Let $$f(x) = \frac{1}{2} x_1^2 + \frac{1}{2} x_2^2 – \log(1 + x_1
+ x_2)$$ be defined on $x_i \in (a, b]$, where $a$ is possibly $0$, $b > 0$.
I wish to show that the function is strongly convex on the domain of $f$ and find the strong convexity parameter $\mu$ associated with it.
One way is by computing the Hessian. By a short calculation, it can be shown
$$\nabla f(x) = \left( x_1 – \dfrac{1}{1+x_1+x_2}, x_2 – \dfrac{1}{1+x_1+x_2} \right)$$
And
$$\nabla^2 f(x) = \begin{bmatrix} 1 + \dfrac{1}{(1+x_1+x_2)^2} & \dfrac{1}{(1+x_1+x_2)^2} \\ \dfrac{1}{(1+x_1+x_2)^2} & 1 + \dfrac{1}{(1+x_1+x_2)^2} \end{bmatrix}$$
Now I need to show that
$$y \nabla^2 f(x) y \geq \mu \|y\|^2$$ for all $y$ in the domain of $f$
where $\mu$ is the parameter of strong convexity.
$$y \nabla^2 f(x) y = (1 + \dfrac{1}{(1+x_1+x_2)^2}) \|y\|^2 + \dfrac{2}{(1+x_1+x_2)^2} y_1 y_2$$
Since $y$ is a positive vector, therefore
$$y \nabla^2 f(x) y \geq (1 + \dfrac{1}{(1+x_1+x_2)^2}) \|y\|^2$$
and
$$\mu = 1+ \dfrac{1}{(1+b+b)^2}$$
Is my analysis correct?
Not totally confident about the $y \nabla^2 f(x) y$ step because most of the references assume $y \in \mathbb{R}^2$ (not just in the domain), which means the inequality will not hold.
Best Answer
Let $\Bbb L := \{ {\rm x} \in \Bbb R^2 \mid 1 + x_1 + x_2 = 0\}$. Let
$${\rm H} ({\rm x}) := \nabla^2 f({\rm x}) = \begin{bmatrix} 1 + \dfrac{1}{(1+x_1+x_2)^2} & \dfrac{1}{(1+x_1+x_2)^2} \\ \dfrac{1}{(1+x_1+x_2)^2} & 1 + \dfrac{1}{(1+x_1+x_2)^2} \end{bmatrix} = {\rm I}_2 + g({\rm x}) {\Bbb 1_2}{\Bbb 1_2}^\top$$
where
$$g({\rm x}) := \dfrac{1}{(1 + x_1 + x_2)^2} \geq 0$$
for all ${\rm x} \in \Bbb R^2 \setminus \Bbb L$. We would like to find a $\mu > 0$ such that ${\rm H} ({\rm x}) \succeq \mu \, {\rm I}_2$, or,
$${\rm H} ({\rm x}) - \mu \, {\rm I}_2 = (1 - \mu) \, {\rm I}_2 + g({\rm x}) {\Bbb 1_2}{\Bbb 1_2}^\top \succeq {\rm O}_2$$
for all ${\rm x} \in \Bbb R^2 \setminus \Bbb L$. Hence, we conclude that $\color{blue}{\mu \in (0,1]}$.