sum of strongly convex function and convex function with different domains

Question

Suppose $f(\theta)=g(\theta)+h(b)$ where $\theta=(a, b)^\intercal$, $g(\cdot)$ is strongly convex and $h(\cdot)$ is convex. Note, $h(\cdot)$ may be non-differentiable. Is $f(\theta)$ strongly convex?

Answer 1

For $g$ to be strongly convex, we have $\tilde{g}(x) := g(x) - \tfrac m2 x^T I x = g(x) - \tfrac m2 \|x\|_2^2$ such that $\tilde{g}$ is convex for some $m>0$, i.e., $\nabla^2 \tilde{g} \succeq 0$. But since $h$ is convex, we have $\nabla^2 h \succeq 0$ too. So, since the sum of convex functions is convex (i.e., looking at the second derivatives), we have that $g$ is strongly convex because $\nabla^2 (\tilde{g} + h) = \nabla^2 (g+h) - \nabla^2 \left(\tfrac m 2 x^T I x\right) \succeq 0 \implies \nabla^2 (g + h) \succeq \frac{m}{2}I$.

Answer 2

I will use the notation $\theta=(\theta_1, \theta_2)$ and $\Delta=(\Delta_1, \Delta_2)$.

Strong convexity of $g$ implies $$g(\theta + \Delta) - g(\theta) - \langle \nabla g(\theta), \Delta \rangle \ge \frac{1}{2} \|\Delta\|^2.$$ Convexity of $h$ implies $$h(\theta_2 + \Delta_2) - h(\theta_2) - h'(\theta_2) \Delta_2 \ge 0.$$

Noting that $\nabla f(\theta) = \nabla g(\theta) + (0, h'(\theta_2))$, we have $$f(\theta + \Delta) - f(\theta) - \langle \nabla f(\theta), \Delta \rangle \ge \frac{1}{2} \|\Delta\|^2.$$