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Use the centered second modulus of smoothnessThe relevant modulus properties are ${\omega }_2(f,t)\le (1+t/\delta {)}^2{\omega }_2(f,\delta )$ for $t\ge 0$, $\delta >0$, and ${\omega }_2(f,\delta )\le {\delta }^2\Vert f^{\prime \prime }{\Vert }_{\infty }$ when $f^{\prime \prime }$ is continuous. The latter follows from the second-difference integral formulaThe Fejér kernel is even, nonnegative and has integral one. Averaging the convolution at $t$ and $-t$ therefore givesApply the second modulus of smoothness bound and $(1+s{)}^2\le 2(1+s^2)$:For $|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$, while ${\sin }^2(nt/2)\le 1$. ThusThe removable value at $t=0$ causes no problem. With $\delta =n^{-1/2}$, the Fejér second-modulus approximation bound isIf $f^{\prime \prime }$ is continuous, this also yields $\Vert {\sigma }_{n}f-f{\Vert }_{\infty }\le (1+{\pi }^2)\Vert f^{\prime \prime }{\Vert }_{\infty }/n=O(n^{-1})$.

Squaring the finite geometric sum in the Fejér kernel gives its Fourier seriesFor a fixed positive integer $k$ and $n>k$, convolution therefore gives ${\sigma }_n(e^{ikx})=(1-k/n)e^{ikx}$. Taking the real part,The ratio to $1/n$ is the nonzero constant $k$, even though the function is infinitely differentiable. This Fejér saturation on a Fourier mode disproves an $o(n^{-1})$ estimate for all $C^2$ periodic functions.

The Fejér kernel for the printed normalization isIts endpoint coefficients at $\pm N$ are zero. It is nonnegative, and integrating its Fourier series gives ${\int }_{\mathbb{T}}{F}_N=1$. Hence ${\tilde{S}}_N(f)=f\ast F_N$ andThus the absolute constant can be $C=1$. Positivity of the Fejér kernel, rather than mere boundedness of its individual Fourier coefficients, is the reason this holds.

Take $n\ge 1$. In the normalization used here, the Dirichlet kernel has the finite expansionAveraging a finite number of the integral formulas for the Fourier partial sums is legitimate by linearity of the integral. Hence the Fejér sum is convolution withTo obtain the nonnegative form of the Fejér kernel, expand a squared geometric sum:The coefficient $n-r$ counts pairs of indices whose difference is $r$. Dividing by $2n$ and summing the geometric progression givesAt $u\in 2\pi \mathbb{Z}$ the ratio has its continuous limiting value $F_n(u)=n/2$. Thus this is a continuous, nonnegative trigonometric polynomial, not a kernel with genuine singularities.

Integration over a full period kills every nonconstant cosine term, soTranslation invariance of integration over the circle consequently gives, for every $x$,The first step is the integral triangle inequality, the second uses nonnegativity, and the last uses the mass just computed. Taking the supremum norm provesIn particular, Fejér summation is a uniform-norm contraction. The factor $\pi$, rather than $2\pi$, is essential for the half-normalized Dirichlet kernel and Fejér kernel in this problem.

There is a minor range issue in the printed bound: for $\epsilon >1$ and positive $C$, the upper bound is below one. We prove the intended result for $0<\epsilon \le 1$; a valid formulation for every $\epsilon >0$ replaces the upper bound by $\max \{1,{\epsilon }^{-C}\}$. In fact the argument below gives $C=32$. All auxiliary estimates are proved here.

For $0<\epsilon \le 1/2$, let $m=\lceil {\epsilon }^{-2}\rceil$, and use the Fejér kernelExpanding the square proves the identity and nonnegativity. The finite geometric series formula and $|\sin \pi t|\ge 2\Vert t\Vert$ give $F_m(t)\le (4m\Vert t{\Vert }^2{)}^{-1}$ off the integers. In particular $F_m(t)\le 1/4$ whenever $\Vert t\Vert \ge \epsilon$.

Suppose, for a contradiction, that no $1\le n\le N$ has $\Vert \alpha n^2\Vert <\epsilon$. Summing the Fejér kernel along the quadratic sequence and separating its constant term givesThe coefficients on the right sum to $m-1$. Also $|S_{-k}|=|S_k|$. Hence some $1\le k<m$ has $|S_k|\ge \eta N$, where $\eta =1/(2m)$.

We need only an elementary Van der Corput inequality for finite scalar sequences. For $|z_n|\le 1$, extended by zero outside $[N]$, each term of $S={\sum }_n{z}_n$ occurs in exactly $H$ windows of length $H$. Applying the Cauchy-Schwarz inequality to the window sums and expanding their squares gives, for $H\le N$,Consequently $|S{|}^2\le 2N^2/H+(4N/H){\sum }_{h=1}^{H-1}|C_h|$. Take $H=\lceil 8{\eta }^{-2}\rceil$. If $|S|\ge \eta N$, some $1\le h<H$ must have $|C_h|\ge {\eta }^{2}N/8$; otherwise the displayed upper bound is less than $3{\eta }^2{N}^2/4$.

For the large quadratic exponential sum just found, put $z_n=e(k\alpha n^2)$. Its quadratic exponential sum has multiplicative derivativeThus $C_h$ is a finite geometric series. Its absolute value is at most $1/(2\Vert 2kh\alpha \Vert )$, unless that distance is zero, in which case the desired estimate is automatic. It follows thatSet $n_{\ast }=2kh$. The distance to the nearest integer satisfies $\Vert \ell x\Vert \le \ell \Vert x\Vert$ for a positive integer $\ell$, by multiplying a nearest integer to $x$. ThereforeThis proves the needed quantitative quadratic recurrence once $N$ is chosen polynomially in ${\epsilon }^{-1}$.

For explicit bookkeeping, $m\le 2{\epsilon }^{-2}$ and $H\le 33m^2$. Choose $N=\lfloor {\epsilon }^{-32}\rfloor$. For $0<\epsilon \le 1/2$ we have $N>33792{\epsilon }^{-11}\ge 8mH/({\eta }^2\epsilon )$ and $N\ge H$, while $2mH\le 528{\epsilon }^{-6}\le N$. Hence $n_{\ast }\le N$ and $\Vert \alpha n_{\ast }^2\Vert <\epsilon$, contradicting our supposition. The estimates have substantial slack even at $\epsilon =1/2$.

For $1/2<\epsilon \le 1$, simply take $n=1$, since $\Vert \alpha \Vert \le 1/2<\epsilon$ and $1\le {\epsilon }^{-32}$. For the intended range, the conclusion isFor $\epsilon >1$, the same choice $n=1$ proves the corrected all-range version.

Averaging the given Dirichlet kernels gives the normalized Fejér kernelThe geometric-series identity, or summing the sine terms, proves the second equality. At multiples of $2\pi$ use its removable value $n/(2\pi )$. Therefore $K_n\ge 0$, andis a positive linear operator on continuous functions. The Fourier partial sums, viewed as orthogonal projections, fix $1$, so ${\sigma }_{n}1=1$ and $\int K_n=1$. They also fix $\sin x$ and $\cos x$ for degrees at least one, whereas $s_0$ annihilates them. ConsequentlyAll three periodic Korovkin theorem tests converge uniformly. Hence Fejér sums converge uniformly for every continuous periodic function:

Averaging the integral representations of the Fourier partial sums givesThe finite trigonometric sum isSince $D_k(u)=\sin ((k+1/2)u)/(2\sin (u/2))$, the Fejér kernel is thereforeand

The displayed square shows that $F_n\ge 0$. Every Fourier partial sum preserves the constant function, so ${\sigma }_n(1)=1$. Substituting $f=1$ in the integral formula givesNonnegativity then yields

Because the Fejér kernel has normalized integral one,For $f\in \mathrm{Lip}\alpha$ this impliesOn $0\le t\le \pi$, the inequalities $\sin (t/2)\ge t/\pi$ and $|\sin (nt/2)|\le n|\sin (t/2)|$ giveSplitting the integral at $1/n$ givesFor $0<\alpha <1$, both terms are $O(n^{-\alpha })$. For $\alpha =1$, the second is $O((\log n)/n)$. Uniformly in $x$,where the constants absorb the Lipschitz continuity constant $M$.

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 Fejér kernel is nonnegative, and preservation of constants givesBy evenness,Put $\delta =n^{-1/2}$. Use ${\omega }_2(f,|t|)\le (|t|/\delta +1{)}^2{\omega }_2(f,\delta )$ together with $F_n(t)\le n/2$ and $F_n(t)\le C/(n{t}^2)$. Splitting at $1/n$ and $\delta$ shows that the remaining weighted integral is uniformly bounded, so

For a universal constant $C>0$ and $0<\epsilon \le 1$, every real $\alpha$ admits $1\le n\le {\epsilon }^{-C}$ with distance to the nearest integer of $\alpha n^2$ less than $\epsilon$. A Fejér kernel detects failure of recurrence as a large quadratic exponential sum. The Van der Corput inequality for finite scalar sequences then produces a short linear near-return, whose suitable multiple gives the quadratic return. To include $\epsilon >1$, use the bound $\max \{1,{\epsilon }^{-C}\}$.

If a periodic integrable function has all Fourier coefficients zero, it vanishes almost everywhere. Its Fejér sums vanish, whereas the Fejér kernel is an approximate identity and converges to that function in the integrable norm. Applied to a periodized squared Fourier transform, this detects orthogonality of integer translates.