} + \lambda \mathbf{y} \right) + 1} \\ &= \sqrt{ \left( (1-\lambda)\mathbf{x}^TQ +\lambda \mathbf{y}^TQ \right) \left( (1-\lambda) \mathbf{x} + \lambda \mathbf{y} \right) + 1 } \\ &= \sqrt{ (1 - \lambda)^2 \mathbf{x}^T Q \mathbf{x} + (1-\lambda)\lambda \left( \mathbf{x}^T Q \mathbf{y} + \mathbf{y}^T Q \mathbf{x} \right) + \lambda^2 \mathbf{y}^T Q \mathbf{y} + 1} \\ . . . \end{align} $$

Rusty on this stuff at the moment, I'm unable to take it from here, I'm afraid.