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Cesàro summation assigns a value to a series when the arithmetic means of its partial sums approach a finite limit of a sequence. Here we use the first-order method, often written $(C,1)$. For a real or complex series beginning at index zero, defineIt is Cesàro summable to $L$ when ${\sigma }_N\to L$. The averaging is applied to the partial sums $s_n$, rather than to the original terms $a_n$. This distinction is essential: the terms can average to zero while their accumulated sum keeps growing. ^[ref]

For Grandi's series $1-1+1-1+\cdots$, the partial sums alternate between one and zero. The series is not a convergent series, but the two values occupy asymptotically equal proportions of the averaging window. In the declared index convention,so its first-order Cesàro sum is $1/2$. This names a particular summation method; it does not turn the oscillating partial sums into an ordinarily convergent sequence.

Why is this useful? An oscillating approximation can retain stable information when its cumulative average settles. Because the Cesàro theorem for convergent sequences preserves an existing limit of a sequence, this adds some new cases without changing the sum of any ordinarily convergent series. For Fourier series, averaging the Fourier partial sums gives a Fejér sum; the existing Fejér kernel and approximation articles explain its effect on reconstruction. Averaging does not guarantee success for every oscillating sequence, as the bounded partial sums need not be Cesàro summable example shows.

One historical motivation was the multiplication of series. In 1918, A. F. Andersen recalled Ernesto Cesàro's product theorem and cited his 1890 publication at the start of a study of products involving Cesàro summable series. This connects the theory with preserving useful operations on sums, as well as with assigning a value to an isolated example. His paper treats several orders of summability; the examples here use only $(C,1)$.^[ref]

There are continuous functions whose Fourier series diverges at a prescribed point. In contrast, the Fejér kernel is nonnegative and has integral one, so Fejér summation is a uniform-norm contraction. The resulting Fejér sums need not be positive: for the constant function $f=-1$, every sum is $-1$.

The normalized Fejér kernel satisfies $0\le F_n(t)\le \min \{n/(2\pi ),\pi /(2n{t}^2)\}$ on $0<|t|\le \pi$. Splitting its first absolute moment at $1/n$ gives $\int |t|F_n(t)dt\le 1/(2\pi n)+(\pi /n)\log (\pi n)$. The chaining inequality for the modulus of continuity, $\omega (f,t)\le (1+t/\delta )\omega (f,\delta )$, then bounds the Fejér sum error by $\omega (f,\delta )[1+{\delta }^{-1}\int |t|F_n]$. Taking $\delta =\log n/n$ proves the stated bound for $n\ge 2$ and hence uniform convergence for continuous periodic functions. The logarithm vanishes at $n=1$, so the displayed estimate is not intended at that index.

A triangular taper majorizes a fixed fraction of the summation interval. Its Fourier kernel is the Fejér kernel, bounded by $C\min (m,(m\Vert \theta {\Vert }^2{)}^{-1})$. For separated sample points, split each row sum at distance $1/m$: the plateau and square-decay tail each contribute $O({\delta }^{-1})$, while the diagonal contributes $O(m)$. Bounding the resulting quadratic form proves the dual sieve with constant $C(N+{\delta }^{-1})$. Operator norm duality gives the primal sieve.

For fixed positive integer $k$ and $n>k$, the Fourier coefficient of the Fejér kernel at frequency $k$ multiplies the mode by $1-k/n$. The displayed error is exact. Consequently the order $n^{-1}$ cannot be replaced by $o(n^{-1})$ for all twice continuously differentiable periodic functions, even though each fixed Fourier mode is infinitely differentiable.

The even positive Fejér kernel of total mass one gives ${\sigma }_{n}f-f=\frac{1}{2}\int F_n(t)[f(\cdot +t)-2f+f(\cdot -t)]dt$. The scaling property of the second modulus of smoothness bounds this by ${\omega }_2(f,\delta )(1+{\delta }^{-2}\int t^2{F}_n(t)dt)$. On $[-\pi ,\pi ]$, $\sin (|t|/2)\ge |t|/\pi$ gives $\int t^2{F}_n(t)dt\le {\pi }^2/n$. Choose $\delta =n^{-1/2}$. For twice continuously differentiable functions, the second-difference integral formula makes the error $O(n^{-1})$.

The $n$th Fejér sum is the arithmetic mean of the first $n$ Fourier partial sums. Equivalently, it is the convolution of the function with the Fejér kernel.

The Fejér kernel is nonnegative and has unit mass in its convolution normalization. The integral triangle inequality therefore bounds the supremum norm of a Fejér sum by that of the original function. In the half-kernel convention $F_n=(2n{)}^{-1}|{\sum }_{j=0}^{n-1}{e}^{ijt}{|}^2$, its integral is $\pi$ and the convolution prefactor is $1/\pi$.

The positive normalized Fejér kernel has a $1/(n{t}^2)$ tail. Split the error integral at $\delta =n^{-\alpha }$ and use the chaining inequality $\omega (f,t)\le (1+t/\delta )\omega (f,\delta )$. Its tail contributes at most a constant times $n^{\alpha -1}(1+\alpha \log n)\omega (f,\delta )$. The factor is uniformly bounded for each fixed $\alpha <1$, proving the displayed estimate. Its constant need not remain bounded as $\alpha$ tends to one.

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. First compute the Fourier coefficients of the periodized sum. Absolute integrability givesThus Fubini's theorem justifies termwise integration. For integer $m$, the change of variable $x=2\pi j+\tau$ and $e^{2\pi imj}=1$ giveReindexing a convergent periodization gives periodicity; the integral argument always gives its periodic representative almost everywhere. This proves the Fourier-coefficient content of the Poisson summation formula.

There is a genuine qualification to the printed pointwise claim: the periodization of an integrable function need not agree pointwise with its Fourier series under the stated hypotheses alone. Set $F(0)=1$ and $F(x)=0$ for $x\ne 0$. This is absolutely integrable, and on one closed period the periodization has only two possibly nonzero summands, so the series is uniformly convergent. Yet $f(0)=1$, while $\hat{F}(m)=0$ for every $m$. The claimed pointwise equality at zero is therefore false. This counterexample also works for Riemann integrability, not only for an almost-everywhere convention.

The coefficient identity gives a rigorous general replacement: the weighted Fejér sumsconverge to the periodization in the integral norm, and at every continuity point. Indeed they are its convolution with the normalized Fejér kernel; that kernel is nonnegative, has integral one, and has integral tending to zero outside every fixed neighbourhood of zero. For convergence in the L1 norm, split the convolution error into small translations, controlled by continuity of translation in the integral norm, and the remaining tail. The same split gives pointwise convergence at a continuity point. If the periodization is piecewise continuously differentiable, the ordinary Fourier series has the usual midpoint limit; if it is continuous and its coefficients are absolutely summable, the displayed unweighted formula holds pointwise. These are sufficient interpretations of the intended identity.

For the specified exponential example all the needed Fourier series convergence properties do hold. Split its Fourier transform at zero to obtainIts periodization is continuous, with uniformly geometrically bounded tails, and the sampled coefficients are absolutely summable. The absolutely summable coefficients define a uniformly convergent Fourier series with the computed coefficients. Its continuous sum equals the continuous periodization: their difference has every Fourier coefficient zero, hence is zero in the L2 norm by Fourier completeness, and continuity upgrades that to equality everywhere. Therefore its ordinary Poisson summation formula is valid. Evaluate it at zero:Thus the fully justified requested sum is

The original PDF's kernel has an incorrect normalization. With ordinary $dt$, its integral is $\pi$, not one. Already at $n=1$ the printed kernel is the constant $1/2$. For $f\equiv 1$, the printed operator therefore returns $\pi$, while $\omega (f,\delta )=0$ for every $\delta$. Thus the literal inequality with the printed kernel is false. The Fejér kernel representing the stated average of Fourier partial sums isThe ratio at zero is interpreted by its limit. To verify the normalization and the averaging property, use the finite geometric sum:Fourier orthogonality gives unit integral; the displayed Fourier coefficients are exactly the multipliers of the Fejér sum. This also proves nonnegativity.

For the correctly normalized Fejér sum, subtraction of $f(x)$ givesTake $0<\delta \le 1$ and split at $|t|=\delta$. The inner part is at most $\omega (f,\delta )$, because the kernel is nonnegative and has mass one. The strict inequality in the definition of the modulus of continuity causes no endpoint issue: continuity gives the same bound at distance exactly $\delta$.

On $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$, soDivide an arc of length $|t|$ into at most $1+|t|/\delta$ shorter arcs. The triangle inequality then bounds $|f(x-t)-f(x)|$ by $(1+\pi /\delta )\omega (f,\delta )$. ConsequentlyChoosing $\delta =n^{-1/2}$ proves the requested fractional-scale Fejér approximation bound for the intended normalization, for example with the absolute constant $1+\pi +{\pi }^2$:The same proof applies to complex-valued continuous functions, since the estimates use absolute values.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

Subtracting the two initial averages of Fourier partial sums givesbecause $2n{\sigma }_{2n-1}={\sum }_{j=0}^{2n-1}{s}_j$ and $n{\sigma }_{n-1}={\sum }_{j=0}^{n-1}{s}_j$. The positive normalized Fejér kernel gives the uniform-norm contraction of Fejér summation, namely $\Vert {\sigma }_{m}f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$. The triangle inequality therefore gives

Periodicity and the Hölder condition give $|f(x-t)-f(x)|\le M|t{|}^{\alpha }$ for $|t|\le \pi$. The positive normalized Fejér kernel therefore yieldsThe exponential-sum representation of the kernel bounds $F_{n-1}(t)\le n/2$. Also $\sin (t/2)\ge t/\pi$ for $0\le t\le \pi$, so $F_{n-1}(t)\le {\pi }^2/(2n{t}^2)$. Splitting at $1/n$ and using $0<\alpha <1$ givesHence the Hölder error bound for Fejér summation isThe constant depends on the Hölder exponent and $M$, but not on $n$ or $f$ beyond its Hölder seminorm bound.

Use the central second difference ${\Delta }_t^{2}f(x)=f(x+t)-2f(x)+f(x-t)$ and define the second modulus of smoothness byWe first need its scaling inequality. If $q$ is a positive integer, translation by $h$ gives the telescoping identityThe weights sum to $q^2$. For $t>0$, choose $q=\lceil t/\delta \rceil$ and $h=t/q\le \delta$; thenThe same bound covers $t=0$ trivially.

In the paper's normalization, the Fejér kernel is even and nonnegative, with ${\pi }^{-1}{\int }_{-\pi }^{\pi }{F}_n=1$. Pairing the positive and negative translations expresses the Fejér sum error asUse the scaling bound and $(1+s{)}^2\le 2(1+s^2)$ to obtainFor $|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$. Bounding the numerator sine by one givesThe bound at $t=0$ holds by continuity. Choosing $\delta =n^{-1/2}$ proves the Fejér second-modulus approximation bound with an explicit universal constant:Finally, for twice continuously differentiable periodic $f$, two applications of the fundamental theorem of calculus give the second-difference integral formulaThe triangular weight has integral $h^2$, so ${\omega }_2(f,\delta )\le {\delta }^2\Vert f^{\prime \prime }{\Vert }_{\infty }$. Therefore

For a continuous $2\pi$-periodic function, define the Fourier coefficients and Fourier partial sum byThe Fejér sum is the arithmetic average of the first $n$ Fourier partial sums:Thus $s_n$ has degree at most $n$, whereas ${\sigma }_n$ has degree at most $n-1$. These indexing conventions matter in the identity in (b). Throughout, $m$ is a positive integer; when $n=0$, any term $n{\sigma }_n$ is interpreted as zero without introducing ${\sigma }_0$.

For later estimates, the Fejér kernel in the PDF's normalization isThe apparent singularities are removable. The squared modulus makes $F_n$ nonnegative, and integrating its finite exponential expansion leaves only the $n$ diagonal terms, so ${\pi }^{-1}\int F_n=1$. In particular Fejér summation is a uniform-norm contraction:

Use the periodic modulus of continuitySubdividing a displacement $t$ into at most $\lceil |t|/\delta \rceil$ steps of size at most $\delta$ proves the chaining inequalityFor $|t|\le \delta$, the sharper bound $\omega (f,\delta )$ holds directly.

Set $\delta =n^{-\alpha }$, where $0<\alpha <1$ and $n\ge 1$. Positivity and unit normalized mass of the Fejér kernel giveThe portion $|t|\le \delta$ is at most $\omega (f,\delta )$. For $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$ and $|\sin (nt/2)|\le 1$, soApply this bound and the chaining inequality to the two tails:Let $\beta =1-\alpha >0$. Calculus gives ${\sup }_{u\ge 1}{u}^{-\beta }\log u=1/(e\beta )$, attained at $u=e^{1/\beta }$, and $u^{-\beta }\le 1$. Thus a possible explicit constant for the fractional-scale Fejér approximation bound isThe constant depends only on $\alpha$, not on $f$ or $n$. Since a continuous periodic function is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every $\alpha <1$; a crude global-oscillation bound would not prove the whole requested range.

The Stone theorem for the circle group says that a strongly continuous unitary representation $U$ of $\mathbb{T}=\{z:|z|=1\}$ has an orthogonal decompositionEquivalently, $U_z={\sum }_n{z}^n{P}_n$ strongly, with mutually orthogonal projections $P_n$ summing strongly to the identity. Define these projections using normalized Haar measure:Haar invariance gives $U_w{P}_n=w^n{P}_n$. Changing $z$ to $z^{-1}$ in the adjoint integral proves $P_n^{\ast }=P_n$. Applying the defining integral to $P_m\xi$ gives $P_n{P}_m={\delta }_{nm}{P}_m$ by orthogonality of the characters $z^k$. Thus $P_n$ are orthogonal projections, and their ranges are exactly the indicated character subspaces.

It remains to prove completeness of these subspaces. The Fejér kernel has the finite expansionIt is nonnegative, has normalized integral one, and its integral outside any fixed neighborhood of the identity tends to zero, since the denominator in the usual geometric-sum formula is bounded away from zero there. Strong continuity therefore givessplit the integral into that neighborhood, where $\Vert U_z\xi -\xi \Vert$ is small, and its complement, where it is bounded by $2\Vert \xi \Vert$. The integral on the left is ${\sum }_{|n|<N}(1-|n|/N)P_n\xi$. Hence the closed span of all $H_n$ is $H$. Orthogonality then gives the unweighted strong sum and the stated representation formula.

Conversely, any orthogonal decomposition indexed by the integers gives this representation; continuity follows first on finite sums and then on all vectors by the uniform unitary norm bound. With $z=e^{it}$ its generator isThis diagonal operator is self-adjoint and $U_{e^{it}}=e^{itA}$. Thus the circle theorem is the integer-spectrum version of the one-parameter theorem.

Write $e(t)=e^{2\pi it}$. We first prove the needed cancellation of quadratic exponential sums, rather than invoke equidistribution without proof. The number $\sqrt{2}$ is irrational: writing it as a reduced fraction would make its numerator even from $u^2=2v^2$, and then its denominator even as well, a contradiction. Thus $2hr\sqrt{2}$ is irrational for nonzero integers $h,r$.

Fix $h\ne 0$ and set $z_n=e(h\sqrt{2}{n}^2)$ for $1\le n\le L$, extending it by zero otherwise. For integers $L\ge H\ge 1$, each term appears exactly $H$ times in the shifted sum, soFor any complex numbers $w_1,\dots ,w_J$, nonnegativity of ${\sum }_j|w_j-J^{-1}{\sum }_i{w}_i{|}^2$ gives $|{\sum }_j{w}_j{|}^2\le J{\sum }_j|w_j{|}^2$. Apply this special Cauchy-Schwarz inequality to the displayed outer sum, and then expand the inner squared modulus. This givesFor fixed $H$ and large $L$, each correlation is the geometric serieswhose absolute value is at most $2/|1-e(2hr\sqrt{2})|$, independently of $L$. This follows by multiplying the geometric sum by $1-e(2hr\sqrt{2})$; its denominator is nonzero by irrationality. Divide the inequality by $L^2$ and let $L\to \infty$, keeping $H$ fixed. It gives a limit superior at most $1/H$. Taking arbitrarily large $H$ provesThis is the quadratic instance of the Van der Corput inequality for finite scalar sequences, with its proof supplied here.

To turn cancellation into the particular approximation, put $\epsilon =10^{-8}$ and choose an integer $M>2/{\sin }^2(\pi \epsilon )$. The Fejér kernel is the explicit finite trigonometric polynomialThe expansion follows by counting pairs of indices with difference $h$. Its constant coefficient is one, so the cancellation just proved givesIf $\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}\ge \epsilon$, the geometric-series formula givesFor all sufficiently large $L$, the displayed average exceeds $1/2$. Consequently at least one $n\in \{1,\dots ,L\}$ satisfiesIt is a positive integer, as required. Every auxiliary cancellation and kernel identity used in this existence proof has been derived; a smooth cutoff or an unproved Weyl criterion is unnecessary.

Use the modulus of continuity on the circle, $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. Splitting a displacement into steps of length at most $\delta$ and applying the triangle inequality givesThe Fejér kernel is nonnegative and has integral one, soWe therefore need its first absolute moment. The inequalities $|\sin (nt/2)|\le n|\sin (t/2)|$ and $\sin (|t|/2)\ge |t|/\pi$ on $[-\pi ,\pi ]$ implySplitting at $1/n$ givesTaking $\delta ={\delta }_n=(\ln n)/n$ in the preceding error bound proves the Fejér first-moment approximation bound:The constants do not depend on $f$ or $n$. By the Heine-Cantor theorem, a continuous function on the compact circle is uniformly continuous, hence $\omega (f,{\delta }_n)\to 0$ as ${\delta }_n\to 0$. Thus the Fejér sums converge uniformly to $f$. The index restriction is necessary: at $n=1$ the printed scale is zero, while ${\sigma }_{1}f$ is the mean of $f$ and generally differs from it.