I am trying to solve exercise 2.9 in chapter VI of Revuz-Yor: Let $P_t$ be the semigroup of a Brownian motion. We put $f(x)=|x|$. Show that $$L_t=\lim_{h\to0}\frac{1}{h}\int_0^t \left(P_hf(B_s)-f(B_s)\right) \text{d}s\text{ a.s.}$$ where $L_t$ is the local time in zero. There is a hint: write the occupation times formula for the function $P_hf-f$. Thanks to the Markov property of Brownian motion: $$\int_0^t (P_hf(B_s)-f(B_s)) \text{d}s=\int_0^t (\mathbb{E}_{B_s}(|B_{s+h}|)-|B_s|)\text{d}s=\int_0^{t} (\mathbb{E}_{B_s}(|B_s+B'_h|)-|B_s|)\text{d}s$$ where $B'$ is a Brownian motion independent from $B$. The occupation times formula for the function $P_hf-f$ allows to write $$\int_0^t (\mathbb{E}_{x}(|B_s+B'_h|)-|B_s|)\text{d}s=\int_{\mathbb{R}} (\mathbb{E}(|x+B'_h|)-|x|)L^x_t\text{d}x$$ From there I am stuck. Of course we can write $$\int_{\mathbb{R}} (\mathbb{E}(|x+B'_h|)-|x|)L^x_t\text{d}x=\int_{\mathbb{R}} \left(\frac{1}{\sqrt{2\pi h}}\int_{\mathbb{R}}(|x+y|)-|x|)e^{-\frac{y^2}{2h}}\text{d}y\right)L^x_t\text{d}x$$ But no change of variable seems to ease the calculation of the integral included in the parenthesis on the right, which I would ideally like to show it converges towards $\frac{1}{2}\mathbf{1}_{\{|x|<h\}}$. I also thought writing de $\mathbb{E}(|x+B'_h|)$ as the expectation of a folded normal variable $$\mathbb{E}(|x+B'_h|)=\sqrt{\frac{2h}{\pi}}\exp(-\frac{x^2}{2h})-x[1-2\Phi(\frac{x}{\sqrt{h}})]$$ where $\Phi$ is the distribtion function of a standard normal distribution. Letting $h$ tend to 0 we get $$\lim_{h\to0}\frac{1}{h}\int_0^t \left(P_hf(B_s)-f(B_s)\right) \text{d}s=\lim_{h\to0}\int_{\mathbb{R}} \left(\frac{|x|-x}{h} \right)L^x_t\text{d}x$$ Using again the occupation times formula we get $$\lim_{h\to0}\frac{1}{h}\int_0^t \left(B_s\mathbf{1}_{\{B_s<0\}} \right)\text{d}s$$
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