The probability that a linear Brownian motion will hit a curve Summary
I am trying to estimate the probability that a standard linear Brownian motion will hit some curve. To make things a bit simple, I can assume that the curve is a graph of a function, that is is positive at $t=0$, that it is bounded from left by $0$ and by right by some $T>0$, that it is continuous, or even differentiable, and many other nice curvish features that may help making this question more feasible.
Formalizing
Let $\{B(t)\mid t\ge 0\}$ be a standard linear Brownian motion, and let $f:[0,T]\to\mathbb{R}$ be infinitely-differentiable (in $(0,T)$) real function with $T>0$ and $f(0)>0$. Let $A_f$ be the event "$\exists t\in(0,T]):\ B(t)=f(t)$", that is, the Brownian motion "hits" the graph of the function $f$.
The question is as follows: given $f$, what is $\mathbb{P}\left(A_f\right)$?
Attempt
All I could do is solve this for $f\equiv c>0$; in that case, if we define $M(t)=\max\{B(s)\mid 0\le s\le t\}$ we have that
$$\mathbb{P}\left(A_f\right)=\mathbb{P}\left(M(T)\ge c\right)$$
and by reflection principle, the last probability equals
$$2\mathbb{P}\left(B(T)\ge c\right)$$
and that can be solved using straight-forward normal cdf.
However, even for non-constant linear $f$'s that trick won't do; and $f(x)=1/x$ (with something at $0$, bounded from the right by some $T$) seems much harder. This is where I stop and post a question.
 A: This is very much only a partial answer but the problem intrigued me and it wouldn't fit in a comment, so just see it as inspiration for others who have more time maybe.
Suppose the curve is piecewise linear, i.e. there exist times $$0 = t_0 < t_1 < ... $$ and values $f_k \in \mathbb{R}$ with $f_0 > 0$ such that
$$
f(t) = \sum_{k=0}^\infty \chi_{[t_k, t_{k+1}]}(t) \cdot \underset{:= g_k}{\underbrace{(f_k + (t - t_k) \frac{f_{k+1} - f_k}{t_{k+1} - t_k})}}.
$$
Then
$$
\mathbb{P}(\exists t > 0: B_t - f(t) = 0) = \mathbb{P}\left(\bigcup_{k = 0}^\infty \{\exists t \in [t_k, t_{k+1}]: B_t - g_k(t) = 0\} \right).
$$
Define
$$w^k_t = B_{t + t_k} - B_{t_k}$$
We know that $W^k$ is a BM.
Define furthermore
$$c_k = \frac{f_{k+1} - f_k}{t_{k+1} - t_k}.$$
Then for $t \in [0, t_{k+1} - t_k]$
$$
B_{t + t_k} - g_k(t + t_k) = B_{t_k} - f_k + W^k_t - t c_k
$$
Then
$$
\mathbb{P}(\exists t \in [t_k, t_{k+1}]: B_k - g_k = 0 ) = \mathbb{P} (\exists t \in [0,t_{k+1} - t_k]: B_{t_k} - f_k + W^k_t - t c_k)
$$
$$ = \mathbb{E} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]: B_{t_k} - f_k + W^k_t - t c_k} \right]
= \mathbb{E} \left[ \mathbb{E} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]: B_{t_k} - f_k + W^k_t - t c_k} \vert \sigma(B_{t_k})\right] \right]
$$
$$
= \mathbb{E}_{- f_k} \left[ \mathbb{E} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]: B_{t_k} + W^k_t - t c_k} \vert \sigma(B_{t_k})\right] \right]
$$
Now we use the Strong Markov property (or maybe the weak one suffices here?)
$$
= \mathbb{E}_{- f_k} \left[ \mathbb{E}_{B_{t_k}} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]:W^k_t - t c_k} \right] \right]
$$
$$
= \mathbb{E}_{- f_k} \left[ \mathbb{P}_{B_{t_k}} \left( T_0^{c_k} \leq t_{k+1} - t_k \right) \right],
$$
where $T^k_0$ is the hitting time of $0$ for a Brownian motion with drift $-c_k$. This can be easily computed as the density is known, you should be able to find it somewhere online.
For clarity let me rewrite which process starts where
$$
= \mathbb{E}_{B_0 = - f_k} \left[ \mathbb{P}_{W^k_0 = B_{t_k}} \left( T_0^{c_k} \leq t_{k+1} - t_k \right) \right].
$$
So this quantity should be computable. But the problem is that the term
$$
\mathbb{P}\left(\bigcup_{k = 0}^\infty \{\exists t \in [t_k, t_{k+1}]: B_t - g_k(t) = 0\} \right)
$$
is not easily reduced to these probabilities. However from here you can maybe start making estimates.
Then to get to smooth functions you approximate with a piecewise linear function and also make some estimates.
