Consider the open unit disk $\mathbb{D} \subset \mathbb{R}^2$, and consider a Brownian motion in the plane starting at the origin. Let $\gamma : \mathbb{R} \to \mathbb{R}^2$ be a smooth planar curve such that $\gamma (s) \in \mathbb{D}$ for some $s$. Let $\text{dist} (0, \gamma) = r \in (0, 1)$, that is, the curve is always at a distance $\geq r$ from the origin. Now, fix a (small) positive real number $t$, and let $B(t)$ denote the position of the Brownian particle at time $t$.
My question is, can the probability that the particle strikes the curve $\gamma$ within time $t$ (that is, $B(s) \in \gamma, s \leq t$ be arbitrarily low? I am guessing not, but not sure. In other words, I am guessing that the probability has a lower bound independent of $\gamma$. Any help will be appreciated.
Well I guess one could answer your question with the law of the supremum of $B_t$ over $t \in (0, \epsilon)$, for any $\epsilon > 0$. Actually due to the reflexion principle, this is known and we have (I'll do the proof in dimension 1 but it exactly the same in dimension 2). Let $S_t = \sup_{t\in (0, \epsilon)} \{ B_t \}$, then we have : $$ \mathbb{P} (S_t \geq a ) = \mathbb{P}(|B_t| \geq a). $$
So now, the probability of hitting the curve of a function $f : \mathbb{R} \rightarrow \mathbb{R_+}$ (say) bounded from below by a constant $c$ is :
$$ \mathbb{P} (B_t \in \mathcal{C}_f \textrm{ for some $t$ before time $\epsilon$} ) \leq \mathbb{P}(S_\epsilon \geq c) $$
where $\mathcal{C}_f$ stands for the "curve" of $f$. Now playing on $\epsilon$ and $c$ should give your answer (you know the law of $|B_\epsilon|$ which looks like a Dirac in 0 when $\epsilon$ goes to 0).