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Let $f(x)$ be a convex function of $x$ in a given positive interval. Also assume $f(x)\geq 0$ everywhere in that interval. Is the function $g(x)=xf(x)$ convex in that interval?. Is it possible that $g(x)$ is monotonically increasing or monotonically decreasing.

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  • $\begingroup$ To address the last part of your Question, can $g(x)$ be monotone, yes, this is possible for $g(x) = xf(x)$ of the form you specify. For example, if $f(x)= x^2$ then $g(x) = x^3$ is monotone increasing on every (positive) interval. But perhaps you have another "possibility" in mind. $\endgroup$ – hardmath Jul 14 '14 at 13:30
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N0. Let $f\colon(0,\infty)\to(0,\infty)$ be defined by $f(x)=x^{-p}$, $0<p<1$. $f$ is convex, since $$ f''(x)=-p(p-1)x^{p-2}>0\quad\forall x>0. $$ Howevwe $$ x\,f(x)=x^{1-p} $$ is concave.

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  • $\begingroup$ does this mean that nothing can be said about the nature of $g(x)$ in general ? $\endgroup$ – dineshdileep Jan 28 '14 at 17:22
  • $\begingroup$ @dineshdileep Yes. $\endgroup$ – Julián Aguirre Jan 28 '14 at 17:41
  • $\begingroup$ Thanks for the reply. I request you to go through this question I am struggling with math.stackexchange.com/questions/654817/… $\endgroup$ – dineshdileep Feb 8 '14 at 4:11
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Let $f:[0,1]\to\mathbb R$ be defined by $f(x)=1-x$. This function is convex. But, $g(x)=x-x^2$, which is not convex. To get a monotonically increasing $g$, take $f(x)=1$. To get a monotonically decreasing $g$, take $f:[1,2]\to\mathbb R$, defined by $f(x)=\frac{2-x}x$.

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