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The original PDF prints $O(n^{\alpha })$, with a positive exponent. That premise is insufficient; the intended decay rate is $O(n^{-\alpha })$. Indeed every bounded continuous periodic function has $E_n(f)\le \Vert f{\Vert }_{\infty }=O(n^{\alpha })$. For an explicit counterexample, on $[-\pi ,\pi ]$ setand extend periodically. This is continuous, but $\omega (h,\delta )\ge 1/\log (e/\delta )$ for $0<\delta <1$, which is not $O({\delta }^{\alpha })$ for any $\alpha >0$. Thus the sign correction is required in the actual printed question, not just in its converted TeX.

For real continuous $2\pi$-periodic $f$, write $E_n$ for best uniform approximation by degree-at-most-$n$ trigonometric polynomials and $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. A quantitative inverse theorem for trigonometric approximation iswhere $C$ is universal. Here is a direct reason for the sum. Choose best approximants $p_j$ and $2^r\le n<2^{r+1}$. WriteThe Bernstein inequality for trigonometric polynomials givesFor $j\ge 2$, monotonicity of best errors bounds $2^{j+1}{E}_{2^{j-1}}$ by $8{\sum }_{\nu =2^{j-2}}^{2^{j-1}-1}{E}_{\nu }$. The first terms are bounded by a fixed multiple of $E_0$. Also $2E_{2^r}\le 4n^{-1}{\sum }_{\nu =0}^n{E}_{\nu }$. Nowproves the theorem, for example with a sufficiently large absolute constant $C=20$.

Suppose, as intended, $E_{\nu }(f)\le A{\nu }^{-\alpha }$ for $\nu \ge 1$, after enlarging $A$ to include any finite exceptional indices, where $0<\alpha <1$. Since ${\sum }_{\nu =1}^n{\nu }^{-\alpha }\le n^{1-\alpha }/(1-\alpha )$, the inverse theorem givesFor $0<\delta \le 1$ take $n=\lfloor 1/\delta \rfloor$. Then $1/(n+1)<\delta \le 1/n$, so monotonicity of the modulus of continuity and $n^{-\alpha }\le 2^{\alpha }{\delta }^{\alpha }$ give $\omega (f,\delta )\le 2^{\alpha }{C}_f{\delta }^{\alpha }$. For $\delta >1$, the separate bound $\omega (f,\delta )\le 2\Vert f{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }{\delta }^{\alpha }$ covers the remaining scales. Therefore the corrected conclusion isNow put $\beta =\log 2/\log 3$. The given best approximant to the Weierstrass function leaves the uniformly convergent tail ${\sum }_{k=m+1}^{\infty }{2}^{-k}\cos (3^{k}x)$. Its supremum norm is exactly $2^{-m}$, attained at $x=0$; henceThe corrected inverse theorem gives $\omega (g,\delta )=O({\delta }^{\beta })$. To show that this order is sharp, set $h_m=\pi /3^{m+1}$. Every term of $g(0)-g(h_m)$ is nonnegative, and for $k\ge m+1$ the cosine equals $-1$ because $3^{k-m-1}$ is odd. ThusGiven small $\delta >0$, select $m$ with $h_m\le \delta <3h_m$. Then $\omega (g,\delta )\ge 4(3\pi {)}^{-\beta }{\delta }^{\beta }$. Combining the estimates gives the precise small-scale orderThe exact approximation error can also be verified independently of the supplied fact: at $x_j=j\pi /3^{m+1}$ the tail has alternating values $(-1{)}^j{2}^{-m}$. There are enough such points in one period for the trigonometric Chebyshev alternation theorem, since $n<3^{m+1}$. This is the positive lacunary trigonometric series mechanism behind the best partial sums.

For $f\in C(\mathbb{T})$, define $E_j(f)={\inf }_{T\in {\mathcal{T}}_j}\Vert f-T{\Vert }_{\infty }$, 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 $C$ satisfiesFor 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 the requested two rates:The logarithm is the endpoint contribution of the harmonic sum and cannot generally be removed, as the following Weierstrass function shows.

The Weierstrass M-test gives a continuous periodic Weierstrass function, since $\sum a^{-k}<\infty$. For $5^m\le n<5^{m+1}$, the given best approximant leaves the tailIts supremum norm is at most the sum of its coefficients, and at $x=0$ all cosine terms equal one, attaining that bound. ConsequentlyPut $\alpha =\ln a/\ln 5$. Then $a^{-m}=(5^m{)}^{-\alpha }$ and $n/5^m\in [1,5)$, giving $E_n(g)\asymp n^{-\alpha }$, with constants depending only on $a$.

For $1<a<5$, part (a) gives $\omega (g,1/n)\le C_a{n}^{-\alpha }$. The first Jackson theorem for periodic approximation, $E_n(g)\le C\omega (g,1/n)$, supplies the matching lower bound. Hence the modulus of continuity has the precise orderAt $a=5$, the inverse theorem for trigonometric approximation gives the upper bound $C\delta \ln (1/\delta )$, and part (c) proves the matching lower bound. Thus the critical Weierstrass modulus isTo pass from scales $1/n$ to all small $\delta$, use monotonicity of the modulus of continuity and adjacent reciprocals, which are comparable to $\delta$.

A standard Weierstrass function is a uniformly convergent lacunary cosine series, here parameterized by $a>1$ and an integer frequency multiplier $b>1$. Its continuity follows from the Weierstrass M-test. In the range $1<a<b$, its modulus of continuity has the fractional-power scale ${\delta }^{\log a/\log b}$ for the positive odd-frequency model; the critical case $a=b$ can have the larger scale $\delta \log (1/\delta )$. These examples distinguish small approximation error from ordinary Lipschitz continuity.