# Math Genius: \$text{lim inf}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}}leq text{lim sup}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}}\$

Let $$g(t) = E[X_{t}^{2}]$$, the starting point is the following set of inequalities which hold for all $$t geq 0$$:
begin{align} – int _{0}^{t} 2g(s)ds + t leq g(t) leq int _{0}^{t} 2g(s)ds + t end{align}

Then, we need to prove the following:begin{align} 0 < text{lim inf}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}}leq text{lim sup}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}} < infty end{align}

An obvious first step is integrating the above inequalities, and Fubini will probably also be useful, since then you can exchange the expectation and integrals. Furtermore, $$X_{t}$$ is also a submartingale, but I am not sure if that is relevant here.

However, even with the above knowledge I still can not picture where the limsup and liminf in the expression come from.

I would really appreciate any help.

EDIT: As per TheBridges’ request, I also give the SDE of $$X_{t}$$ below, though I am not sure if it is relevant towards answering my question.
begin{align} dX_{t} = |X_{t}|dt + dW_{t}, X_{0} = 0 end{align}

To derive the first inequalities at the top, I applied Ito to rewrite $$X_{t}^2$$, and together with the SDE of $$X_{t}$$ you can fairly easily derive these inequalities. But it is still unclear as to how they should be used to derive the second set of inequalities.