= Solution
Here the imaginary-time <Heisenberg picture> position operator is $\bar x(\tau)=e^{\tau H_0}\hat x e^{-\tau H_0}$, where $H_0=H(0,0)$. Bosonic <time ordering> $T_\tau$ places the larger imaginary time to the left without an exchange sign. The brackets denote the normalized <thermal expectation value>
$$
\langle O\rangle_\beta=\frac{\operatorname{Tr}(e^{-\beta H_0}O)}{Z_0},
\qquad Z_0=\operatorname{Tr}e^{-\beta H_0}.
$$
The trace is cyclic and $H_0$ is time independent, so the <two-point correlation function> depends only on the time difference and extends periodically with period $\beta$. This periodicity is a property of the thermal <correlation function>, not the operator identity $\bar x(\tau+\beta)=\bar x(\tau)$.
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