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

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!

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 $$_Q=x^{T}Qy$$ towards the end.

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.

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.