Inverse of norm of Brownian motion is a semi-martingale.

Question

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.

Answer 1

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).