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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 $-1$.

A weakly stationary process has a time-series spectral density precisely when its spectral measure of a stationary time series is absolutely continuous with respect to Lebesgue measure. Absolute summability of its autocovariances is sufficient and yields a continuous Fourier-series density. It is not necessary for existence; for a general integrable density, Fejér sums give an $L^1$ recovery formula.

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.

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 $1/\pi$.

For a periodic Hölder continuous function of Hölder exponent $0<\alpha <1$ and Hölder seminorm at most $M$, the degree-$n-1$ Fejér sum has the displayed supremum norm error bound. In the half-kernel normalization $F_{n-1}=(2n{)}^{-1}{\sin }^2(nt/2)/{\sin }^2(t/2)$, use $F_{n-1}(t)\le \min \{n/2,{\pi }^2/(2n{t}^2)\}$. Splitting the convolution error integral at $1/n$ gives the valid constant $C_{\alpha }=1/[\pi (\alpha +1)]+\pi /(1-\alpha )$. At the endpoint $\alpha =1$ this argument instead gives $O(M\log n/n)$.

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.

Use degree-indexed Fourier partial sums and Fejér sums, as in this paper. For a $2\pi$-periodic continuous function, putEquivalently,These are trigonometric polynomials of degree at most $n$. Some treatments index the Fejér sum by the number of terms rather than its degree; the degree convention here is why the formulas below contain ${\sigma }_{2n-1}$ and ${\sigma }_{n-1}$.

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

A real weakly stationary process has finite second moments, a time-independent mean $\mu$, and autocovariance function ${\gamma }_k=\mathrm{Cov}(X_t,X_{t+k})$ depending only on the lag. Use the angular-frequency conventionThe time-series spectral density is nonnegative and integrates to ${\gamma }_0$. The displayed Fourier series is ordinarily justified by ${\sum }_k|{\gamma }_k|<\infty$. More generally, weak stationarity gives a spectral measure of a stationary time series, not necessarily a spectral density. If a spectral density exists without this summability, its Fourier inversion can be understood using Fejér sums, which converge in $L^1$ and at its Lebesgue points. Thus spectral density existence should not be inferred from weak stationarity alone.

A white noise process $WN(0,{\sigma }^2)$ has mean zero and $\mathrm{Cov}({\epsilon }_t,{\epsilon }_s)={\sigma }^2{1}_{\{t=s\}}$. Neither independence nor a normal distribution is needed for this second-order definition. Its autocovariance function is zero at nonzero lags, soFor a real absolutely summable filter $(c_r)$, the series defining $Y_t$ has mean-square convergence: its tails have $L^2$ norm at most $\Vert X_0{\Vert }_2{\sum }_{\text{tail}}|c_r|$. Its mean is $\mu {\sum }_r{c}_r$, andThe sum is absolutely convergent, since $|{\gamma }_X(j)|\le {\gamma }_X(0)$ by the Cauchy-Schwarz inequality. It depends only on $k$, proving that the output is a weakly stationary process. Substitute the spectral integral for ${\gamma }_X$ and interchange the sums and integral, justified by $f_X\in L^1$ and ${\sum }_{r,s}|c_r{c}_s|<\infty$. This givesBecause the coefficients are real, the two factors are complex conjugates. Hence the spectral density transformation under a linear filter isFor complex coefficients the same formula holds with the conjugate-covariance convention.

Applying this result to the order-two moving-average model yieldsEquivalently, its nonzero autocovariances are ${\gamma }_0={\sigma }^2(1+{\theta }_1^2+{\theta }_2^2)$, ${\gamma }_1={\sigma }^2{\theta }_1(1+{\theta }_2)$ and ${\gamma }_2={\sigma }^2{\theta }_2$, together with ${\gamma }_{-k}={\gamma }_k$.

For the order-one moving-average process, invert its filter by a geometric series. The required coefficients areThey are absolutely summable for $|\theta |<1$. Direct cancellation giveswith mean-square convergence. Thus the inverse filter recovers exactly the original white noise process, including its variance; it is an invertible time-series representation.

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:

A strongly continuous unitary representation of the circle group decomposes as $H={\widehat{⨁}}_{m\in \mathbb{Z}}{H}_m$, with $U_z\xi =z^m\xi$ on $H_m$. Its positive energy unitary representation of the circle condition is that the occurring integers are bounded below. Equivalently $U_{e^{it}}=e^{itL}$ with a self-adjoint generator bounded below. An integer character can shift the lowest weight to zero.

Let $M=S^{\prime \prime }$. Because $S=S^{\ast }$, this is a Von Neumann algebra, and rotation invariance makes $U_z{M}^{\prime }{U}_z^{\ast }=M^{\prime }$. Joint irreducibility implies $M^{\prime }\cap \{U_z{\}}^{\prime }=\mathbb{C}I$: a nonscalar self-adjoint element of that joint commutant would have a nontrivial invariant spectral projection gives a reducing subspace.

For $T\in M^{\prime }$, take its strong Fourier coefficients $T_k={\int }_{\mathbb{T}}{z}^{-k}{U}_{z}T{U}_z^{\ast }dz$. They lie in $M^{\prime }$ and satisfy $U_z{T}_k{U}_z^{\ast }=z^k{T}_k$. Both $T_k^{\ast }{T}_k$ and $T_k{T}_k^{\ast }$ are rotation-invariant elements of $M^{\prime }$, hence scalar. If $T_k\ne 0$, both scalars equal $\Vert T_k{\Vert }^2$, so $T_k/\Vert T_k\Vert$ is a unitary. It shifts energy by $k$. For $k\ne 0$, applying it or its adjoint to a nonzero lowest-energy vector produces a vector below the lowest energy, a contradiction. Thus $T_k=0$ for every $k\ne 0$, while $T_0$ is scalar.

Fejér sum averaging of the strongly continuous orbit $U_{z}T{U}_z^{\ast }$ converges strongly to $T$. Its Fourier coefficients have just shown every such average is $T_0$, so $T=T_0$ is scalar. In particular every bounded operator commuting with $S$ commutes with every $U_z$. In fact the hypotheses force the stronger conclusion $S^{\prime }=\mathbb{C}I$.

For a real weakly stationary process, extend its autocovariance by ${\gamma }_{-k}={\gamma }_k$. The exact existence of a time-series spectral density condition is that its spectral measure of a stationary time series be absolutely continuous with respect to Lebesgue measure. Absolute summability ${\gamma }_0+2{\sum }_{k\ge 1}|{\gamma }_k|<\infty$ is a useful sufficient condition, not a necessary one.

We use the conventional angular-frequency density on $[-\pi ,\pi ]$, restricted to $[0,\pi ]$ by symmetry. Under the absolute-summability condition, the Fourier series and its inverse relation areThus $2{\int }_0^{\pi }f={\gamma }_0$. If a one-sided density is instead normalized to integrate to the full variance, use $g=2f$ and omit the factor two in the inverse formula. This is the positive-frequency spectral normalization convention difference. With merely an integrable spectral density, the inverse relation remains valid; one must not assume pointwise convergence of the unweighted Fourier series. Its Fejér sums recover the density in $L^1$:

If $X$ and $Y$ are independent stationary processes, the cross covariances vanish, soLinearity of the inverse Fourier series relation, or addition of the spectral measures of a stationary time series, gives $f_Z=f_X+f_Y$. Independence can in fact be weakened to zero cross-covariances at every lag.

For the ARMA$(1,1)$ representation, away from an uncancelled unit root the transfer function givesThe usual causal case has $|\varphi |<1$; the same expression holds for the two-sided stationary solution when $|\varphi |>1$. We work in the nondegenerate case $v,w>0$ and $\varphi \ne \pm 1$; cancellations and zero-noise cases are obtained by the appropriate reduced representation or limits. Since independent white noise of variance $w$ contributes $w/(2\pi )$, putThenThe white-noise addition to an ARMA(1,1) process problem is therefore the factorizationDefineBoth are positive under the stated nondegeneracy conditions. An invertible choice isThese satisfy $|\alpha |<1$, $\lambda (1+\alpha {)}^2=P$, and $\lambda (1-\alpha {)}^2=Q$. HenceIn particular, it is the autoregressive coefficient $\varphi$ that enters the terms from the added observation noise.

To obtain a representation of the actual $Z$, defineThe stable inverse exists because $|\alpha |<1$, and spectral filtering gives $f_{\xi }=\lambda /(2\pi )$. Thus $\xi$ is weak white noise, andIts variance can equivalently be writtenWhen $\alpha \ne 0$, also $\lambda =(v\theta -w\varphi )/\alpha$; when $\alpha =0$, necessarily $C=0$ and $\lambda =A$, so the latter quotient should not be used. The model may reduce in order through cancellation. Boundary limits with $P=0$ or $Q=0$ give $|\alpha |=1$ and need not be stably invertible; the displayed stable reconstruction applies to the nondegenerate case above.

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.

The indexing of the Fejér sums givesSubtracting removes precisely the initial $n$ Fourier partial sums. ThusFor $n=0$, omit the second term, so that no undefined ${\sigma }_0$ is needed. Apply the triangle inequality and the uniform-norm contraction of Fejér summation estimate to getHence the operator norm of the de la Vallée Poussin sum is at most $1+2n/m$.

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: