# How the Brownian motion escapes the minimum? A question based on the Lecture notes from Zeitouni on RWRE

In the "Lectures on Probability Theory and Statistics Ecole d’Eté de Probabilités de Saint-Flour XXXI - 2001" in page 249 one reads

$$A_n^{J,\delta}=\left\{\begin{array}{ll}\omega\in\Omega:\!\!\! & \overline{b} ^n =\overline{b} ^n_\delta, \text{ any refinement }(a,b,c)\text{ of }(\overline{a} _\delta^n,\overline{b}_\delta^n,\overline{c}_\delta^n)\text{ with}\\ & b\neq\overline{b}^n\text{ has depth }<1-\delta,|\overline{a}^n_\delta|+|\overline{c}_\delta^n|\leq J, \\ & \min_{t\in[\overline{a}^n,\overline{b}^n]\left\backslash\right. [\overline{b}^n-\delta,\overline{b}^n+\delta]}W^n(t)-W^n(\overline{b}^n)>\delta^3\end{array}\right\}$$

$$\text{then}$$ $$\text{it}$$ $$\text{is}$$ $$\text{easy}$$ $$\text{to}$$ $$\text{check}$$ $$\text{by}$$ $$\text{the}$$ $$\text{properties}$$ $$\text{of}$$ $$\text{Brownian}$$ $$\text{motion}$$ $$\text{that}$$

$$\lim_{\delta\to0}\lim_{J\to\infty}\lim_{n\to\infty}P(A^{J,\delta}_n)=1.\tag{2.5.2}$$

To check 2.5.2 I need to check $$P(\min_{t\in[a^n,c^n]\backslash [b^n -\delta, b^n+\delta]} W_t - W_b > \delta^3) \xrightarrow[\delta \to 0]{} 1$$ where here $$W_t$$ is a Brownian motion. and $$b$$ is the minimum of the Brownian motion in the interval $$[a,c]$$

In words, I want to prove that the minimum of the brownian motion reached at $$b$$ is isolated and that nowhere on an interval $$[a,c]$$ it get's closer than $$\delta^3$$ to this minimum.

I understand that the Brownian motion behaves in a scale that is almost $$\sqrt{\delta}$$. But the typical results do not hold for special points like the minimum, once typically, say, at 0, one has that the Brownian motion visits infinitely often the positive and the negative numbers, a oscillatory behavior that certainly does not apply for the minimum.

How do I prove this result?

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