= Solution
Use $\hat u^n(\theta)=\sum_m u_m^ne^{-im\theta}$, initially for finite-support sequences and then by <Plancherel theorem> for $\ell^2$. A shift by $k$ multiplies the transform by $e^{ik\theta}$. Hence the recurrence gives
$$
A(\theta)\hat u^{n+1}=B(\theta)\hat u^n,\qquad \boxed{\hat u^{n+1}=H(\theta)\hat u^n=H(\theta)^{n+1}\hat u^0.}
$$
Here the last equality is understood with time-zero data, and the scheme is assumed well-defined, so the denominator is nonzero or any removable singularity is filled continuously.
If $|H|\leq1$, Plancherel gives $\|u^n\|_2\leq\|u^0\|_2$ for every $n$. Conversely, if the continuous amplification factor has $|H(\theta_0)|>1$, it exceeds $1+\delta$ on an interval of positive length. Choose a nonzero initial Fourier transform supported in that interval. Then $\|u^n\|_2\geq(1+\delta)^n\|u^0\|_2$, contradicting a uniform stability bound. Therefore \b[uniform power stability holds exactly when]
$$
\boxed{|H(\theta)|\leq1\quad(-\pi\leq\theta\leq\pi).}
$$
Without continuity the exact condition is an essential-supremum bound, since isolated Fourier frequencies have zero measure. The fixed recurrence here uses a bound uniform over all time steps; mesh-dependent finite-time growth bounds are a different stability convention.
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