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Let $\mathrm{ Lip} (M)$ denote the space of all functions on $[0,T]$ with Lipschitz constant and $L^\infty$ norm bounded by $M$. Let $(B_t)_t$ be a Brownian motion defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. Does the following lemma hold?

Lemma: For all $M>0$ and almost all $\omega\in\Omega$, there is a constant $C=C(M,\omega)$, such that for all $\epsilon>0$,

$$\sup_{g\in \mathrm{ Lip}(M)} \int_0^T 1_{|B_t+g(t)|<\epsilon}dt<C\epsilon.$$

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  • $\begingroup$ The lemma follows from this paper: arxiv.org/pdf/math/0002012v1 (see also later erratum). Theorem 1.2 is an informal restatement of Theorems 3.6 and 3.8, and $L^g$ of Theorem 1.2 is replaced by a version $\tilde{L}^g$ in Theorem 3.6. For my application of the lemma, it is important to take supremum over all $g$ (i.e. not allow for ignoring single 'bad' functions $g$), since $g$ will be adapted to $B$. However, it is sufficient to prove the lemma for $g$ in a countable dense subset of Lip$(M)$, so the discussion in the proof of Theorem 3.6 should be sufficient to imply the lemma. $\endgroup$
    – Nina Holden
    Commented Dec 31, 2014 at 8:26
  • $\begingroup$ I looked more closely at the paper by Bass and Burdzy referred to above, and I don’t think anymore that the lemma of my initial question follows from their paper. The paper only shows that the local time $L^g$ has a version $\tilde{L}^g$ which is bounded over all $g\in\mathrm{Lip}(M)$, not that the local time $L^g$ itself is bounded. The lemma is therefore still not proved. $\endgroup$
    – Nina Holden
    Commented Jan 16, 2015 at 17:49

2 Answers 2

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From scaling relations, we may assume that $T=M=1$.

**** The argument I sketch below only gives a weaker result than in the OP, so the question is still open. Thanks to Nina for pointing this out. ****

Step 1: Introduce the random variable $$Z_\epsilon=\sup_x \int_0^\epsilon 1_{|B_t-x|<2\epsilon} dt \leq \sup_{j} \int_0^\epsilon 1_{|B_t-j\epsilon|<4\epsilon}dt, $$ and set $M_\epsilon=Z_\epsilon/\epsilon^{3/2}$. Using the heat kernel and the second inequality above, we obtain that $EM_\epsilon\leq C_1$ and $EM_\epsilon^2\leq C_2$. (This requires a detailed computation, which I am skipping here, so this needs to be double checked. There may be a log correction.)

Step 2: Let $M_\epsilon^{(i)}$ be independent copies of $M_\epsilon$. By conditioning on the Brownian motion at times $(i-1)\epsilon$ and forgetting about the Lipschitz condition between the intervals, you get that the random variable in the OP is stochastically dominated by $$\epsilon^{1/2} \cdot \epsilon \sum_{i=1}^{1/\epsilon} M_\epsilon^{(i)}=: \epsilon^{1/2} S_{\epsilon}\,.$$

Step 3: We have that $ES_\epsilon=C_1$ and therefore, by Markov's inequality and the estimate on $EM_\epsilon^2$, $$P(S_\epsilon>2C_1)\leq P(S_\epsilon-ES_\epsilon\geq C_1)\leq \epsilon \frac{C_2}{C_1^2}\,.$$

Step 4: By interpolation, it is enough to consider the sequence $\epsilon_j=2^{-j}$ and apply Borel-Cantelli to conclude the estimate $\epsilon^{1/2}$ in the right hand side of the OP.

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  • $\begingroup$ Thanks Ofer. But why do we have $EM_\epsilon\leq C_1$? Defining the Brownian motion $\widehat{B}$ by $B_t=\epsilon \widehat{B}_{t\epsilon^{−2}}$ and change of variables $s=t\epsilon^{-2}$, we get $M_\epsilon=\sup_x \int_0^{ϵ^{-1}} 1_{|\widehat{B}_s−x|<2}ds$, which diverges when $\epsilon\rightarrow 0$. $\endgroup$
    – Nina Holden
    Commented Dec 25, 2014 at 15:38
  • $\begingroup$ I did not check step 1 carefully, and you are right, it is flawed. I think that it can be rescued (working with longer intervals), I'll check and will let you know. $\endgroup$
    – ofer zeitouni
    Commented Dec 25, 2014 at 17:10
  • $\begingroup$ Thanks! I think the logarithmic correction is not necessary for the moments. Defining the Brownian motion $\tilde{B}$ by $\tilde{B}_t=\epsilon^{-1/2}B_{\epsilon t}$, and letting $L_t^x$ denote local time at $x$ at time $t$, we have $\epsilon^{-3/2}Z_\epsilon= \epsilon^{-1/2}\sup_x\int_0^1 1_{|\tilde{B}_s-x|<2\epsilon^{1/2}}\,ds \leq 4\sup_x L_1^x(\tilde{B})$. The $p$th moment of the right-hand side is bounded by $C_p$, see e.g. "(Semi-) martingale inequalities and local times" by Barlow and Yor $\endgroup$
    – Nina Holden
    Commented Dec 28, 2014 at 1:13
  • $\begingroup$ I posted a comment above with an already published paper implying the lemma. The paper use some similar ideas to the proof sketched above: The number of rectangles of side length $\sim N^{-1}$ containing both $-g$ and $B$, is bounded by $cN^{1/2+\epsilon}$ with high probability, and this is used to prove uniform continuity of $g\mapsto L^g(B)$ on a countable dense subset of $\mathrm{Lip}(M)$. $\endgroup$
    – Nina Holden
    Commented Dec 31, 2014 at 8:46
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I first attempted to make this a comment, but lacked the reputation for it. The following is a cryptic sketch which you should consider:

Choose some $g$ which is $Lip(M)$ and define $$W_t := B_t + g(t).$$ Under an explicit change of measure, i.e. under Girsanov, $W_t$ becomes a Brownian motion.

We can therefore find a specific $C$ for each $g,$ and we know how to bound the moments of $C$ using the explicit change of measure. (Since the local time of a BM is distributed as its maximum below zero), and these will be computed using the measure change, so the moment bounds will be in terms of $M.$

Do this for a dense set of $g$ and pass to a limit?

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  • $\begingroup$ Thanks very much for the reply. It turned out the result of Bass and Burdzy was sufficient. We first prove continuity of $f\mapsto \int_0^T 1_{B_t\geq f(t)}dt$, where $f\in Lip(M)$ and we use the supremum norm on $Lip(M)$. Then we prove Lipschitz continuity of this map on a dense countable set of $Lip(M)$ by using the result of Bass and Burdzy. $\endgroup$
    – Nina Holden
    Commented Nov 4, 2015 at 14:38

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