= Solution
For $f\in C(\mathbb T)$, define $E_j(f)=\inf_{T\in\mathcal T_j}\|f-T\|_\infty$, where $\mathcal T_j$ is the space of degree-at-most-$j$ <trigonometric polynomials>. The first <inverse theorem for trigonometric approximation> states that a universal constant $C$ satisfies
$$
\boxed{\omega(f,1/n)\le\frac Cn\sum_{j=0}^n E_j(f),\qquad n\ge1.}
$$
For perspective, take best approximants at dyadic degrees and telescope their differences. Each difference between consecutive approximants has <supremum norm> at most twice the earlier error, and the <Bernstein inequality for trigonometric polynomials> multiplies that bound by its degree when differentiating. The resulting dyadic sum is bounded by the displayed sum because $E_j$ decreases with $j$; the remaining approximation error is also bounded by an average of earlier errors.
Now suppose $E_j(f)\le K j^{-\alpha}$ for large $j$, absorbing finitely many earlier terms into a constant. If $0<\alpha<1$, comparison with an <integral> gives $\sum_{j=1}^n j^{-\alpha}=O(n^{1-\alpha})$. If $\alpha=1$, the same comparison gives $\sum_{j=1}^n j^{-1}=O(1+\ln n)$. Substitution proves \b[the requested two rates]:
$$
\boxed{\omega(f,1/n)=
\begin{cases}
O(n^{-\alpha}),&0<\alpha<1,\\
O(\ln n/n),&\alpha=1\quad(n\ge2).
\end{cases}}
$$
The logarithm is the endpoint contribution of the harmonic sum and cannot generally be removed, as the following <Weierstrass function> shows.
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