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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 a real continuous $2\pi$-periodic function $f$, the trigonometric Chebyshev alternation theorem says that $t^{\ast }\in {\mathcal{T}}_n$ is its best uniform approximation if and only if there are $2n+2$ distinct cyclically ordered points in one period,and a sign $\epsilon \in \{1,-1\}$ such thatThe best approximant exists and is unique. Alternation is cyclic: the last and first displayed extrema also have opposite signs across the closing interval. If the error is zero, $f=t^{\ast }$ and the characterization holds trivially.

The sufficiency can be seen from zero counting. A strictly improving trigonometric polynomial would have a difference from $t^{\ast }$ with the alternating nonzero signs at these points, forcing at least $2n+2$ zeros on the circle. A nonzero member of ${\mathcal{T}}_n$ has at most $2n$ zeros: multiplying its Laurent expression by $e^{inx}$ gives an ordinary polynomial of degree at most $2n$ in $e^{ix}$. This contradiction is the alternation mechanism used in the next part.

The coefficient sum is finite, so the Weierstrass M-test makes the cosine series uniformly convergent and $f$ continuous. Put $Q=5^{m+1}$ and $A_m={\sum }_{k=m+1}^{\infty }{a}_k$. The indicated partial sum belongs to ${\mathcal{T}}_n$ because its largest frequency is $5^m\le n$. Its tail satisfiesand equality holds at $x=0$. Thus its supremum norm error is exactly $A_m$.

At the $2Q$ points $x_j=j\pi /Q$, $0\le j<2Q$, every omitted frequency is an odd integer multiple of $Q$. Indeed for $k\ge m+1$,Uniform convergence permits summing these values, givingThese are cyclically alternating extrema. Since $n<Q$, there are $2Q\ge 2n+2$ of them, so the trigonometric Chebyshev alternation theorem proves that the partial sum is the unique best approximant. Consequently the positive lacunary trigonometric series hasThe positivity of every tail coefficient makes the same tail attain its full coefficient sum with alternating signs; a triangle bound alone would not establish optimality.

Use real-valued continuous functions and real trigonometric polynomials. The trigonometric Chebyshev alternation theorem says that $p\in {\mathcal{T}}_n$ is the unique best uniform approximation to $f$ if and only if there are distinct points, ordered within one period,Thus the required number of alternating extrema is $2n+2$. When the error is zero, the assertion is interpreted trivially and $p=f$ is unique. The count reflects $\dim {\mathcal{T}}_n=2n+1$ and the fact that a nonzero degree-$n$ trigonometric polynomial has at most $2n$ zeros in one period. In particular a strictly better candidate would force its difference from $p$ to have alternating signs at these points and hence at least $2n+2$ zeros, counting the closing interval around the circle. The trigonometric Chebyshev alternation theorem also gives necessity and uniqueness.

The Weierstrass M-test gives uniform convergence and continuity of the series. For the proposed trigonometric polynomial, set $q=5^{m+1}$ and $A={\sum }_{k=m+1}^{\infty }{a}_k$. The tail satisfiesAt $x_j=j\pi /q$, for $j=0,\dots ,2q-1$, the integer $5^k/q$ is odd for every omitted frequency. Thus $\cos (5^k{x}_j)=(-1{)}^j$ and $r(x_j)=(-1{)}^{j}A$. In particular $\Vert r{\Vert }_{\infty }=A$.

Since $n<q$, these $2q$ points include at least $2n+2$ consecutive alternating extrema in a single period. Also $5^m\le n$, so $t_n\in {\mathcal{T}}_n$. The trigonometric Chebyshev alternation theorem proves that the proposed partial sum is the unique best approximant, withThis is an instance of positive lacunary trigonometric series: all omitted odd frequency ratios align their extrema.

For an odd integer $q\ge 3$ and positive summable coefficients $a_k$, the Weierstrass M-test makes $f(x)={\sum }_{k\ge 0}{a}_k\cos (q^{k}x)$ continuous. Its partial sum over $q^k\le n$ is the unique best uniform approximation by degree-at-most-$n$ trigonometric polynomials, for $n\ge 1$. If $Q$ is the first omitted frequency, at $x_j=j\pi /Q$ every omitted cosine equals $(-1{)}^j$ because all frequency ratios are odd. Thus the tail attains alternating extrema equal to its coefficient sum at $2Q\ge 2n+2$ points. The trigonometric Chebyshev alternation theorem proves the formula. For $a_k=r^{-k}$, $r>1$, the largest decay exponent is $\log r/\log q$.