Equivalent definitions of Brownian local time

Question

To keep this question simple, we only consider the local time of a standard Brownian motion $B_t$ at zero, we denote it as $L_t^0$. Generally, we define $$\tag{1} L_t^0(B)=\lim_{\varepsilon\to\infty}\dfrac{1}{2\varepsilon}\int_0^t1_{(-\varepsilon,\varepsilon)}(B_s)\,ds. $$
Another definition is given by $$L_t^0(B)= |B_t|-\int_0^t\text{sgn}(B_s)\,dB_s\tag{2}.$$
We can approximate $|B_t|$ pointwise by $\sqrt{\varepsilon+B_t^2}$ by taking $\varepsilon\to0$. Applying Ito formula to $\sqrt{\varepsilon+B_t^2}$ we get, $$\sqrt{\varepsilon+B_t^2}=\sqrt{\varepsilon}+\int_0^t\dfrac{B_s}{\sqrt{\varepsilon+B_s^2}}\,dB_s+\dfrac{1}{2}\int_0^t\dfrac{\varepsilon}{(\varepsilon+B_s^2)^{3/2}}\,ds. $$
If we take the limit, the above formula suggests that $$L_t^0(B)=\lim_{\varepsilon\to0}\dfrac{1}{2}\int_0^t\dfrac{\varepsilon}{(\varepsilon+B_s^2)^{3/2}}\,ds\tag{3}.$$
My question is can we show that by direct computation that equation (3) and (1) are the same?

Answer 1

It may depend on what you mean by "the same". By the occupation time formula for the Brownian local time $$ \int_0^t{\epsilon\over (\epsilon+B_s^2)^{3/2}}\,ds = \int_{-\infty}^\infty {\epsilon\over(\epsilon+x^2)^{3/2}}L_t^x\,dx=\int_{-\infty}^\infty {1\over (1+u^2)^{3/2}}L^{\sqrt{\epsilon}u}_t\,du. $$ Because $$ \int_{-\infty}^\infty {1\over (1+u^2)^{3/2}}\,du=2, $$ sending $\epsilon$ to $0$ on the right side of (3) therefore yields $L^0_t$ because of the continuity in $x$ of $L^x_t$.