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For a periodic function use the modulus of continuityOn the interval, take the supremum over pairs of points whose distance is at most $\delta$. The mean value theorem applied to cosine, whose derivative has absolute value at most one, givesBoth cosine values lie in $[-1,1]$, so for $|h|\le \delta$,Taking the supremum provesThe cosine substitution for polynomial approximation is also a linear isometry in the supremum norm, since cosine maps a full period onto $[-1,1]$; its image consists of even continuous periodic functions.

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.

For the indicated function the cosine substitution givesThe absolute-value function is Lipschitz continuous with constant one, and so is sine. Their composition therefore satisfiesUse the sharper intermediate estimate from the preceding part, rather than the ordinary interval modulus of continuity:

The usual inverse theorem for trigonometric approximation cannot hold verbatim with the ordinary interval modulus of continuity. It would implyBut comparison with the endpoint $1$ giveswhich is of order $n^{-1/2}$ and contradicts that bound as $n\to \infty$. This is an endpoint obstruction to an algebraic inverse approximation theorem. The algebraic approximation rate measures smoothness after cosine substitution; cosine compresses distances quadratically near the endpoints. A valid algebraic inverse theorem must account for that endpoint geometry rather than using the unchanged periodic formulation.

Use the same dyadic slope martingale $G_n$ and dyadic filtration as in (c). Each $G_n$ is integrable, since it takes finitely many finite values. On each dyadic cell, the absolute slope is $2^n$ times the absolute endpoint increment. Consequently the hypothesis in the PDF is exactlyThus $(G_n)$ is uniformly integrable. It is also bounded in L1 norm: choose a finite ${\lambda }_0>0$ at which the supremum of the tails is finite, and use $\Vert G_n{\Vert }_1\le {\lambda }_0+{\sup }_j{V}_j(f,{\lambda }_0)$. The uniformly integrable martingale convergence theorem supplies $g\in L^1([0,1])$ with $G_n\to g$ in L1 norm.

The functions $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$ are again the dyadic linear interpolations of $f$. Since $f$ is continuous on a compact interval, it is uniformly continuous, and $\Vert f_n-f{\Vert }_{\infty }\le {\omega }_f(2^{-n})\to 0$, where ${\omega }_f$ is its modulus of continuity. On the other hand, the integral of $G_n-g$ is uniformly bounded in absolute value by $\Vert G_n-g{\Vert }_1\to 0$. Hence the dyadic slope-tail criterion for absolute continuity givesNo boundedness of $g$ is asserted here; the tail condition permits integrable densities that are unbounded.

At the cusp of $f_0(t)=|t|$, positivity and evenness of the Fejér kernel giveSince $\sin (t/2)\le t/2$,Summing the supplied lower bound over $I_k$, $1\le k\le n^{\prime }=\lfloor n/2\rfloor$, yieldsThe harmonic series satisfies ${\sum }_{k=1}^{n^{\prime }}{k}^{-1}\ge c^{\prime }\log n$, so

The modulus of continuity of the periodic function $f_0$ obeys $\omega (f_0,1/n)=1/n$. If a universal Jackson-type estimateheld for all continuous periodic $f$, it would give $O(1/n)$ for $f_0$, contradicting the lower bound. Thus

The Poisson kernel on the circle is nonnegative, has normalized integral one, and its mass outside any fixed arc about zero tends to zero as $r\uparrow 1$. Split the convolution error into a short arc, controlled by the modulus of continuity, and its complement, controlled by the vanishing kernel mass. This proves uniform convergence for every continuous periodic function. Its Fourier multiplier is $r^{|n|}$, so the convolution is the Abel sum of the Fourier series.

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.