Forgery theorem: the Brownian motion stays close to any curve with positive probability

Question

In a paper I am reading the authors claim that, if $B$ is a standard BM in $\mathbb{R}$ and $f\in C([0,1],\mathbb{R})$, then for any $\epsilon>0$ $$ \mathbb{P}(\sup_{t\in [0,1]}|B_t-f(t)|<\epsilon)>0. $$ They do not give a proof for this result so I assume it is somewhat well-known, but I don't seem to be able to prove it. Looking online, I found that this is Exercise $1.8$ in the book of Morters and Peres (https://people.bath.ac.uk/maspm/book.pdf). They give the following hint: Given $f\in C[0,1]$ and $\epsilon>0$ there exists n such that the function $g\in C[0,1]$, which agrees with $f$ on the dyadic points in $D_n$ and is linearly interpolated inbetween, satisfies $\sup|f(t)−g(t)|<\epsilon$. Then use Lévy’s construction of Brownian motion and the fact that normal distributions have full support to complete the proof.

The hint however seems not to work. In facts, when I approximate $B$ with an appropriate piecewise linear process $S^{(n)}$ with nodes at $D_n$, I find that $$ \forall \alpha>0\quad\forall \epsilon>0\quad \exists N\quad \forall n\geq N \quad \mathbb{P}(\sup_{[0,1]}|S^{(n)}_t-B_t|<\epsilon)>1-\alpha. $$ Hence (assuming $f$ piecewise linear with nodes at $D_n$) $$ \mathbb{P}(\sup_{[0,1]} |S^{(n)}_t-f(t)|<\epsilon)=:C(\epsilon,\alpha)>0. $$ Combining I get $$ \mathbb{P}(\sup_{[0,1]} |B_t-f(t)|<\epsilon)\geq C(\epsilon,\alpha)-\alpha, $$ which however might very well be negative depending on the behavior of $C(\epsilon,\alpha)$.

Any idea on how to fix things?

Answer 1

There was part of the hint that you ignored, namely using Levy's construction. To add some detail, the approximation of $S^{n}$ by BM is done recursively in each dyadic interval, rather than globally. For fixed $\epsilon>0$, it suffices to prove that if $n$ is large enough, then for any real $r$ and any $a \in (-\epsilon/2, \epsilon/2)$, the event $$A:=\{ |a+B(2^{-n})-r2^{-n}|<\epsilon/2 \quad \text{and} \quad \forall t \in [0,2^{-n}], \quad |a+B(t)-rt|<\epsilon,\},.$$ has positive probability. This follows easily from Levy's construction, or alternatively from the relation between Brownian motion and Brownian bridge.

See also the references in my previous answer The Wiener measure of an open set