# How much time does a function spend above or below its average value around a point?

Given a locally integrable function $$f: \mathbb R \to \mathbb R$$, define $$K: \mathbb R \times \mathbb R+ \to \mathbb R$$ by

$$K(x, r) := \begin{cases} 1, & \text{if }f(x) > \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ -1, & \text{if }f(x) < \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ 0,& \text{if }f(x) = \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x \end{cases}$$

Define $$U: \mathbb R \to [-1, 1]$$ and $$L: \mathbb R \to [-1, 1]$$ by

$$\begin{split} U(x) &:= \limsup_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t\\ L(x) &:= \liminf_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t \end{split}$$

Intuitively, $$U$$ and $$L$$ represent the weighted proportion of time a function spends above or below its average value in an infinitesimal neighbourhood of a point.

i) Is it true that for any locally integrable function, $$U = L$$ a.e?

ii) Is it true that $$U = 1, 0, or -1$$ a.e.?

We choose $$f(t)=W_t$$ with $$W_t$$ a Brownian on $$\mathbb{R}$$. Because one can calculte $$U(0)$$ and $$L(0)$$ from the Brownian motion restricted to any neighbourhood of $$0$$, we can apply Blumenthal’s 0-1 law: There exists $$c_1,c_2\in \mathbb{R}$$ such that $$\mathbb{P}(U(0)=c_1)=1$$ and $$\mathbb{P}(L(0)=c_2)=1$$. By symetry we have $$c_1 =-c_2$$. Moreover for any $$0 there exists $$\epsilon$$ such that for all $$r>0$$ $$\mathbb{P}\Big(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \geq a\Big)>\epsilon$$ Therefore $$c_1\geq a$$ and we conclude that $$c_1=1$$ and $$c_2=-1$$.
Finally by translation invariance of the Brownian motion we have for all $$t\in \mathbb{R}$$ $$\mathbb{P}(U(t)=1)=\mathbb{P}(L(t)=-1)=1$$
• How do you show that for each $a\in(0,1)$ there is some $\epsilon[>0?]$ such that for all $r>0$ $P(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \ge a)>\epsilon$? Also, even if your have $P(U(t)=1)=P(L(t)=-1)=1$ for all $t$, how do you deduce from this that $P(U(t)=1\ \forall t)=1$? Apr 29, 2020 at 18:36
• @losif Pinelis: The Brownian motion is scale invariant ($f^{(r)}(t):=\frac{1}{\sqrt{r}}W_{rt}$ is also a brownian motion). So $K(0,t)$ have the same law as $K(0,r^{-1}t)$. So $\mathbb{P}(\frac{1}{2r}\int K(0,t)dt\geq a)$ is independant of $r$. Apr 30, 2020 at 7:16
• @losif Pinelis : The second point you mention is not true for all t but almost all $t$. Calling $\Omega$ the probability space of the brownian motion then $\int_{-L}^L\int_{\Omega} 1_{U(t)=1}dt d\mu=2L=\int_{\Omega} \int_{-L}^L1_{U(t)=1}dt d\mu$. So with probability 1, for almost all $t\in [-L,L]$ $U(t)=1$ Apr 30, 2020 at 7:20