Assuming $f(x)$ is not a linear function, given a concave decreasing function $f(x)$, I want to find whether $g(x)=f(x)\cdot x$ is convex or concave for strictly positive $x$.
However, I'm having a trouble proving it mathematically.
Since $f(x)$ is concave, $$ f((1-\alpha)x_1 +\alpha x_2)\geq (1-\alpha)f(x_1)+\alpha f(x_2)$$
Then $$g(x)=f(x)\cdot x$$ and $$f((1-\alpha)x_1 +\alpha x_2)\cdot ((1-\alpha)x_1+\alpha x_2)\geq (1-\alpha)f(x_1)+\alpha f(x_2)\cdot ((1-\alpha)x_1+\alpha x_2)$$ since $((1-\alpha)x_1+\alpha x_2)$ is positive and $f(x)$ is concave. And then I'm stuck and don't know how to go further. Could anyone help please? Thank you in advance.
Best Answer
Proof: Let us prove that $x\mapsto x f(x)$ is concave on $(0, \infty)$.
We have to prove that, for $x_1, x_2 > 0$ and $t\in [0, 1]$, $$\lambda x_1f(x_1) + (1-\lambda)x_2 f(x_2) \le (\lambda x_1 + (1-\lambda)x_2)f(\lambda x_1 + (1-\lambda)x_2).$$ Since $f$ is concave, we have $$\lambda f(x_1) + (1-\lambda) f(x_2) \le f(\lambda x_1 + (1-\lambda)x_2).$$ Thus, it suffices to prove that $$\lambda x_1f(x_1) + (1-\lambda)x_2 f(x_2) \le (\lambda x_1 + (1-\lambda)x_2)[\lambda f(x_1) + (1-\lambda) f(x_2)]$$ which is written as $$\lambda (1-\lambda) (x_1 - x_2)(f(x_1) - f(x_2)) \le 0. \tag{1}$$ Since $f$ is non-increasing, we have $(x_1 - x_2)(f(x_1) - f(x_2)) \le 0$. So, (1) is true. (Q. E. D.)