= Solution
Insert a <Fourier mode> $U_m^n=G^ne^{im\theta}$. With $r=1-2\mu$ and $\eta=e^{i\theta}$, the <amplification polynomial> is
$$
G^2-r(1-\eta)G-\eta=0.
$$
Its two roots have product $-\eta$, of modulus one. Thus if both roots are bounded by one, both must have modulus exactly one. Set $G=e^{i\theta/2}\lambda$. The equation becomes
$$
\lambda^2+2ir\sin(\theta/2)\lambda-1=0,
\qquad
\lambda_\pm=-ir\sin(\theta/2)\pm\sqrt{1-r^2\sin^2(\theta/2)}.
$$
When $|r|<1$, both roots have modulus one and their separation obeys $|G_+-G_-|\geq2\sqrt{1-r^2}>0$ for every frequency. The companion <matrix> has bounded entries and can be diagonalized with the two <eigenvectors> $(G_+,1)^T,(G_-,1)^T$. The uniform separation bounds the inverse <eigenvector> <matrix>. Hence every <matrix> power is bounded uniformly in $n$ and $\theta$, and the <Parseval identity> proves <L2 norm> <stability> for the two starting levels on the whole grid.
If $|r|>1$, frequencies near $\theta=\pi$ give one root outside the unit circle, so the scheme is unstable. If $|r|=1$, the Nyquist <polynomial> is $(G-r)^2$, a repeated unit root. The corresponding companion <matrix> has a nontrivial <Jordan block> and solutions containing $nr^n$. Thus the endpoints are unstable as well. On the whole line the individual Nyquist plane wave is not square-summable, but Fourier packets localized arbitrarily close to that frequency realize the unbounded power norm by continuity. Therefore the <uniform stability of a shifted two-level advection scheme> criterion is
$$
\boxed{0<\mu<1.}
$$
In particular $\mu=1/2$ is stably exact with compatible starts, whereas $\mu=1$ is formally exact but unstable under arbitrary perturbations. Checking only the moduli of the roots and retaining both endpoints would miss this essential distinction.
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