Certain transformation of Brownian motion is a submartingale

Question

I have a question about a proof in Protter. Let $B$ Brownian motion and $u$ a harmonic (subharmonic) function. Then $u(B)$ is a local martingale (submartingale). I was able to show the case of local martingale by myself. The case for subharmonic functino should be similar as Protter suggest. We have by Itô

$$u(B_t)=u(B_0)+\int_0^t\nabla u(B_s)dB_S+\frac{1}{2}\int_0^t \Delta u(B_s)ds$$

Using subharmonicity of $u$ we have:

$$u(B_t)\ge u(B_0)+\int_0^t\nabla u(B_s)dB_S$$

How does this imply the submartingale property of $u(B)$?

Answer 1

Fix $a<b$. After having applied Ito's formula, we get $$u(B_b)-u(B_a)=\int_a^b\nabla u(B_s)dB_s+\frac 12\int_a^b\Delta u(B_s)ds\geq \int_a^b\nabla u(B_s)dB_s.$$ Therefore, writing the definition of stochastic integral and taking the condition expectation with respect to $\sigma(B_a)$, we can see that the conditional expectation of the integral is $0$. Indeed, it's the limit of a sum of terms of the form $E[\nabla(B_{t_i})(B_{t_{i+1}}-B_{t_i})\mid B_a]$. As $\nabla u(B_{t_i})$ is independent of $B_a$, we get the wanted result.

Answer 2

This may be wrong, but I believe you just take the expected value of both sides for the last equation above and the integral vanishes and you'll be left with the definition of a submartingale.