0
$\begingroup$

Let $\{L_x(t):(t,x)\in [0,T]\times\mathbb R\}$ be the local time of a Brownian motion $(B_t)_{t\in [0,T]}$, I know that the map $x\mapsto L_x(t)$ is $\alpha$-Hölder for $\alpha<1/2$ (uniformly in time?) continuous but by no means this implies differentiability.

Let $\xi\in S(\mathbb R)$ then using the occupation time formula we have that

\begin{align*} \int_0^t\xi'(B_s)ds=\int_{\mathbb R}\xi'(y)L_{y}(t)dy<\infty,\; \xi\in S(\mathbb R) \end{align*} and the right hand side equals $-\langle \xi, \partial_x L_x(t)\rangle$ where the angle brackets denote the dual pairing between the space $S(\mathbb R)$ and its dual. Thus we can conclude that the local time $L_x(t)$ has a spatial derivative that is a tempered distribution.

Can we say something else regarding the regularity of this object?

Thanks in advance.

$\endgroup$
7
  • 1
    $\begingroup$ Have you read the Wikipedia page for local times? Section 3 answers your question. $\endgroup$ Commented Oct 18, 2021 at 10:50
  • 1
    $\begingroup$ @Chaos: What kind of other regularity results are you thinking of? $\endgroup$ Commented Oct 18, 2021 at 10:53
  • $\begingroup$ Thanks! @MartinHairer Do you mean the Ray–Knight theorem? How can I use it could you mind giving me some hint? $\endgroup$
    – Chaos
    Commented Oct 18, 2021 at 11:12
  • $\begingroup$ @MateuszKwaśnicki Well, I was hoping that the derivative was actually a function $\endgroup$
    – Chaos
    Commented Oct 18, 2021 at 11:13
  • $\begingroup$ @Chaos: No, the local time is as irregular as the Wiener process itself. $\endgroup$ Commented Oct 18, 2021 at 11:57

0

You must log in to answer this question.