11
$\begingroup$

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 $\epsilon>0$,

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

$\endgroup$
  • $\begingroup$ 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 Dec 31 '14 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 Jan 16 '15 at 17:49
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Dec 25 '14 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 Dec 25 '14 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 Dec 28 '14 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 Dec 31 '14 at 8:46
0
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Nov 4 '15 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.