# $C^1$ 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?

• I view the "close" vote as particularly harsh given the fact that the question is about math, not trivial, is clearly stated, and the person asking it quite new on MSE. Dec 2, 2019 at 13:34
• I have an example from Protter's book on stoachsitc integation and Differtal Equation at Theorem 71 chapter IV section 7 page 217, but it is not fully $C^1$ on $\mathbb R$ only on $\mathbb R \{0\}$. There you take $F(x)=|x|^{\alpha}$ with $\alpha\in (0,1/2)$ and any continuous (local) martingale $X$ starting at 0. The proof is completely non trivial (this is for your fan who wants to close this good question). Can someone fill the gap with an example fully $C^1$ ? Dec 3, 2019 at 13:45

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.

• Thank you so much!!! Dec 6, 2019 at 4:27
• Still the best of the best ... Dec 6, 2019 at 17:51