Looking for a function such that...

FYI this was problem B5 in the 2010 Putnam exam, so you can find it here: http://amc.maa.org/a-activities/a7-problems/putnamindex.shtml

They had a pretty succinct solution. Suppose $f$ is strictly increasing. Then for for any $y_0$ you can define an inverse funciton $g(y)$ for $y>y_0$ such that $x=g(f(x))$. Differentiating, we get $1=g'(f(x))f'(x)=g'(f(x))f(f(x))$, so that $g'(y)=\dfrac{1}{f(y)}$. We know that $g$ obtains arbitrarily large values since it is the inverse function of $f$ and $f$ is defined for all $x$, which means $g(z) - g(y_0) = \displaystyle \int_{y_0}^zg'(y)dy = \int_{y_0}^z\frac{dy}{f(y)}$ must diverge as $z\rightarrow\infty$.

Now all we have to do is show that $f$ is bounded below by a function that causes the integral to converge. For $x>g(y_0)\equiv g_0$, we have $f'(x)>g_0$, so we can assume that for some $\beta$ and $x$ large enough, $f(x)>\beta x$. Iterating this argument, we get that $f(x)>\alpha x^2$ for some $\alpha$ and $x$ large enough. So we can assume that $f(x)$ is asymptotically greater than $\alpha x^2$. But then the integral above converges, contradicting that $g(z$) is unbounded as $z\rightarrow\infty$. Thus, we conclude that $f$ cannot be strictly increasing.

Another partial answer...

First of all call $L^+, L^-$ the limits at $\pm\infty$ (which exist by monotonicity). The result I prove is the following:

$\textbf{Partial result.}$ Assume $L^+<\infty$ then $f=0$

We have the following 'obvious' remarks

1. If $f$ is a solution then $f\in C^{\infty}(\mathbb{R})$.

2. Since composition of monotone functions are monotone we have that $f'\geq0$ and is monotone increasing. This also implies $f$ convex.

Now, a continuity argument together with the second remark imply that $L^->-\infty$ (otherwise there is a point with $f'(x)<0$). Now the FTC gives for $x<a$

$$ f(a)-f(x)=\int_x^{a} f'(t)dt $$ and letting $x\to -\infty$ and the monotone convergence theorem ($f'\geq0$) we conclude $f'\in L^1(-\infty,0)$. In a similar way we can get that $f'\in L^1(0,\infty)$ whenever $L^+<\infty$. All this together then yields: If $L^+<\infty$ then $f'\in L^1(\mathbb{R})$ and $f'$ continous and monotone, and so

$$ \lim_{x\to \pm\infty} f'(x)=0 $$

which in turn gives $f'=0$ and so $f$ is constant, but the only possible constant solution is $0$.

Now, in case $L^+=\infty$: Since $f(x)\to \infty$ when $x\to \infty$, we have that $f'$ also blows up. Pick $a\in (0,\infty)$ such that $f'(a)>1$ and $f(a)>1$, then for $x>2a$ we have (integrate twice the inequality $f\geq0$)

$$ f(x)\geq xf'(a)+f(a) \geq x $$ Plugging this in the equation gives $f'(x)\geq f(x)$. Gronwall's inequality then gives

$$ f(x)\geq e^{x-a} $$ Plugging this in the equation gives again

$$ f'(x)\geq f(e^{x-a}) \geq e^{e^{x-a}-a} $$ this we can integrate, and iterate the procedure, but I don't see a helpful estimate being easy to obtain this way.