Show that a process is no semimartingale _Hello everyone! 
I got a little question about how to show that the process $X_t:=|B_t|^{\frac{1}{3}}$ is NOT a semimartingale.
So far I tried to apply Ito. Since if $X_t$ was a semimartingale so is $F(X_t)$ for every $C^2$ function $F$. Unfortunately I found no appropriate function. I tried to use $F(x)=x^3$ but, trying to show that $Y_t:=|B_t|$ is no semimartingale is in fact the same problem.
The only thing I found out so far, is that $X_t$ is NO local martingale (since if it was, it would be positive, starting in 0, hence being a supermartingle, hence being convergent, hence having a contradiciton).
Thank you!
 A: Suppose that $X_t := |B_t|^{\frac{1}{3}}$ is a semimartingale. Then, we have by Itô's formula
$$|B_t| = X_t^3 = 3 \int_0^t X_s^2 \, dX_s + 3 \int_0^t X_s \, d \langle X \rangle_s. \tag{1}$$
Moreover, Tanaka's formula shows
$$L_t = \int_0^t 1_{\{B_s=0\}} dL_s = - \int_0^t 1_{\{B_s=0\}} \text{sign}(B_s) \, dB_s + \int_0^t 1_{\{B_s=0\}} \, d|B|_s$$
where $(L_t)_{t \geq 0}$ denotes the local time of $(B_t)_{t \geq 0}$ at $0$. Using Itô's isometry and Fubini's theorem, it is not difficult to show
$$\int_0^t 1_{\{B_s=0\}} \text{sign}(B_s) \, dB_s = 0$$
Consequently, we find by $(1)$
$$\begin{align*} L_t &= \int_0^t 1_{\{B_s=0\}} \, d|B|_s =3 \int_0^t X_s^2 1_{\{B_s=0\}} \, dX_s + 3 \int_0^t X_s 1_{\{B_s=0\}} \, d \langle X \rangle_s = 0\end{align*}$$
as $\{X_s=0\} = \{B_s=0\}$. Hence, again by Tanaka's formula,
$$|B_t| = \int_0^t \text{sign}(B_s) \, dB_s.$$
Since the right-hand side defines a one-dimensional Brownian motion (and the left-hand side is, obviously, not a Brownian motion since it is positive), we reach a contradiction. Consequently, $|B_t|^{1/3}$ is not a semimartingale.
Remark The proof is compiled from Philip E. Protter: Stochastic integration and differential equations, Theorem 71, Chapter IV. There, the result is shown for any continuous local martingale.
