 Solution

Use ${\hat{u}}^n(\theta )={\sum }_m{u}_m^n{e}^{-im\theta }$, 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 givesHere 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 $|H|\le 1$, Plancherel gives $\Vert u^n{\Vert }_2\le \Vert u^0{\Vert }_2$ 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 $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, contradicting a uniform stability bound. Therefore uniform power stability holds exactly whenWithout 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.