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We may define a semimartingale as: $$X_{t}=X_{0}+M_{t}+A_{t}$$ Where $X_0$ is $\mathcal{F}_0$ measurable, $M$ is a continuous local martingale $M_0 = 0$ and $A$ is an adapted continuous finite variation process with $A_0 = 0$. The Itô's formula reads: $$df(t,X_t) =\frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial X_s}\,dX_t+\frac{1}{2}\frac{\partial^2f}{\partial X_s^2}d[X]_t$$

$[X]_t$ is a quadratic variation and hence a finite variation process, is $\frac{\partial^2f}{\partial x^2}d[X]_t$ also a finite variation process?

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    $\begingroup$ @Calculon I am not sure what you mean. The first ($dt$, which is missing) and last ($d[X]_t$) parts of the formula give finite variation processes, while the $dX_t$ part is a semimartingale. This is the nice thing about semimartingales, they can be integrated, and are stable by composition. $\endgroup$
    – Raoul
    Commented Jan 3, 2021 at 11:22

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Since $[X]_t$ is of finite variation, $\int f_{xx}(s, X_s)d[X]_s$ is interpreted in the Lebesgue-Stieltjes sense. This integral itself produces finite variation functions. You can see that by decomposing $f_{xx}(s, X_s)$ into positive and negative parts and $d[X]_s$ as the difference of two nondecreasing functions.

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