I have moved to https://sites.google.com/temple.edu/xinli/

And ResearchGate https://www.researchgate.net/profile/Xinli_Yu

I have moved to https://sites.google.com/temple.edu/xinli/

And ResearchGate https://www.researchgate.net/profile/Xinli_Yu

By the previous result in Property of two dimensional Brownian motion(part I) we have shown that in two dimensional Brownian motion we have $latex P_x(S_r<S_R)=\frac{\log(R)-\log(x)}{\log(R)-\log(r)}$.

Now we fix $latex R$, let $latex r\to 0$, and let $latex S_0=\inf{t>0:B_t=0}$, then for $latex x\neq 0$, we have $latex P_x(S_0<S_R)\le \lim_{r\to\infty}P_x(S_r<S_R)=0$.

This is true for arbitrary $latex R$. We also know that $latex S_R\to\infty$ as $latex R\to\infty$ by continuity of Brownian motion.

We have $latex P_x(S_0<\infty)=0$ for all $latex x\neq 0$.

Now consider $latex x=0$, by Markov property we have $latex P_0(B_t=0$ for some $latex t\ge \epsilon)=E_0[P_{B_\epsilon}(T_0<\infty)]=0$ for all $latex \epsilon>0$, so $latex P_0(B_t=0$ for some $latex t>0)=0$.

By definition of $latex S_0=\inf{t>0:B_t=0}$, we have proved that $latex P_x(S_0<\infty)=0$ for all $latex x$.

Thus, the two dimensional or higher dimensional Brownian motion will not hit 0 at a positive time even if it starts there.

**Theorem: **

As $latex t\to \infty$, $latex |B_t|\to\infty$ a.s. when $latex d\ge 3$

**Proof:**

For $latex d\ge 3$, by previous result we have $latex P_x(S_r<S_R)=\frac{R^{2-d}-|x|^{2-d}}{R^{2-d}-r^{2-d}}$.

Since two dimensional Brownian motion does not hit points then similarly three dimensional brownian motion does not hit the line $latex {x:x_1=x_2=0}$. If fix $latex r$ and let $latex R\to\infty$ we can get $latex P_x(S_r<\infty)=(r/|x|)^{d-2}<1$ when $latex |x|>r$.

Let $latex A_n=|B_t|>n$ for all $latex t\ge S_{n^3}$, by previous result $latex S_{n^3}$ is almost surely finite.

By the strong Markov property $latex P_x(A_n^c)=E_x(P_{B(S(n^3)}(S_n<\infty))=(\frac{1}{n^2})^{d-2}$.

Then by Borel Cantelli, when $latex d\ge 3$, $latex \sum_n P_x(A_n^c)=\sum_n (\frac{1}{n^2})^{d-2}<\infty$.

Thus $latex P_x(\limsup_{n\to \infty}A_n^c)=0$.

So we have $latex P_x(\limsup_{n\to \infty}A_n)=1$, which is $latex P_x(A_n$ as $latex n\to\infty) =1$.

Thus we have proved As $latex t\to \infty$, $latex |B_t|\to\infty$ a.s.

**Theorem**:

Two-dimensional Brownian motion is **recurrent **in the sense that if $latex G$ is any open set, then $latex P_x(B_t\in G i.o.)=1$

**Proof**:

**First step**:

We try to find the appropriate harmonic function for d-dimensional Brownian motion. Since Brownian motion is spherical symmetric, we can let $latex \phi(x)=f(|x|^2)$ and find the $latex f$ such that $latex \Delta \phi=0$. We can use $latex |x|^2$ rather than $latex |x|$ for easier differentiation, then we have

$latex D_i f(|x|^2)=f'(|x|^2)2x_i$ and $latex D_{ii}f(|x|^2)=f”(|x|^2)=f”(|x|^2)4x_i^2+2f'(|x|^2)$

So $latex \Delta \phi(x)=\Delta f(|x|^2)=\sum_i [f”(|x|^2)4x_i^2+2f'(|x|^2)]=4|x|^2f”(|x|^2)+2df'(|x|^2)$

Letting $latex y=|x|^2$, then we have $latex 4yf”(y=0)+2df'(y)$,

if $latex y>0$, $latex f”(y)=\frac{-d}{2y}f'(y)$.

Taking $latex f'(y)=Cy^{-d/2}$ as a solution, by choosing the appropriate constant, we have

$latex \phi(x)=\log|x|$ when $latex d=2$, and $latex \phi(x)=|x|^{2-d}$ when $latex d\ge 3$

**Second step**:

Now we can use the Theorem (basically is Mean Value property of harmonic function applied on Brownian Motion) on $latex d\ge 2$.

Let $latex S_r=\inf{t:|B_t|=r}$ which is the first time the Brownian motion hitting the ball with radius $latex r<R$. Since $latex \Delta\phi=0$ in $latex G={x:r<|x|<R}$, and is continuous on $latex \overline G$, the Theorem (basically is Mean Value property of harmonic function applied on Brownian Motion) implies that $latex \phi(x)=E_x\phi (B_\tau)$, where $latex \tau=\inf{t:B_t\notin G}$.

So we have $latex \phi(x)=E_x\phi(B_\tau)=\phi(r)P_x(S_r<S_R)+\phi(R)(1-P_x(S_r<S_R))$ where $latex \phi(r)$ is short for the value of $latex \phi(x)$ on $latex {x:|x|=r}$.

Solving this we can get $latex P_x(S_r<S_R)=\frac{\phi(R)-\phi(x)}{\phi(R)-\phi(r)}$

**Third step**:

In $latex d=2$, we have $latex P_x(S_r<S_R)=\frac{\log(R)-\log(x)}{\log(R)-\log(r)}$

If we fix $latex r$ and let $latex R\to\infty$, the right hand side goes to 1.

So $latex P_x(S_r<\infty)=1$ for any $latex x$ and any $latex r>0$.

By Markov property, $latex P_x(|B_t|=r$, for some $latex t>s)=E_x(P_{B_s}(S_r<\infty))=1$

For any open set G centered at the origin, we can always choose $latex r$ such that $latex B_r(0)\in G$, thus by $latex P_x(|B_t|=r$, for some $latex t>s)=E_x(P_{B_s}(S_r<\infty))=1$, we can conclude the $latex P_x(B_t\in G i.o.)=1$.

Since Brownian motion has the translation invariant property, for any open set $latex G$, this is also true.

I am a PHD student in the department of Mathematics of Temple University.