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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.

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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.

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  • $\begingroup$ Thank you - not sure why I didn't get a notification, but just saw your answer! $\endgroup$ Commented Jan 24 at 22:27

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