I want to show that for $u(t,x)$ which is a polynomial in $t$ and $x$ such that $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}=0$$ we have $u(t,B_t)$ is a martingale where $B_t$ stands for the standard Brownian motion.
Durrett's "Probability theory and examples" shows that $ E_x u(t, B_t) $ is constant in $t$ and concludes right away that $u(t,B_t)$ is a martingale. How is this possible?
Any help is appreciated.
Your self-answer is unsatisfactory.
Also: In OP, what do you mean by "a polynomial $u(t,x)$ in $t,x$ such that $$ \frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}"\quad\dots\quad ? $$ The correct statement is that any function $u(t,x)$ (regardless if polynomial or not) that satisfies the Kolmogorov backward equation $$ \frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}=0 $$ ensures that $u(t,B_t)$ is a (local) martingale.
Proof. Use Ito's formula $$ u(t,B_t)=u(0,0)+\underbrace{\int_0^tu_x(s,B_s)\,dB_s}_{\text{local martingale}}+ \underbrace{\int_0^tu_t(s,B_s)\,ds+\frac12u_{xx}(s,B_s)\,ds}_{0}\,. $$ It is fairly easy to see that that local martingale does in general not have independent increments.