# local martingale and submartingale

Consider a local continuous martingale $$M_t$$ for $$t \in [0,T]$$. Taking a localizing $$\{\tau_k\}$$ yields that $$(M^2 - \langle M \rangle)(\tau_k \wedge t)$$ is martingale. How can I show that $$M^2(\tau_k \wedge t)$$ is a submartingale then.

I know that the quadratic variation is increasing and it holds:

$$\mathbb{E}[(M^2 - \langle M \rangle)(\tau_k \wedge t) | F_t]= (M^2 - \langle M \rangle)(\tau_k \wedge s)$$ How can I isolate $$M^2$$ from there?

By Jensen's inequality and the fact that $$(M_t)_{t \in \mathbb{R}^+}$$ is a local martingale: $$E[M^2_{t \wedge \tau_n}|\mathcal{F}_s]\geq (E[M_{t \wedge \tau_n}|\mathcal{F}_s])^2=M_{s \wedge \tau_n}^2$$ The claim follows.
You have $$\Bbb E[M^2_{\tau_k\wedge t}-\langle M\rangle_{\tau_k\wedge t}\mid\mathcal F_s]= M^2_{\tau_k\wedge s}-\langle M\rangle_{\tau_k\wedge s}.$$ Now move the $$\Bbb E[\langle M\rangle_{\tau_k\wedge t}\mid\mathcal F_s]$$ term to the right side: \eqalign{ \Bbb E[M^2_{\tau_k\wedge t}\mid\mathcal F_s] &= M^2_{\tau_k\wedge s}+\Bbb E[\langle M\rangle_{\tau_k\wedge t}\mid\mathcal F_s]-\langle M\rangle_{\tau_k\wedge s}\cr &= M^2_{\tau_k\wedge s}+\Bbb E[\langle M\rangle_{\tau_k\wedge t}-\langle M\rangle_{\tau_k\wedge s}\mid\mathcal F_s]\cr &\ge M^2_{\tau_k\wedge s},\cr } where the second equality is true because $$\langle M\rangle_{\tau_k\wedge s}$$ is $$\mathcal F_s$$-measurable, and the final inequality is true because $$u\mapsto\langle M\rangle_{\tau_k\wedge u}$$ is non-decreasing.