Let $f:\mathbb{R}^m\to\mathbb{R}:=f(x_1,x_2,\dots x_m)={x_1}^{2k_1}{x_2}^{2k_2}\dots {x_m}^{2k_m}, k_i \in \mathbb{N}.$ Is there a way to show that $f$ is strictly convex?
Clearly $0$ is the unique minima.
One way to proceed would be to show that the factors $g_i(x_1,x_2,\dots x_m):={x_i}^{2k_i}$ are strictly convex, which I believe is true. But then the problem is that, the product of two convex function isn't necessarily convex.But then, is it true that:
product of two strictly convex, nonnegative functions (as in our case) is strictly convex? If true (which I'm not sure about), this can be applied to prove the result.
No, it's not true. Consider the 2d case so $f(x,y) = x^{2k_x}y^{2k_y}$. Then consider $v_1=(1,0)$ and $v_2=(0,1)$; $f(v_1) = 0 = f(v_2)$ but $$f(0.5 v_1 + 0.5 v_2) = f((0.5,0.5))= (0.5)^{2k_x}(0.5)^{2k_y}> 0 = 0.5 f(v_1) + 0.5 f(v_2)$$
Usually multiplying components of a vector will break convexity.