= Solution
Write $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_m)$ and let $|\Lambda|=\max_i|\lambda_i|$. The stated <Euclidean norm> gives
$$
\|f_n\|^2=\sum_i|\lambda_i|^{2n}|(f_0)_i|^2
\le |\Lambda|^{2n}\sum_i|(f_0)_i|^2=|\Lambda|^{2n}\|f_0\|^2.
$$
Taking nonnegative square roots proves \b[$\|f_n\|\le|\Lambda|^n\|f_0\|$]. At $n=0$ this is equality. This is a contraction bound in the transformed coordinates; an arbitrary nonorthogonal $S$ does not give the same unit-constant bound for the original error <norm>.
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