Showing a function is convex/concave given another function is concave

Question

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.

Answer 1

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.)

Answer 2

If $f$ is twice differentiable and concave then $f''(x)\leq 0$. And $f$ decreasing implies $f'(x)\leq 0$. Setting $g(x)=xf(x)$, we have \begin{align*} g''(x)=2f'(x)+xf''(x)\leq 0 \end{align*} for all $x>0$. It follows that $g$ is concave for all $x>0$.

If the assumption of $f$ twice differentiable is too stringent for your purposes let me know and I can remove my answer.

Answer 3

If $f(x)$ is derivable, then the condition maybe be written as $$\frac{f'(x_1)-f'(x_2)}{x_1-x_2}\le 0$$and $$f'(x)\le 0.$$ Then prove $$\frac{(x_1f(x_1))'-(x_2f(x_2))'}{x_1-x_2}\le 0$$ Since $(xf(x))'=f(x)+xf'(x)$, \begin{align*} &\frac{(x_1f(x_1))'-(x_2f(x_2))'}{x_1-x_2}\\ &=\frac{f(x_1)-f(x_2)}{x_1-x_2}+\frac{x_1f'(x_1)-x_2f'(x_2)}{x_1-x_2}\\ &=\frac{f(x_1)-f(x_2)}{x_1-x_2}+\frac{x_1f'(x_1)-x_1f'(x_2)+x_1f'(x_2)-x_2f'(x_2)}{x_1-x_2}\\ &=\frac{f(x_1)-f(x_2)}{x_1-x_2}+x_1\frac{f'(x_1)-f'(x_2)}{x_1-x_2}+f'(x_2) \end{align*} By the condition, it is easy to see the three terms are all nonnegative, so $xf(x)$ is concave.

But it is under the assumption that $f'(x)$ exist. Although the concave function is derivable but some coutable points, it is still not complete.