Let $\sigma: \mathbb R \times \mathbb R \to \mathbb R$ be a Lipschitz continuous function bounded below by some $M > 0$.


Let $W$ be a standard Brownian motion, and let $X$ be the solution to the SDE


$$dX\_t = \sigma(t, X\_t) dW\_t$$


with $X\_0 = 0$.


**Question:** Fix $T > 0$. Does there exist, for every $\varepsilon, h > 0$ a $\delta > 0$ such that


$$\mathbb P\left(\int\_{0}^T \mathbf 1\_{[-\delta, \delta]} (X\_s) ds > h\right) < \varepsilon?$$



A bound for the occupation time of a diffusion
I think so. The martingale $X\_t$ is a time-changed Brownian motion: $X\_t = B\_{A\_t}$, where $$A\_T = \int\_0^T \sigma^2(t, X\_t) dt$$ and $B\_t$ is some Brownian motion. Now write
$$ \int\_0^T \mathbb 1\_{[-\delta,\delta]}(X\_t) dt = \int\_0^{A\_T} \mathbb 1\_{[-\delta,\delta]}(B\_s) dA^{-1}\_s = \int\_0^{A\_T} \frac{\mathbb 1\_{[-\delta,\delta]}(B\_s)}{\sigma^2(A^{-1}\_s, B\_s)} ds \leqslant \int\_0^{A\_T} \frac{\mathbb 1\_{[-\delta,\delta]}(B\_s)}{M^2} ds . $$
If $\sigma(t, x)$ is bounded above by some number $N$, then we get
$$ \int\_0^T \mathbb 1\_{[-\delta,\delta]}(X\_t) dt \leqslant \int\_0^{N^2 T} \frac{\mathbb 1\_{[-\delta,\delta]}(B\_s)}{M^2} ds , $$
and the desired result follows directly from the corresponding statement for the Brownian motion. The general case requires a more delicate bound on $A\_T$. I do not have time now to work out the details, but if need be, I will get back to this when I am more available.


