I was reading Chapter 3 of Brownian Motion and Stochastic Calculus by Karatzas and Shreve and Exercise (6.6) asks us to prove the following(I also include the screenshot of definition 6.3)
I tried unsucessfully to reach a contradiction which according to the author is an easy proof but I have had no luck. Can anybody help me here?
Here is a proposition, I'm really not sure 100% it's ok, so let's wait for reviews from the audience of the forum before accepting this solution :
So let's fix $\omega \in \Omega^*$ where $\Omega^*$ is such that $L_t(x)$ continuous. Now we have by definition :
$$\Gamma_t[B=C.h,\omega] =\int_{C.h}L_t(x)dx$$ (where $B=C.h$ is the ball of center $W_t(\omega)$ and radius $C.h$ omitting dependency in $\omega$ and everything here being deterministic).
We also have : $$\Gamma_{t+h}[C.h,\omega] =\int_{C.h}2.L_{t+h}(x)dx$$
Now using the fact that :
$$\lim\limits_{h\to 0}\frac{1}{h}(\Gamma_{t+h}[C.h]-\Gamma_{t}[C.h]) =\lim\limits_{h\to 0} \frac{1}{h}(\int_t^{t+h}1_{W_s \in C.h}ds)= \lim\limits_{h\to 0} \frac{h}{h} = 1 $$ Remember that by absurd reasoning the hypothesis $W_s-W_t<C.h, \forall s\in (t-h,t+h)$ and $h<\delta$ holds at $\omega$ so that $1_{W_s \in C.h}=1 \forall s\in (t,t+h)$.
We also have :
$$\lim\limits_{h\to 0} \frac{1}{h}(\Gamma_{t+h}[C.h]-\Gamma_{t}[C.h])= \lim\limits_{h\to 0} \frac{2}{h}(\int_{C.h} L_t(x) - L_{t+h}(x) dx) $$
But by joint-continuity of $L$ in t and x, for any $\epsilon>0$ there exist $\eta_1, \eta_2$ so that we can have $\forall x,y \in B(W_t(\omega), \eta_1)$ and $0<h<\eta_2$ :
$$L_{t}(x)-\epsilon <L_{t+h}(y)<L_{t}(x)+\epsilon $$
So taking $h< \eta_2\wedge \eta_1/C$ we get :
Using this in the preceding expression leads us to :
$$-2.C\epsilon=-\frac{2}{h}(\int_{C.h} \epsilon dx)<\frac{2}{h}(\int_{C.h} L_t(x) - L_{t+h}(x) dx)<\frac{2}{h}(\int_{C.h} \epsilon dx)= 2.C\epsilon$$
Implying as $\epsilon$ is arbitrary that $\lim\limits_{h\to 0} \frac{1}{h}(\Gamma_{t+h}[C.h]-\Gamma_{t}[C.h])=0$, which highlights the required contradiction.