Let's say I got two functions $f(z)$ and $g(y)$ and these are strictly decreasing functions, so $f(0) \gt f(1) \gt f(2) \gt \dots$ and $g(0) \gt g(1) \gt g(2) \gt \dots$. I then define a third function $h(x, y)$ which is a sum of $f$ and $g$:
$$h(x, y)= f(x-y) + g(y)$$, where $x \ge y$.
For example: $$h(10, 2) = f(8) + g(2)$$
Question
Can we, with only knowing this about these 2 functions $f$ and $g$, state that their convolution function $h$ is a convex function? If yes, how? And if no, why not? Do we need to know something about the derivatives of functions $f$ and $g$?
$h(x, y) = f(x-y) + g(y)$ is convex if and only if both $f$ and $g$ are convex.
The “if“ is a straight-forward calculation: $$ h(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \\ = f(\lambda(x_1 -y_1) + (1-\lambda)(x_2 - y_2)) + g(\lambda y_1 + (1-\lambda) y_2) \\ \le \lambda f(x_1-y_1) + (1-\lambda) f(x_2 - y_2) + \lambda g(y_1) + (1-\lambda) g(y_2) \\ = \lambda h(x_1, y_1) + (1-\lambda) h(x_1, y_2) \, . $$
The “only if“ follows from $f(x) = h(x,0)-g(0)$ and $g(y) = h(y, y) - f(0)$.
The monotony of $f$ and $g$ is irrelevant here.