I have a question that is simple but I was unable to answer it.
Given a function $f(x): \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ that is convex. Is the function $g(x): \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ with $g(x):= f(x) \cdot x $ also convex?
Thanks a lot in advance!
Thanks a lot for the fast reply. Maybe it's true when I say my specific circumstances:
$f(x) = \arcsin\left(\dfrac{a}{x-a}\right)+\arccos\left(\dfrac{2x^2-2x-2a^2-2a+1}{2\left(x^2-x-a^2+a\right)}\right)$ where $a$ is a constant with $\frac{1}{4} \leq a \leq \frac{1}{2}$ and $x \geq 2$.
Furthermore, I consider the function $g(x) = (2x-1)$.
I already checked that the second derivative of $f(x)$ in the interval $[2,\infty)$ is positive which means that $f(x)$ is convex in $[2,\infty)$.
Is it now true that $f(x) g(x)$ is convex in $[2,\infty)$?
Let $f(x)=e^{-x}$. Then $f''>0$ so $f$ is convex. Let $g(x)=xf(x)$. Then $g'(x)=-xe^{-x}+e^{-x}$ and $g''(x)=xe^{-x}-2e^{-x} <0$ if $x<2$. Hence $xf(x)$ is not convex.