currently I try to solve the following exercise that is presented in Le Gall's book: Brownian Motion and Stochastic Calculus (Exercise $7.27$), which states as follows:
Let $f : \mathbb C \to \mathbb C$ be a holomorphic and non-constant function. Using the conformal invariance of Brownian Motion proof that $\{f(z) : z \in \mathbb C\}$ is dense.
Before presenting my proof, here are the main ingredients I need:
- Conformal invariance: Let $h : \mathbb C \to \mathbb C$ be holomorphic non-constant function. For every $t \geq 0$ set $$ C_t = \int^t_0 |h^{\prime}(B_s)|^2 ds. $$ Take any $z \in \mathbb C$ then there exists a Brownian motion $\Gamma$ starting from $h(z)$ under $P_z$ such that $$ h(B_t) = \Gamma_{C_t} \quad \forall t \geq 0 \quad P_z \text{ a.s.} $$
- Hitting Time Lemma: Let $x \neq 0$ and denote the following stopping time $$ U_{\epsilon} = \inf\{t \geq 0 : |B_t| = a\} $$ then for any $\epsilon > 0$ in dimension $2$ we have $$ P_x(U_{\epsilon} < \infty) = 1, $$ which means that no matter where we let two dimensional Brownian motion start, we are almost surely to hit the boundary of a ball of radius $\epsilon$ in finite time.
Let me demonstrate my proof:
Proof: Suppose that $\{f(z) : z \in \mathbb C\}$ is not dense, then there exists $w \in \mathbb C$ and some $\epsilon > 0$ such that $\{f(z) : z \in \mathbb C\} \cap \overline{B(w,\epsilon)} = \emptyset$. Now take $g(z) = \frac{z-w}{\epsilon}$, which is entire and non-constant (note that the composition $g \circ f$ is still entire) then $$ \big\{g \circ f(z) : z \in \mathbb C\} \cap \overline{B_1(0)} = \emptyset. $$ Now using the first theorem we know that taking $h = g \circ f$ and $z = g \circ f(0)$ that there exists two dimensional Brownian motion given by $$ h = g \circ f(B_{t}) \quad t \geq 0 \quad P_{z} \text{ a.s.} $$ and starting from $h \circ f(0) \in \mathbb C \setminus \overline{B_1(0)}$. Then using the second lemma for $\epsilon = 1$ is and since $h \circ f(0) \neq 0$ we directly deduce that $$ P_z(U_{1} < \infty) = 1, $$ thus there exists an $s \geq 0$ such that $$ g \circ f(B_s) \cap \partial B_1(0) \neq \emptyset, $$ which yields a contradiction.
Questions: First of all is my proof correct? I also believe that taking the composition with this conformal map is not strictly necessary, however this gives a nice expression to work with the ball of radius $1$ and thus a cleaner write for applying the second lemma. Does anybody know an alternative proof?
Thanks in advance.
I think your solution is correct, although you could maybe go faster: since as you said it holds that
$$\mathbb{P}_z(|B_t - w|<\varepsilon \text{ for some } t>0)=1 \quad \forall \; z,w \in \mathbb{C}, \varepsilon >0,$$
and $(f(B_t))_{t\geq 0}$ is a (time-changed) Brownian motion, then $f(\mathbb{C})$ not being dense in $\mathbb{C}$ would directly be a contradiction to the fact above.
Maybe you would be interested to know that this fact admits an easy proof relying on conformal invariance of Brownian motion (the classical approach would exploit the Strong Markov Property). Indeed up to taking an affine transformation it is sufficient to show that for all $r \in \mathbb{R}_{>1}$,
$$\mathbb{P}_r(|B_t|<1 \text{ for some } t>0)=1.$$
By continuity it is quite clear that $\lim_{r \downarrow 1}\mathbb{P}_r(|B_t|<1 \text{ for some } t>0)=1$. But now,
$$\begin{align*} \mathbb{P}_{r^{1/n}}(|B_t|<1 \text{ for some } t>0)&=\mathbb{P}_{r^{1/n}}(|B_t^n|<1 \text{ for some } t>0) \\ &=\mathbb{P}_{r}(|B_t|<1 \text{ for some } t>0) \end{align*}$$
where the last equality holds since $f(z)=z^n$ is entire. By letting $n \to \infty$ you obtain the claim.