$C^1$ functions of continuous semimartingale Given any continuous semimartingale $X_t$ and any $C^1$ function $F:\mathbb{R}\to\mathbb{R}$ is $F(X)_t$ still a semimartingale?
 A: Here is an example where $F(X)$ is not a semimartingale. Let $X$ be a standard Brownian motion, and
$$
F(x)=\begin{cases}
x^3\sin(1/x),&{\rm if\ }x\not=0,\\
0,&{\rm if\ }x=0.
\end{cases}
$$
This is $C^1$, and the second derivative is,
$$
F^{\prime\prime}(x)=-x^{-1}\sin(1/x)+O(1)
$$
(for small $x$) so that the integral of $F^{\prime\prime}$ is infinite in any neighbourhood of the origin. I will show that $F(X)$ is not a semimartingale. Suppose that, on the contrary, $F(X)$ is a semimartingale. By standard decomposition of semimartingales, $F(X)=M+V$ for continuous local martingale $M$ and (locally) finite variation process $V$. If $\theta\colon\mathbb R\to\mathbb R$ is a bounded measurable function which is zero in a neighbourhood of the origin, then Ito's formula can be applied,
$$
\int\theta(X)dM+\int\theta(X)dV=\int \theta(X)F^\prime(X)dX+\frac12\int\theta(X)F^{\prime\prime}(X)dt.
$$
If you are concerned about applying Ito's formula when $F$ is not $C^2$ at the origin, you can apply it instead to $g(x)F(x)$ where $g$ is smooth, equal to $1$ on the support of $\theta$ and $0$ in a neighbourhood of $0$, to obtain the same result.
Using the uniqueness of the semimartingale decomposition, 
\begin{align}
\int_0^t\theta(X)dV&=\frac12\int_0^t\theta(X)F^{\prime\prime}(X)dt\\ &=\frac12\int_{-\infty}^{\infty}\theta(x)F^{\prime\prime}(x)L^x_tdx
\end{align}
where $L^x$ is the local time of $X$ at $x$, which is continuous with compact support in $x$ and $L^0_t > 0$ (almost surely) at positive times $t$. Hence, setting $\theta(x)={\rm sgn}(F^{\prime\prime}(x))1_{\{\lvert x\rvert > \epsilon\}}$ for positive $\epsilon$,
$$
\int_0^t\theta(X)dV=\frac12\int_{-\infty}^{\infty}1_{\{\lvert x\rvert > \epsilon\}}\lvert F^{\prime\prime}(x)\rvert L^x_tdx.
$$
Letting $\epsilon$ decrease to zero, the left hand side is bounded by the variation of $V$, but the right hand side almost surely goes to infinity, giving a contradiction.
