= Solution
For $x-Sx$, choose the consecutive shifts $j_r=r$, $1\leq r\leq k$. The average telescopes:
$$
\frac1k\sum_{r=1}^k(\alpha_{n+r}-\alpha_{n+r+1})=\frac{\alpha_{n+1}-\alpha_{n+k+1}}k.
$$
Its absolute value is at most $2\|x\|_\infty/k$. Taking the <limit superior> and then the infimum over $k$ proves \b[$q(x-Sx)\leq0$].
For the requested functional, use the allowed sublinearity of $q$. On convergent sequences, every finite shifted average has the same limiting value, so $q$ equals the ordinary limit there. Hahn-Banach extends this limit to a linear $\psi$ satisfying $\psi\leq q\leq p$. Apply domination to $x-Sx$ and to $(-x)-S(-x)$ to obtain respectively $\psi(x-Sx)\leq0$ and $-\psi(x-Sx)\leq0$. Hence
$$
\boxed{\psi(x)=\psi(Sx),\qquad\psi(x)\leq\limsup_n\alpha_n.}
$$
This is a <Banach limit>; the two-sided limit bounds also show positivity, <norm> one, and agreement with ordinary limits.
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