= Solution
For $x\ne0$, the <Strong Markov property> at the first step gives the discrete mean-value identity
$$
v(x)=\frac16\sum_{|e|=1}v(x+e).
$$
At $x=0$, $v(0)=1$ while the same average is at most one. Thus $v$ is a bounded superharmonic function on $\mathbb Z^3$. Conditioning on the <natural filtration> and using the one-step <Markov property> gives
$$
\mathbb E[v(X_{n+1})\mid\mathcal F_n]
=\frac16\sum_{|e|=1}v(X_n+e)
\leq v(X_n).
$$
Therefore $(v(X_n))$ is a nonnegative <supermartingale>.
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