Below is from Rene Schilling's Brownian Motion.
Let $(B_t, \mathscr{F}_t)$ be a BM${^d}$. Recall from Theorem 5.6 that for $u\in C^{1,2}((0,\infty) \times R^d) \cap C([0,\infty) \times R^d)$ satisfying $$|u(t,x)|+|\frac{\partial u(t,x)}{\partial t}| + \sum_{j=1}^d |\frac{\partial u(t,x)}{\partial x_j}|+ \sum_{j,k=1}^d |\frac{\partial^2 u(t,x)}{\partial x_j\partial x_k}|\le c(t)e^{C|x|} \;\;\;\;\;\; (5.5)$$ for all $t>0$ and $x\in R^d$ with some constant $C>0$ and a locally bounded function $c:(0,\infty) \to [0,\infty),$ the process $$M_s^u = u(t-s,B_s) - u(t,B_0) +\int_0^s (\frac{\partial}{\partial t}-\frac{1}{2} \Delta_x) u(t-r,B_r)dr, s\in [0,t),$$ is an $\mathscr{F}_s$ martingale for every measure $P^x$, i.e. for every starting point $B_0=x$ of a Brownian motion.
Here the author refers to Theorem 5.6 for $M_s^u$ being a martingale. However, Theorem 5.6 given below, takes a slightly different form in the parameters. I don't see how we can adjust the parameters in this Theorem to show that $M_s^u$ is a martingale. Does it follow directly from this Theorem? Or perhaps we need to follow through the steps in the proof of this Theorem.
