A Subadditive Function Satisfying an Integrability Property Is there a function $f: [0,\infty) \rightarrow [0,\infty)$ that is 


*

*Subadditive $\left(f(x + y) \leq f(x) + f(y), \forall x,y \in [0,\infty)\right)$,

*Satisfies $\lim_{r \rightarrow \infty} f(r) = \infty$, and

*$\int_{0}^{\infty}\frac{f(r)}{r^{3}}d r < \infty$ ?


It seems to me that we need at least $f(r) \geq r^2$ near the origin, but $r^a$ with $a > 2$ is not subadditive (the issue at $\infty$ aside).
 A: I'll post the thoughts I've come up with, but I'm unable to finish this, also I'm not entirely convinced what I have done thus far is legitimate:
Suppose
$$
F(x) = \int_0^x \frac{f(r)}{r^3} \, \mathrm{d} r
$$
then we have
$$
F'(x)x^3 = f(x)
$$
Now in order to have subadditivity we must have that
$$
F'(x+y)(x+y)^3 \le F'(x)x^3 + F'(y)y^3 \; \forall x,y \in [0, \infty)
$$
Now let $\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}} \in \left[ 0,k \right]$ so that $x_n \to k, y_n \to k$ then we must have
$$
F'(x_n + y_n)(x_n + y_n)^3 \to F'\left( 2k \right) 8 k^3
$$
and
$$
F'(x_n)x_n^3 + F'(y_n)y_n^3 \to F'\left( k \right) k^3 + F'\left( k \right) k^3 = 2 F'\left( k \right) k^3
$$
So that we must have
$$
F'\left( 2k \right) 8 k^3 \le 2 F'\left( k \right) k^3 \iff 4 F'\left( 2k \right) \le F'\left( k \right)
$$
which implies that
$$
4 F' \left( \frac{1}{2^{n-1}} \right) \le F' \left( \frac{1}{2^n} \right)
$$
so that
$$
F' \left( \frac{1}{2^n} \right) \ge 4 F' \left( \frac{1}{2^{n-1}} \right) \ge 4^2 F' \left( \frac{1}{2^{n-2}} \right) \ge \cdots \ge 4^n F' \left( \frac{1}{2^0} \right) = 4^n F' ( 1 ) = 4^n f(1)
$$
So in order for $F'(0)$ to exist we need $f(1) = 0$. I'm unsure how to equate integrability to $F'(0)$ existing (thus being finite). However from the restriction that $f(1) = 0$ we can see that
$$
f(k) = f( (k-1) + 1 ) \le f(k-1) + f(1) = f(k-1) \; \forall k \ge 1
$$
and then specifically when $k$ is an integer
$$
f(2) \le f(1) + f(1) = 0 \implies f(2) = 0 \implies f(k) = 0
$$
But this tells us that $\lim_{r \to \infty} f(r) \neq \infty$ since there's a subsequence that converges to $0$.
I'll delete this if there's some glaring fallacy that I've missed (or fix it if it's small enough). Hope this helps someone!
A: The answer is no.
Assume $1)$ and $3)$ hold.
First observe that there are points $r$ arbitrarily close to $0$ such that $f(r)\leq r^2$. Indeed setting $A:=\{r:f(r)> r^2\}$ we obtain
$$\int_A\frac{1}{r}\leq\int_A \frac{f(r)}{r^3}<\infty$$
therefore $A$ cannot have full measure in any interval $(0,\delta)$, and there must be some point $r_\delta\in(0,\delta)\backslash A$. In particular $f(r_\delta)\leq r_\delta^2< \delta r_\delta$.
By subadditivity $f(x)\leq \delta x$ for every $x$ integer multiple of $r_\delta$. Since $\delta$ could be taken arbitrarily small this contradicts the condition that $\lim_{r\to+\infty}f(r)=+\infty$.
With some additional work it can be shown that under hypotheses $1)$ and $3)$ $f$ must be eventually zero, and I believe it must actually be zero everywhere.
