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

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.
