Why can $\int_0^t f''(X_s) \, d\langle X \rangle_s$ not be a local martingale?

Question

We know from Itos formula, if $X$ is a continuous local martingale and $f$ has two continuous derivatives, we can write $f(X_t)$ as

$$ f(X_t) = \int_0^t f'(X_s) dX_s + \frac{1}{2} \int_0^t f''(X_s) d\langle X \rangle_s + f(X_0) $$

Now my stochastic calculus book says:

"$f(X_t)$ is always a semimartingale but it is not a local martingale unless $f''(x) = 0 $ for all $x$"

For me its clear that $f(X_t)$ is a semimartingale, since the first integral is a continuous local martingale and the second integral is of bounded variation. But my question is, why can $f(X_t)$ only be a local martingale if $f''(x)$ is equal to zero. Why can the second integral not be a continuous local martingale?

Answer 1

This follows from the following theorem:

Theorem: Let $(M_t,\mathcal{F}_t)_{t \geq 0}$ be a martingale such that the sample paths are (almost surely) continuous and of bounded variation. Then $M_t = M_0$ almost surely.

(For a proof see e.g. René Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes.)

Since $$M_t := \int_0^t f''(X_s) \, d\langle X \rangle_s$$ has continuous sample paths and is of bounded variation, this means that $M_t$ cannot be a (locale) martingale (unless it is constant).