I would like to show that $f(x_1,x_2,x_3)=\frac{1}{\sqrt{(x_1^2+x_2^2+x_3^2)}}$ is a local martingale. I know that for $B_t=((B_1)_t,(B_2)_t,(B_3)_t)$ the Itô formula reads $$df(t,B_t)=f_t(t,B_t)dt+df(t,B_t)\cdot dB_t+\frac{1}{2}\Delta f(t,B_t)dt.$$ Now I know $f_t(x)=0$ and $\Delta f(x)=0 for x\neq 0$. Now if the above was true for every $x$, I could use Itô's formula to deduce $$f(B_t)=f(B_0)+\int_0^t \nabla f(B_s)\cdot dB_s$$ which I think is a local martingale. But $f(B_0)$ is not defined and moreover $f$ is not smooth in $0$ so I am not sure I can apply Itô's formula.
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1$\begingroup$ You have to start $B$ away from zero. Then for $\delta> 0$ define $f^\delta$ as a function which equals $f(x)$ for $|x| \geq \delta$ and is smoothly truncated in an $\delta$ neighbourhood of zero. Define the stopping time $T_\delta = \inf\left\{t \geq 0 : \left|B_t \right|\leq \delta\right\}$. Now you can apply Itô's formula to $f^\delta$ and $B^{T_\delta}$. You may now deduce that $f\left(B_t\right)$ is a local martingale by taking the limit as $\delta \to 0$ of $f^\delta\left(B^{T_\delta}\right)$ as $T^\delta\to \infty$ because $B$ almost surely doesn't return to the 0 in dimension 3. $\endgroup$– ShivaCommented Dec 19, 2021 at 13:47
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$\begingroup$ @Shiva, thank you. Isn't $T_{\delta}=0$ since $B_0=0$? What is $B^{T_{\delta}}$? Is it the stopped process $B_{t\wedge T_{\delta}}$? As I understand, you first prove that $B_{t\wedge T_{\delta}}$ is a martingale. But how can you deduce that $f(B_t)$ is a martingale for every $\delta>0$? I am also confused because $f^{\delta}(B^{T_{\delta}})$ seem to converge to $0$ in $0$. $\endgroup$– edamondoCommented Dec 19, 2021 at 14:04
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$\begingroup$ You have start $B$ away from zero, i.e $B_0 = x \neq 0$. Otherwise $f\left(B_0\right)$ is not well defined nor are its derivatives. Yes, $B^{T_\delta}$ is the stopped process, $\left(B_{t\wedge T_\delta}\right)_{t\geq 0}$. It's known that a stopped martingale is still a martingale, see Corollary 3.24 of JF Le Gall's Brownian Motion, Martingales, and Stochastic Calculus. Also do you see that for $T > 0$ and almost all $\omega \in \Omega$, $\sup_{0 \leq t \leq T} \left|B_t\left(\omega\right) - B_{t\wedge T_\delta\left(\omega\right)}\left(\omega\right)\right| \to 0$? $\endgroup$– ShivaCommented Dec 19, 2021 at 14:14
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$\begingroup$ Okay, so actually I did not get why $f^{\delta}(B_{t\wedge T_{\delta}})$ is a martingale implies $B_{t\wedge T_{\delta}}$ is a martingale $\endgroup$– edamondoCommented Dec 19, 2021 at 14:22
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$\begingroup$ It doesn't. The fact that $\left(B_{t}\right)$ is a martingale implies that $\left(B_{t\wedge T_\delta}\right)$. Most books will have a proof of this fact for general continuous or càdlàg martingales. $\endgroup$– ShivaCommented Dec 19, 2021 at 14:25
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As discussed in the comments, your Brownian motion should start away from zero. Itô's lemma implies that the class of semi-martingales is stable under applications of $C^2$ maps. Since $B$ is a martingale (hence a semi-martingale), and since $f$ is $C^2$ away from zero, then $f(B)$ is a semi-martingale provided that we can prove that the hitting time of zero is infinite almost surely. But this result is standard (see e.g. Corollary 2.23 in https://www.stat.berkeley.edu/~aldous/205B/bmbook.pdf).
answered Dec 19, 2021 at 18:11