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The first Jackson theorem for periodic approximation applied to $|\sin \theta |$ gives algebraic best error $O(n^{-1})$ for $\sqrt{1-x^2}$. Its ordinary interval modulus of continuity at $1/n$ is at least $\sqrt{2/n-1/n^2}$. Substituting that error rate into the unmodified inverse theorem for trigonometric approximation would instead bound the modulus by $O(\log (n+1)/n)$, a contradiction. Algebraic inverse estimates must incorporate the endpoint compression of cosine substitution for polynomial approximation.

Define $h(x)=|x|\log (e\pi /|x|)$ on $[-\pi ,\pi ]$, set $h(0)=0$, and extend periodically. Concavity and monotonicity on $[0,\pi ]$ give $\omega (h,\delta )=\delta \log (e\pi /\delta )$ for $0<\delta \le \pi$. Its cosine Fourier coefficients areOne integration by parts gives the formula; the remaining integral is $\log j+O(1)$ by the Dirichlet test. All these coefficients are nonpositive. The de la Vallée Poussin sum $V_n=2{\sigma }_{2n}-{\sigma }_n$ reproduces degree-at-most-$n$ trigonometric polynomials and has operator norm at most three. Hence $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and annihilates those polynomials. Its weights on coefficients are zero up to $n$, nonnegative thereafter, and one from $2n$ onwards, so $|L_n(h)|\ge {\sum }_{j\ge 2n}|a_j|\gtrsim \log n/n$. Thus $E_n(h)\ge |L_n(h)|/4\gtrsim \log n/n$; the first Jackson theorem for periodic approximation proves the matching upper bound. This cusp and the critical Weierstrass modulus have the same first-modulus order but different best-approximation rates.

The first Jackson theorem for periodic approximation states that an absolute constant $C_J$ exists such that, for every continuous $2\pi$-periodic function $g$ and integer $n\ge 1$, there is a trigonometric polynomial $q_n$ of degree at most $n$ satisfyingApply it to the continuous periodic function $g(\theta )=f(\cos \theta )$, which is even. Replace the approximant by its even partSince $g(-\theta )=g(\theta )$, the triangle inequality gives $\Vert g-q_n^{\mathrm{ev}}{\Vert }_{\infty }\le \Vert g-q_n{\Vert }_{\infty }$. This symmetrization cancels all sine terms, so $q_n^{\mathrm{ev}}={\sum }_{j=0}^n{\alpha }_j\cos (j\theta )$. The Chebyshev polynomials satisfy $T_j(\cos \theta )=\cos (j\theta )$ and have degree $j$; the latter follows inductively from $T_0=1$, $T_1=x$ and $T_{j+1}=2x{T}_j-T_{j-1}$. Thereforeis an algebraic polynomial of degree at most $n$, and $p_n(\cos \theta )=q_n^{\mathrm{ev}}(\theta )$. Because cosine maps onto $[-1,1]$, the supremum norms agree exactly:Finally, the definition of the best error, periodic Jackson approximation, and part (a) give the successive boundsThusNeither the constant nor the choice of approximation space depends on $f$. The cosine substitution for polynomial approximation and even symmetrization are what preserve the required degree bound.

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 $\delta$.

Take $a=5$ and $x=\pi /2$. Since $5^k\equiv 1(\mathrm{mod}4)$, every cosine term vanishes at $x$, andSet $h=1/n$ and $m=\lfloor {\log }_{5}n\rfloor$. For $0\le k\le m$, the sine argument lies in $[0,1]$, where $\sin u\ge (\sin 1)u$. The initial part therefore has a common sign and magnitude at least $(m+1)(\sin 1)/n$. The remaining tail has absolute value at mostThe reverse triangle inequality now givesfor sufficiently large $n$, because $m+1>\ln n/\ln 5$. This establishes the sharp logarithmic lower boundThe constant can be reduced to include the finitely many small indices: for $n\ge 5$ the displayed lower bound is positive; for $n=2,3$ the first sine exceeds the tail bound $1/4$, and for $n=4$ the first two sines contribute $\sin (1/4)+\sin (5/4)/5$, exceeding the remaining tail bound $1/20$. At $n=1$ the logarithmic right side is zero.

Here $E_n(g)\asymp 1/n$ but $g$ is not Lipschitz continuous, so a first modulus of continuity bound $O(1/n)$ is sufficient by the first Jackson theorem for periodic approximation but is not necessary. Nor is the weaker necessary bound $O(\ln n/n)$ sufficient: an explicit comparison is the logarithmic cusp with slow trigonometric approximationextended periodically. On $[0,\pi ]$ it is increasing and concave, so its largest increment over distance $\delta$ is from zero, giving $\omega (h,\delta )=\delta \ln (e\pi /\delta )$ for $0<\delta \le \pi$.

For a concrete approximation lower bound, its cosine Fourier coefficients satisfyIndeed, integration by parts once replaces $h$ by $h^{\prime }(x)=\ln (\pi /x)$; integrating that against sine with primitive $(1-\cos jx)/j$ yields the stated positive integral. Its logarithmic growth follows because ${\int }_1^A\cos u/udu$ stays bounded by the Dirichlet test. All $a_j$ are negative and absolutely summable.

Let $V_n=2{\sigma }_{2n}-{\sigma }_n$, a de la Vallée Poussin sum. Its multiplier is one for $j\le n$, $2-j/n$ for $n<j<2n$, and zero for $j\ge 2n$; the Fejér summation is a uniform-norm contraction property gives $\Vert V_n\Vert \le 3$. Thus the functional $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and vanishes on every degree-at-most-$n$ trigonometric polynomial. Because the remaining multiplier weights are nonnegative and all $a_j<0$,The first Jackson theorem for periodic approximation gives the corresponding upper bound. The same first-modulus order $\delta \ln (1/\delta )$ occurs with both $E_n\asymp 1/n$ and $E_n\asymp \ln n/n$. Therefore first-modulus rates alone do not characterize the endpoint approximation class, unlike the matched direct and inverse rates for $0<\alpha <1$.

The first Jackson theorem for periodic approximation asserts that a universal constant $C$ satisfiesfor every continuous $2\pi$-periodic function $g$, where the infimum is over degree-at-most-$n$ trigonometric polynomials.

Apply it to the even function $g(\theta )=f(\cos \theta )$. If $q_n$ is a trigonometric polynomial approximating $g$, its even parthas no larger error, because $g(-\theta )=g(\theta )$ and the triangle inequality bounds each averaged error by $\Vert g-q_n{\Vert }_{\infty }$. Every even trigonometric polynomial has the formHere the Chebyshev polynomials $T_j$ have algebraic degree $j$, so $p_n$ has degree at most $n$. Surjectivity of cosine givesConversely, every algebraic polynomial of degree at most $n$ yields such an even trigonometric polynomial. Taking infima therefore proves the exact identity $E_n^{\mathrm{alg}}(f)=E_n^{\mathrm{trig}}(\tilde{f})$, not merely an inequality. Combine this with the periodic theorem and the preceding modulus of continuity estimate:Symmetrization, the degree correspondence and the equality of norms justify every step in transferring the Jackson-type estimate.