What follows, up to the horizontal line, is taken from Rogers "Arbitrage with fractional Brownian motion".

Consider an interval $[0,T]$ on which is defined the fractional Brownian motion $B$, and consider its partitions $\pi_n = \{t^n_k = \frac{kT}{n} : 0\le k\le n\},\ n\in\mathbb N$.

Let $p\ge1$, the $p$-variation of $B$ is $$ V_p(B) = \lim_{n\to\infty} \sum_{k=0}^{n-1} |B(t^n_{k+1})-B(t^n_k)|^p = \begin{cases} \infty, & \text{if }\ pH < 1, \\ T\mathbb E[|B(1)|^{1/H}], & \text{if }\ pH = 1, \\ 0, & \text{if }\ pH > 1. \end{cases} $$ If $H>1/2$ we can choose $p\in(1,\frac1H)$ so that $pH<1$, then the $p$-variation is infinite, hence the quadratic variation of $B$ is infinite too.

If $H<1/2$ we can choose $p>2$ so that $pH<1$, then again we obtain that the $p$-variation and the quadratic variation of $B$ are infinite.

In both cases the quadratic variation of $B$ is not finite, hence the fBm is not a semimartingale for $H\ne1/2$.

Could somebody further explain the above reasoning? I don't fully get what has to be proved, is it related to the fact that a semimartingale has to have finite variation? But which variation: quadratic, p-variation or another one?

Moreover, I don't understand how to deduce what the quadratic variation is, given that we know the p-variation. Is it related to the fact that given $p_1<p_2$ then $V_{p_2}\le V_{p_1}$?

Fianlly, what about the case $H=1/2$, in which $B$ is the usual Brownian motion? If we take $p\in(1,2)$ then we are still in the case $pH<1$ and so the $p$-variation is infinite hence the quadratic variation of $B$ is infinite too, contradicting the fact that B is a martingale.

What follows, up to the horizontal line, is taken from Rogers "Arbitrage with fractional Brownian motion".

Consider an interval $[0,T]$ on which is defined the fractional Brownian motion $B$, and consider its partitions $\pi_n = \{t^n_k = \frac{kT}{n} : 0\le k\le n\},\ n\in\mathbb N$.

Let $p\ge1$, the $p$-variation of $B$ is $$ V_p(B) = \lim_{n\to\infty} \sum_{k=0}^{n-1} |B(t^n_{k+1})-B(t^n_k)|^p = \begin{cases} \infty, & \text{if }\ pH < 1, \\ 0, & \text{if }\ pH > 1. \end{cases} $$ If $H>1/2$ we can choose $p\in(1,\frac1H)$ so that $pH<1$, then the $p$-variation is infinite, hence the quadratic variation of $B$ is infinite too.

If $H<1/2$ we can choose $p>2$ so that $pH<1$, then again we obtain that the $p$-variation and the quadratic variation of $B$ are infinite.

In both cases the quadratic variation of $B$ is not finite, hence the fBm is not a semimartingale for $H\ne1/2$.

Could somebody further explain the above reasoning? I don't fully get what has to be proved, is it related to the fact that a semimartingale has to have finite variation? But which variation: quadratic, p-variation or another one?

Moreover, I don't understand how to deduce what the quadratic variation is, given that we know the p-variation. Is it related to the fact that given $p_1<p_2$ then $V_{p_2}\le V_{p_1}$?

Fianlly, what about the case $H=1/2$, in which $B$ is the usual Brownian motion? If we take $p\in(1,2)$ then we are still in the case $pH<1$ and so the $p$-variation is infinite hence the quadratic variation of $B$ is infinite too, contradicting the fact that B is a martingale.