Continuity, Support, and Scaling of Brownian Local Time

Local Time as a Continuous Random Field

Brownian local time records how much occupation accumulates near a specified level. The occupation density formula describes its integral over levels, but several pathwise questions require more: can the density be chosen continuous jointly in time and level, and can it increase at times when Brownian motion is away from the indexed level? Joint continuity makes both questions accessible. It also allows Brownian scaling to be expressed as a relation between entire local-time fields.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). Use the symmetric local-time normalization established for Brownian motion: for every \(t\ge0\) and bounded Borel function \(g\), the occupation density formula is \(\int_0^t g(B_s)\,ds=\int_{\mathbb{R}}g(a)L_t^a\,da\). In this normalization Tanaka’s formula is \(|B_t-a|=|a|+\int_0^t\operatorname{sgn}(B_s-a)\,dB_s+L_t^a\), where \(\operatorname{sgn}(x)=1\) for \(x>0\), \(-1\) for \(x<0\), and \(0\) for \(x=0\). Each stochastic integral here is an Itô integral: its integrand is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on every finite time interval. The sign integrand is bounded, so it satisfies this condition.

Increment Estimates and Joint Continuity

The key is to bound moments of increments in both parameters. We give the estimates on a finite time interval \([0,T]\) and a bounded level interval \([-R,R]\). Constants may depend on \(T\), \(R\), and the moment order, but not on the particular times or levels in these ranges.

Proof. First bound occupation of a level interval \(I\) of length \(\delta\). Put \(Y=\int_0^T\mathbf{1}_{\{B_s\in I\}}\,ds\). Expanding the \(m\)-th power and ordering the time variables gives \(\mathbb{E}Y^m=m!\int_{0<s_1<\cdots<s_m<T}\mathbb{P}(B_{s_j}\in I,\ 1\le j\le m)\,ds_1\cdots ds_m\). The Brownian transition density is bounded by \((2\pi r)^{-1/2}\) over a time gap \(r>0\). Applying this bound to the initial density and each successive transition density, then integrating the level variables over \(I\), yields \(\mathbb{P}(B_{s_j}\in I,\ 1\le j\le m)\le \delta^m(2\pi)^{-m/2}s_1^{-1/2}\prod_{j=2}^m(s_j-s_{j-1})^{-1/2}\). The integral of the product of gap powers over the time simplex is \(T^{m/2}\Gamma(1/2)^m/\Gamma(m/2+1)\), by the Dirichlet beta integral. Therefore \(\mathbb{E}Y^m\le C_{m,T}\delta^m\).

For a spatial increment, subtract Tanaka’s formulas at levels \(a\) and \(b\). The difference of the two absolute-value terms, including their initial values, has magnitude at most \(2|a-b|\). The difference of the sign integrands vanishes outside the interval between \(a\) and \(b\), up to its endpoints, and has magnitude at most \(2\). Brownian motion spends zero Lebesgue time at either fixed endpoint: this follows from the occupation formula because a singleton has zero Lebesgue measure in the level integral. Thus the quadratic variation of the difference stochastic integral is at most \(4Y\), where \(Y\) is the occupation time of the interval between the levels. The Burkholder–Davis–Gundy inequality gives its \(2m\)-th moment at most a constant times \(\mathbb{E}Y^m\), hence at most \(C|a-b|^m\). The deterministic term contributes at most \(C|a-b|^{2m}\), which is bounded by another constant times \(|a-b|^m\) when \(a,b\in[-R,R]\). We have proved \(\mathbb{E}|L_t^a-L_t^b|^{2m}\le C|a-b|^m\).

For a time increment at a fixed level, subtract Tanaka’s formulas at \(t\) and \(s\). The absolute-value difference is bounded by \(|B_t-B_s|\). The stochastic integral over \([s,t]\) has quadratic variation at most \(t-s\), since its sign integrand has magnitude at most \(1\). The normal moment formula and the Burkholder–Davis–Gundy inequality therefore give \(\mathbb{E}|L_t^a-L_s^a|^{2m}\le C|t-s|^m\). The Burkholder–Davis–Gundy inequality is a standard martingale moment inequality; its proof is beyond this tutorial.

Finally, write \(L_t^a-L_s^b=(L_t^a-L_s^a)+(L_s^a-L_s^b)\) and use \(|x+y|^{2m}\le2^{2m-1}(|x|^{2m}+|y|^{2m})\). This proves the stated two-parameter estimate. On a rectangle, \(|t-s|^m+|a-b|^m\le C\|(t,a)-(s,b)\|^m\). Since \(m>2\), the moment exponent \(m\) exceeds the parameter-space dimension \(2\); the standard Kolmogorov continuity theorem gives a continuous version and every Hölder exponent below \((m-2)/(2m)\). The Kolmogorov theorem, like the martingale moment inequality, is a standard result whose proof is beyond this tutorial. Applying the result on a countable increasing sequence of rectangles gives one version continuous on all of \([0,\infty)\times\mathbb{R}\). \(\square\)

The continuity assertion concerns a single version on the whole parameter set, not a separate exceptional null set for every fixed level. This distinction matters when making pathwise claims for all levels or intervals at once.

Where Local Time Can Increase

Joint continuity turns the occupation formula into a support statement about the time measure. For \(0\le u\le v\), subtraction of the occupation formulas at \(v\) and \(u\) gives \(\int_u^v g(B_s)\,ds=\int_{\mathbb{R}}g(x)(L_v^x-L_u^x)\,dx\). The increment \(L_v^x-L_u^x\) is nonnegative for every \(x\). Indeed, it is nonnegative for almost every \(x\) because it is the density of a nonnegative occupation measure; continuity in \(x\) then makes it nonnegative everywhere. Taking rational \(u,v\) first and using continuity in time gives this monotonicity simultaneously for all \(u,v\).

Proof. Fix a path on the full-measure event where the occupation identity and joint continuity hold. Suppose \([u,v]\) is an interval on which \(|B_s-a|\ge\varepsilon\) for all \(s\in[u,v]\), for some \(\varepsilon>0\). Choose a nonnegative continuous function \(g\) supported in \((a-\varepsilon,a+\varepsilon)\) and strictly positive on a neighborhood of \(a\). The path never enters the support of \(g\), so \(\int_u^v g(B_s)\,ds=0\). The occupation identity says this equals \(\int g(x)(L_v^x-L_u^x)\,dx\). Its integrand is nonnegative and continuous. Since \(g\) is positive near \(a\), this forces \(L_v^a-L_u^a=0\); otherwise continuity would make the integral strictly positive. The open set of times where \(B_t\ne a\) is a countable union of intervals with rational interior endpoints on which the path is separated from \(a\) on each compact subinterval. The preceding argument shows that \(L^a\) is constant on each such subinterval. Thus its Stieltjes measure assigns zero mass to \(\{t:B_t\ne a\}\), proving the support claim. \(\square\)

One immediate consequence is that the local-time profile at a fixed time vanishes outside the path range. If \(m_t=\min_{0\le s\le t}B_s\) and \(M_t=\max_{0\le s\le t}B_s\), then \(L_t^a=0\) for \(a<m_t\) or \(a>M_t\). For such a level, the continuous path stays a positive distance from \(a\) throughout \([0,t]\), so the support proposition and \(L_0^a=0\) give the claim. No assertion of vanishing at the endpoint levels is needed.

Brownian Scaling for Local Time

For \(c>0\), define the rescaled process \(\widetilde B_t=c^{-1/2}B_{ct}\). Brownian scaling says that \(\widetilde B\) is again standard Brownian motion. The occupation density normalization determines exactly how its local time relates to that of \(B\).

Proof. For a bounded Borel function \(g\), change variables \(u=cs\) in the occupation integral and then \(x=\sqrt{c}\,y\) in the level integral: \(\int_0^t g(\widetilde B_s)\,ds=c^{-1}\int_0^{ct}g(B_u/\sqrt{c})\,du =c^{-1}\int_{\mathbb{R}}g(x/\sqrt{c})L_{ct}^x(B)\,dx =c^{-1/2}\int_{\mathbb{R}}g(y)L_{ct}^{\sqrt{c}\,y}(B)\,dy\). By the occupation formula for \(\widetilde B\), the left side is \(\int_{\mathbb{R}}g(y)L_t^y(\widetilde B)\,dy\). Thus the two continuous densities agree almost everywhere in \(y\); because both are continuous, they agree for every \(y\). This proves the pathwise relation for the corresponding continuous versions. Since \(\widetilde B\) has the same law as \(B\), the stated random-field equality in law follows. \(\square\)

Common Pitfall: Confusing the Level and Time Scales

The factor \(\sqrt{c}\) acts on the level because Brownian space scales as the square root of time. The local-time factor is the reciprocal in the pathwise expression for the rescaled process. Losing either factor gives an incorrect occupation density. For instance, omitting the change of variables \(x=\sqrt{c}\,y\) would incorrectly produce \(c^{-1}\), rather than \(c^{-1/2}\), in the density.

The occupation formula also explains why continuity is useful rather than cosmetic. A density is initially identified only up to changes on a set of levels of Lebesgue measure zero. Joint continuity removes that ambiguity: two continuous densities equal almost everywhere must agree at every level. This is what lets the support and scaling statements hold level by level, on a common full-measure event.