If $M_t$ and $M^2_t-t$ are martingales, can we have $\langle M\rangle_t=t$ ? I meet this in the proof of Levy' theorem.
The levy's characterization of brownian motion can be stated as follows:
Suppose that both $M_t$ and $M^2_t-t$ are martingales. Then $M$ is a Brownian notion.
In the proof, we can frist do the Ito formular to $f\in C^2(\mathbb{R})$ which gives us $$ f\left(M_{t}\right)=f(0)+\int_{0}^{t} f^{\prime}\left(M_{s}\right) d M_{s}+\frac{1}{2} \int_{0}^{t} f^{\prime \prime}\left(M_{s}\right) d \langle M\rangle_t $$ where $\langle M\rangle_t$ is the quadratic variation. However, in the lecture notes, the proof says we can obtain that $$ f\left(M_{t}\right)=f(0)+\int_{0}^{t} f^{\prime}\left(M_{s}\right) d M_{s}+\frac{1}{2} \int_{0}^{t} f^{\prime \prime}\left(M_{s}\right) d s. $$ So, why we have $\langle M\rangle_t=t$ here?