Let $\alpha \geq 0$ and $f(x) = x^2$. Can we prove or disprove that the following function is convex?
$$g(x_1, x_2) = f(\max \{\alpha, x_1\} - x_2)$$
My Approach: It is clear that the function $\max \{\alpha, x_1\} - x_2$ is convex. For $x \geq 0$, $f$ is convex and increasing, so combination of these two functions is convex. But this does not seem to work in general because of the region that $\max \{\alpha, x_1\} - x_2$ might go negative.
By definition, $g$ is convex iff for all $x, y\in\mathbb{R}^2$ and $t\in[0, 1]$ we have $g(tx+(1-t)y)\leq tg(x)+(1-t)g(y)$.
Let $x=(0, 1)$ and $y=(2, 3)$. Let $t=0.5$ and $\alpha=1$.
We have $$g(x)=g(0, 1)=(\max\{1, 0\}-1)^2=0$$ and $$g(y)=g(2, 3)=(\max\{1, 2\}-3)^2=1$$ Therefore $$tg(x)+(1-t)g(y)=0.5$$ However, $$g(tx+(1-t)y)=g(1, 2)=(\max\{1, 1\}-2)^2=1>0.5$$ so the function is not convex.