I have a function $$f(x)=\frac{-1}{2(1+x^Tx)}$$ and I need to find out for what values of $x \in \mathbb{R}^n$ this function is convex and strongly convex.
I found the gradient $$\nabla f(x)=\frac{x}{(1+x^Tx)^2}$$
and the Hessian matrix $$\nabla^2 f(x)=\frac{1}{(1+x^Tx)^2} I-\frac{4xx^T}{(1+x^Tx)^3}$$
How can I find when my Hessian $\nabla^2f(x)\succeq 0$ and $\nabla^2f(x)\succeq \mu I$?
For $x, z \in \mathbb R^n$ and we have \begin{align*} z^T \nabla^2 f(x) z & = \frac{\| z \|_2^2}{(1 + \| x \|_2^2)^2} - \frac{4 \| z^T x \|_2^2}{(1 + \| x \|_2^2)^3} \\ & \ge \left(\frac{1}{(1 + \| x \|_2^2)^2} - \frac{4 \| x \|_2^2}{(1 + \| x \|_2^2)^3}\right) \| z \|_2^2 \\ & = \left(\frac{1 - 3 \| x \|_2^2}{(1 + \| x \|_2^2)^3}\right) \| z \|_2^2. \end{align*} This term is nonnegative if and only if $\frac{1 - 3 \| x \|_2^2}{(1 + \| x \|_2^2)^3} \ge 0$, that is, $\| x \|_2 \le \frac{1}{\sqrt{3}}$, so on the ball centered at the origin with radius $\frac{1}{\sqrt{3}}$, $f$ is convex.
For strong convexity, let $m \in [0, 1)$ be fixed. We want to find out for which values of $x$, the function $$ x \mapsto \frac{1 - 3 \| x \|_2^2}{(1 + \| x \|_2^2)^3} $$ is lower bounded by $m$. Note that it is upper bounded by 1, so we only consider $m \in (0, 1)$ in the sequel.
With $y := \| x \|_2^2 \ge 0$ this can be rewritten as $$ 1 - m - 3 m y - 3 y - 3 m y^2 - m y^3 \ge 0, $$ which (as $m \in (0, 1)$) is, according to WolframAlpha equivalent to $$ 0 < \| x \|_2^2 \le \frac{1}{m}\left(2 m^2 + \sqrt{4 m^4 + m^3}\right)^{\frac{1}{3}} - \left(2 m^2 + \sqrt{4 m^4 + m^3}\right)^{-\frac{1}{3}} - 1. $$ When plugging in $m = \frac{1}{2}$ into the right hand side, the upper bound is approximately $0.1$, so that for $\| x \|_2^2 \le \frac{1}{10}$ the function is $\frac{1}{2}$-strongly convex.
