Let $f$ be a non-negative, increasing (at least weakly) smooth function defined on $[0,1]$, where $f(0) = 0$ and $f(1) = 1$. Is it true that $g := xf$ is convex?
We have $g(x) = xf(x)$ and so $g'(x) = f(x) + xf'(x)$, and $g''(x) = 2f'(x) + xf''(x)$. This is obviously greater than $0$ if and only if $2f'(x) + xf''(x) > 0$. Since $f$ is increasing, the first term is positive, but the second term isn't if $f$ is concave.
Let $f(x) = 1-(1-x)^2$, $g(x)=xf(x)$ and compute $g''({4 \over 5})$ and show that it is negative.