# Brownian level sets and continuous functions

Let $$V_t$$ and $$W_t$$ be independent standard Wiener processes ($$t\ge 0$$, $$W_t,V_t\in\mathbb R$$).

Let $$C$$ be the event that there is a continuous function $$f$$ such that for all $$s$$, $$t$$, $$W_t=W_s\iff V_{f(t)}=V_{f(s)}.$$ Does $$C$$ have probability 0?

(The question arose in connection with a question by Noah Schweber.)

Call two sequences $$(a_n)$$ and $$(b_n)$$ tail-equivalent if there are $$p$$ and $$q$$ such that $$a_{p+n} = b_{q+n}$$ for every $$n \geqslant 0$$. Write $$W(t)$$ rather than $$W_t$$.

Suppose that $$f$$ with the desired property exists, that both $$W(t)$$ and $$V(t)$$ take every real value, and that $$V(t)$$ is not monotone on any interval (which, of course, happens with probability one). The argument is divided into a number of steps.

1. For some one-to-one function $$\phi$$, we have $$V(f(t)) = \phi(W(t)) .$$ Indeed: for every $$x$$ find an arbitrary $$T(x)$$ such that $$W(T(x)) = x$$, and set $$\phi(x) = V(f(T(x))$$. Then $$V(f(t)) = \phi(x)$$ if and only if $$V(f(t)) = V(f(T(x)))$$, that is, $$W(t) = W(T(x))$$, that is, $$W(t) = x$$.

2. The function $$\phi$$ is in fact strictly monotone. Indeed: for a given $$x \in \mathbb{R}$$, there is $$T(x) \geqslant 0$$ such that $$W(T(x)) = x$$ and for every neighbourhood $$I$$ of $$T(x)$$, the set $$W(I)$$ contains a right neighbourhood of $$x$$. Thus, $$\limsup_{y \to x} \phi(y) \leqslant \limsup_{t \to T(x)} \phi(W(t)) = \limsup_{t \to T(x)} V(f(t)) = V(f(T(x))) = \phi(W(T(x))) = x$$. A similar argument shows that $$\liminf_{y \to x} \phi(y) \geqslant \phi(x)$$. Thus, $$\phi$$ is continuous at $$x$$. Since $$x$$ is arbitrary, $$\phi$$ is continuous, and hence (being one-to-one) strictly monotone.

3. With no loss of generality we assume $$\phi$$ is strictly increasing. The other case is dealt with in a similar manner.

4. Denote $$I(t) = \inf_{s \in [0, t]} W(s)$$ and $$M(t) = \sup_{s \in [0, t]} W(s)$$. Let $$T_0 > 0$$ be chosen arbitrarily (e.g. $$T_0 = 1$$), and define $$T_{2n+1} = \inf\{t > T_{2n} : W(t) < I(T_{2n})\}$$ and $$T_{2n+2} = \inf\{t > T_{2n+1} : W(t) > M(T_{2n+1})\}$$ In other words $$T_{2n+1}$$ is the first time $$W(t)$$ exceeds its current infimum after $$T_{2n}$$, and $$T_{2n+2}$$ is the first time $$W(t)$$ exceeds its current supremum after $$T_{2n+1}$$. Then $$T_n$$, $$W(T_{2n})$$ and $$-W(T_{2n+1})$$ all go to infinity as $$n \to \infty$$. (It is good to make a picture here.)

5. The tail-equivalence class of $$(T_n)$$ does not depend on $$T_0$$. Indeed: suppose that $$T_0' > T_0$$ and $$T_n'$$ is defined in a similar way as $$T_n$$, but with $$T_0$$ replaced by $$T_0'$$. Clearly, $$T_0' \in [T_{2n}, T_{2n+2})$$ for some $$n$$. If $$T_0' < T_{2n+1}$$, then $$T_1' = T_{2n+1}$$ and consequently $$T_k' = T_{2n+k}$$ for $$k > 0$$. If $$T_0' \geqslant T_{2n+1}$$, then either $$T_1' \in [T_{2n+1}, T_{2n+2})$$ and consequently $$T_k' = T_{2n+k}$$ for $$k > 1$$, or $$T_1' \in [T_{2n+3}, T_{2n+4})$$, and consequently $$T_k' = T_{2n+2+k}$$ for $$k > 1$$. (Looking at a picture helps a lot here.) Either way, $$T_n$$ and $$T_n'$$ are tail-equivalent.

6. Recall that $$W(t)$$ exceeds its past supremum at $$T_{2n}$$; that is, there is a sequence $$\epsilon_k > 0$$ convergent to zero, such that $$X(T_{2n} + \epsilon_k) > M(T_{2n})$$ for every $$k$$. Therefore, $$f(T_{2n} + \epsilon_k) \notin f([0, T_{2n}])$$. This means that $$f(T_{2n})$$ is one of the endpoints of $$f([0, T_{2n})$$. A similar argument shows that $$f(T_{2n+1})$$ is one of the endpoints of $$f([0, T_{2n+1}])$$.

7. For the next few items, suppose that $$\phi$$ is unbounded both from below and from above. Choose $$n$$ large enough, so that $$\phi(W(T_{2n})) > \sup_{s \in [0, f(0)]} V(s), \qquad \phi(W(T_{2n+1})) < \inf_{s \in [0, f(0)]} V(s).$$ By item 6, $$f(T_{2n})$$ is one of the endpoints of $$f([0, T_{2n}])$$, and by the above condition, $$f(T_{2n})$$ does not belong to $$[0, f(0)]$$. Therefore, $$f(T_{2n})$$ is the right endpoint of $$f([0, T_{2n}])$$. Similarly, $$f(T_{2n+1})$$ is the right endpoint of $$f([0, T_{2n}])$$. This means that $$f(T_n)$$ is eventually non-decreasing, and for $$n$$ large enough, $$f(T_{2n})$$ is the first time $$V(s)$$ exceeds its past supremum after $$f(T_{2n-1})$$, and $$f(T_{2n+1})$$ is the first time $$V(s)$$ exceeds its past infimum after $$f(T_{2n})$$.

8. Define the sequence $$S_n$$ in a similar way as $$T_n$$, but using $$V(t)$$ rather than $$W(t)$$. The previous item shows that $$f(T_n)$$ and $$S_n$$ are tail-equivalent.

9. Define $$A_n = 1$$ if there are $$s, t \in [T_{2n+1}, T_{2n+2}]$$ such that $$\text{s < t, W(s) > W(T_{2n}) and W(t) < W(T_{2n+1}),}$$ and $$A_n = 0$$ otherwise. (Again, have a look at the picture.) By the strong Markov property, $$A_n$$ is an i.i.d. sequence of (non-trivial) Bernoulli random variables with some parameter $$p \in (0, 1)$$. Define in a similar way $$B_n$$, using $$S_n$$ and $$V(t)$$ rather than $$T_n$$ and $$W(t)$$. Finally, let $$C_n$$ be a similarly defined sequence for $$f(T_n)$$ and $$V(t)$$ rather than $$T_n$$ and $$W(t)$$.

10. By continuity of $$f$$, $$A_n = 1$$ implies $$C_n = 1$$ (but not necessarily vice versa). Therefore, $$C_n \geqslant A_n$$. Furthermore, $$B_n$$ is tail-equivalent to some $$C_n$$. At the same time, independence of $$W(t)$$ and $$V(t)$$ implies that $$B_n$$ and $$C_n$$ are independent.

11. The probability that, given two independent i.i.d. Bernoulli sequences $$A_n$$ and $$B_n$$ (with the same parameter $$p$$), there is a third sequence $$C_n$$ such that $$C_n \geqslant A_n$$ and $$C_n$$ is tail-equivalent to $$B_n$$, is easily found to be zero. This proves that the probability that a function $$f$$ with the desired property exists, and additionally the corresponding $$\phi$$ is unbounded both from below and from above, is necessarily zero.

12. We now turn our attention to the case when $$\phi$$ is bounded from below or bounded from above. Clearly, it is sufficient to consider the case when $$\phi$$ is bounded from below. The argument is here more sketchy, but I will try to fill in the details later.

13. If $$\phi(x) \geqslant c$$ for every $$x$$, then $$f$$ is bounded (for $$f$$ necessarily takes values in a connected component of $$\{s : V(s) \geqslant c\}$$). If we define $$T_n$$ as in item 7, then one can show that the sequences $$f(T_{2n})$$ and $$f(T_{2n+1})$$ are eventually monotone, and one of them is eventually increasing, and the other eventually decreasing. (Otherwise, the path of $$V_s$$ would have an infinite number of oscillations of a fixed size over a finite time horizon, a contradiction with continuity.) With no loss of generality we consider the case where $$f(T_{2n})$$ is increasing for $$n \ge N$$, and $$f(T_{2n+1})$$ is decreasing for $$n \ge N$$.

14. Since $$V(s)$$ is not monotone on any interval, it attains a local extremum somewhere in the interior of $$f([T_{2N+1}, T_{2N+2}])$$. Let $$V(s_0) = y_0$$ be this local extremum, and let $$y_0 = \phi(x_0)$$. What we have found above implies that $$W(t)$$ attains a local extremum equal to $$x_0$$ in every interval $$[T_{2n+1}, T_{2n+2}]$$, $$n = N, N+1, \ldots$$ However, with probability one, the local extrema of $$W(t)$$ are all distinct. Thus, the probability that a function $$f$$ with the desired property exists, and additionally the corresponding $$\phi$$ is bounded both from below or from above, is necessarily zero.

The desired result follows.

• Ah, of course you're right. I somehow thought that $f$ must map level sets of $W_t$ into the level sets of $V_t$ corresponding to the same value. I will think about the fix. – Mateusz Kwaśnicki Nov 8 at 19:55
• I have heavily edited the answer. I did not have time to think carefully about the details, so I will not be surprised if it is still wrong. And in any case it is rather sketchy; sorry for that. – Mateusz Kwaśnicki Nov 8 at 22:39
• I have added some details to the first two claims without a rigorous proof. – Mateusz Kwaśnicki Nov 8 at 23:46
• Have you seen the edit at the bottom (the "Easy-to-see 1")? Essentially: $\phi$ is monotone, because it is continuous; and it is continuous, because $\phi(W_t)$ is continuous, and $W_t$ is sufficiently "generic" with probability one. (In fact it is sufficient to know that $W_t$ takes every real value, but that would take longer to write.) – Mateusz Kwaśnicki Nov 9 at 8:45
• There was a problem with the third "easy-to-do". I added additional details, but some are still missing. – Mateusz Kwaśnicki Nov 9 at 21:00