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Thank you. I suspect that for $p>2$ it may be different, because the absence of eigenvectors does not follow from $\tau(S^n)/\|S\|^n\to 0$ (at least the way it does for $p=2$).
For $|x|\le 1$ the function $a(x)$ is analytic by the assumption (1), and $a(x^{-1})$ is analytic because numerator and denominator are analytic and the latter is not zero.
