Convexity of $x^2f(x)$ Given a function $f$ which is decreasing and convex on $(0,\infty)$, is it possible to find a simple condition on $f$ such that 
\begin{equation}
2f(x) + 4xf^\prime(x) + x^2f^{\prime\prime}(x) \geq 0.
\end{equation}
By integrating I worked out that this is equivalent to requiring that $x^2f(x)$ is convex but I would like to express it as a condition on $f$ only. Any ideas? Thanks.
 A: As $f(x)$ is convex, we have
$$
f''\geq 0
$$
As $f(x)$ is decreasing, we also have
$$
f'< 0
$$
These are applying when $x>0$. As we seek a sufficient condition, let us simply require that
$$
\frac12f(x)+xf'(x)\geq 0
$$
which, given that $f''\geq 0$, means that the given inequality will be satisfied if this inequality is satisfied. Dividing by $\sqrt{x}$, we have
$$
\frac1{2\sqrt{x}}f(x)+\sqrt{x}f'(x)\geq 0
$$
This reveals the required substitution - let $\sqrt{x}f(x)=g(x)$. This allows us to state our inequality as
$$
g'(x)\geq 0
$$
This may then be stated in a different way, as
$$
g(y)\geq g(x) \qquad \text{when} \qquad y> x
$$
Substituting back to our original function, we have
$$
\sqrt{y}f(y)\geq \sqrt{x}f(x) \qquad \text{when} \qquad y> x
$$
This is a sufficient condition - if this condition is satisfied, then the required inequality will be satisfied, too. It does not necessarily mean that all functions that satisfy the required inequality will also satisfy this condition.
Now for a necessary condition: Suppose that there is a value of $x=a$ for which $f(a)=0$, and $f'(a)<0$ (and thus, as it is decreasing, all values $x>a$ will have $f(x)<0$). Now, we wish to show that there exists an $x$ such that $x^2f(x)<(x-a)a^2f'(a)$ (that is, a value of $x$ such that $x^2f(x)$ lies below the tangent line drawn at $x=a$, thus making the function non-convex).
Suppose otherwise - that is,
$$
x^2 f(x)\geq(x-a)a^2f'(a)
$$
Now, we have that
$$
f(x)\geq \left(\frac1x-\frac{a}{x^2}\right)a^2f'(a)
$$
But
$$
\lim_{x\to\infty} \left(\frac1x-\frac{a}{x^2}\right)a^2f'(a) = 0
$$
Therefore,
$$
\lim_{x\to\infty} f(x) \geq 0
$$
But $f(x)$ is decreasing - therefore, $f(x)\geq 0$ again. This contradicts the assumption that there exists an $a$ such that $f(a)=0$, and thus, if $f(a)=0$ for some $a$, there must exist a point $x>a$ at which the function is concave.
This is a necessary condition - in order for the required inequality to be satisfied, this condition must be satisfied, but it does not guarantee that the required inequality be satisfied.
