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A continuous complex-valued function can have uniformly bounded Fourier partial sums which fail to converge at one point. Normalize harmonic sine polynomials by their harmonic coefficient sums, modulate them into disjoint high-frequency intervals and choose their degrees so that the complete norms are summable. At zero, complete blocks vanish while their midpoint prefixes have a common nonzero value. The frequency-separated Fourier block series gives a uniform bound on every partial sum as well as two distinct subsequential values.

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]

For a compact circle set with normalized Lebesgue measure at most $\exp (-8\pi M/\epsilon )$, where $M\ge 1$ and $0<\epsilon \le 1$, the displayed bounds can be realized by a trigonometric polynomial with spectrum in a positive-frequency interval arbitrarily far from zero. Take the logarithm of a positive-real-part Schwarz integral on the unit disk of a small-mean cutoff, truncate after radial dilation, and modulate its bounded imaginary part. A prefix isolates one analytic half and thus reveals the large real part. The separation between small function norm and large Fourier partial sums drives the Kahane-Katznelson divergence theorem.

This average of Fourier partial sums reproduces every degree-at-most-$n$ trigonometric polynomial. With ${\sigma }_r=r^{-1}{\sum }_{j=0}^{r-1}{s}_j$,The bound follows because Fejér summation is a uniform-norm contraction. For $n=0$, the term involving ${\sigma }_0$ is omitted.

For an integrable periodic function with one-sided limits $L_{+},L_{-}$ at $t$, the displayed condition implies that its classical Fourier partial sums converge to $(L_{+}+L_{-})/2$. To prove it use the Dirichlet kernel $D_n(s)=\sin ((n+1/2)s)/\sin (s/2)$ and its integral $2\pi$. Subtract the target value and combine the positive and negative halves of the convolution integral. The resulting difference isThe coefficient of the sine is integrable near zero by the Dini condition, and away from zero by integrability of $f$. The Riemann-Lebesgue lemma therefore makes this expression tend to zero. Step functions satisfy the condition because the numerator vanishes in a sufficiently small punctured neighborhood. This is a local Fourier convergence test, distinct from Dini's theorem on uniform convergence of monotone sequences of continuous functions.

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 Fourier partial sum retains the modes whose integer frequency multipliers satisfy $|k|\le n$ in a complex Fourier series:For a real function of period $2\pi$, this is equivalently $a_0/2+{\sum }_{k=1}^n[a_k\cos (kx)+b_k\sin (kx)]$. It is a trigonometric polynomial, even when the original function is discontinuous. For a square wave, retaining $M$ nonzero odd sine modes gives $S_{2M-1}f$; all even-frequency coefficients are zero. The Gibbs phenomenon illustrates why more retained modes need not reduce the maximum error.

If trigonometric polynomials have spectra in increasing disjoint positive-frequency intervals, the displayed summability gives a continuous uniform sum. At any cutoff its Fourier partial sum contains complete earlier blocks, at most one active prefix and no later blocks. Differences across an active prefix therefore isolate that block exactly. Fixed-size excursions occurring in arbitrarily late blocks contradict the Cauchy criterion, even when the complete blocks vanish at the point under consideration.

Every null subset of the circle lies in the divergence set of the Fourier series of some continuous complex-valued function. One can arrange unbounded Fourier partial sums there. The compact-set Fourier amplification lemma, compact batching of a small open set and frequency-separated Fourier block series construct the function by a summable series of small blocks. The result covers nonclosed and dense null sets; it does not require the divergence set to equal the originally specified set.

The classical Fourier partial sums areHere $a_0={\pi }^{-1}\int f$, and functions are extended periodically. One version of the Dini test for Fourier series states that if one-sided limits $L_{+},L_{-}$ at $t$ exist andthen $s_n^f(t)\to (L_{+}+L_{-})/2$. A step function on a finite interval is constant on each side of any given point for sufficiently small distance, so its displayed numerator is zero there. Dini test for Fourier series gives exactly the asserted midpoint of the two one-sided values, including at jumps.

Let $P_n=s_n$ on $L^2$. It is the orthogonal projection onto the span of the displayed trigonometric functions through frequency $n$, hence $\Vert P_n\Vert \le 1$. For a step-function approximation $\psi$,The convergence on a dense set under a uniform operator bound, here applied using the allowed convergence for $\psi$ and density of step functions, proves $P_{n}f\to f$ in $L^2$. Consequently the union of these finite trigonometric spans is dense, which is precisely completeness of the orthonormal system. It also proves Parseval's equality by passing to the limit in the Pythagorean projection identity.

For $f(t)=t^2$, oddness gives $b_j=0$, direct integration gives $a_0=2{\pi }^2/3$, and two integrations by parts give $a_j=4(-1{)}^j/j^2$. ThusDini test for Fourier series justifies this pointwise on the interval, or absolute coefficient summability gives its continuous extension after the $L^2$ identification. At zero it yields ${\sum }_{j\ge 1}(-1{)}^{j-1}/j^2={\pi }^2/12$. Parseval givesTherefore ${\sum }_{j\ge 1}{j}^{-4}={\pi }^4/90$.

With $\hat{f}(k)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(s)e^{-iks}ds$, the classical Fourier partial sum is $s_N^f(t)={\sum }_{|k|\le N}\hat{f}(k)e^{ikt}$. Its Cesàro mean isFor the uniform Fejér theorem, extend the continuous endpoint-matching $f$ periodically. The given Fejér kernel is nonnegative, even and has total integral $\pi$. ThusGiven $\epsilon >0$, uniform continuity supplies $\delta >0$ so the bracket has modulus less than $\epsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\epsilon$. On the remainder, $F_N(u)\le [2(N+1){\sin }^2(\delta /2){]}^{-1}$, and the bracket is at most $2\Vert f{\Vert }_{\infty }$. HenceLetting $\epsilon$ decrease to zero proves uniform convergence of the Fejér sums.

The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt{j}}$ and consider its formal trigonometric series. For $t=\pi p/q$, there is $J_0$ such that $\cos (j!t)=1$ for every $j\ge J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes toThe exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m{)}^2$ and $((2m+1)\pi {)}^2$. Thus the ordinary sums diverge at every $t\in \pi \mathbb{Q}$.

But their genuine Fourier Cesàro means also diverge there. Indeed,Since $|a_j|\le 2$ and ${\sum }_{j=0}^{J}j!\le 2J!$ for $J\ge 3$, its absolute value is at most $4/(J+1)\to 0$, uniformly in $t$. Thereforewhich has the same incompatible limit points. This is an instance of Cesàro sums at factorial cutoffs track ordinary sums. The printed Cesàro-convergence claim is false. Averaging the sparse series by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.

There is a further issue: no $L^1$ function has exactly this Fourier series. Here is a direct proof using strongly lacunary polynomial L1-L2 bound. For a trigonometric polynomial $p$ supported on $\{\pm j!:j\ge 3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi )$ and bounding the three possible pairings givesThe Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$, hence $\Vert p{\Vert }_2\le \sqrt{3}\Vert p{\Vert }_1$.

If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\Vert {\sigma }_N^f{\Vert }_1\le \Vert f{\Vert }_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a polynomial of the type above. Its $L^2$ norms would therefore be uniformly bounded. Orthogonality then forces ${\sum }_{j\ge 3}|a_j{|}^2<\infty$, by fixing any finite coefficient block and letting $N\to \infty$. However,so that sum diverges. This contradiction disproves the stated existence premise.

Even considered only formally, these coefficients rule out an $L^2$ function by Bessel's inequality, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted series either. The asserted $L^1$ function does not exist.

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.

First prove the Baire category theorem in its complete-metric form. Let $X$ be a nonempty complete metric space, let $G_n$ be open dense sets, and let $V$ be any nonempty open set. Choose a closed ball $\overline{B}(x_1,r_1)\subset V\cap G_1$ with $0<r_1<1/2$. Inductively, density and openness permit the choiceFor $m>n$, $x_m$ belongs to the $n$th closed ball, so $d(x_m,x_n)\le r_n$. The centers form a Cauchy sequence and converge to some $x\in X$ by completeness. Each closed ball contains all later centers and therefore contains their limit. Hence $x\in V\cap {\bigcap }_n{G}_n$. Since this holds for every $V$, every countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of closed sets with empty interior; this is the form needed below.

For the Banach-Steinhaus theorem, let $X$ be a Banach space, $Y$ a normed vector space, and $\mathcal{T}$ a family of bounded linear operators from $X$ to $Y$ such that ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x$. DefineEach $E_m$ is closed, as an intersection of inverse images of closed balls under continuous maps, and pointwise boundedness gives $X={\bigcup }_m{E}_m$. The Baire category theorem gives some $E_m$ with nonempty interior. Choose $B(x_0,r)\subset E_m$. If $\Vert h\Vert \le r/2$, both $x_0+h$ and $x_0$ lie in $E_m$, so $\Vert Th\Vert \le 2m$ for every $T$. Applying this to $h=(r/2)u$ with $\Vert u\Vert \le 1$ provesThis proves the Uniform boundedness principle; completeness is required for the domain, but not for $Y$.

Now fix the prescribed point ${\theta }_0$ on the unit circle, parametrized by angles modulo $2\pi$. On the real Banach space $C(\mathbb{T};\mathbb{R})$ with the supremum norm, consider the continuous linear functionals $T_{N}f=S_{N}f({\theta }_0)$, where $S_N$ is the symmetric Fourier partial sum. Summing the finite geometric series gives its Dirichlet kernel representationThe continuous extension at $t=0$ is $2N+1$. The kernel bound gives $\Vert T_N\Vert \le {\Lambda }_N=(2\pi {)}^{-1}\int |D_N|$. Equality holds: the continuous periodic functions defined byhave supremum norm at most one, and dominated convergence gives $T_N{f}_{\delta }\to {\Lambda }_N$ as $\delta \downarrow 0$.

These norms are unbounded. Indeed, evenness and $\sin (t/2)\le t/2$ for $0<t\le \pi$ giveFor the second inequality, integrate over $[(j-1)\pi ,j\pi ]$, where $1/u\ge 1/(j\pi )$ and the integral of $|\sin u|$ is two. The harmonic series diverges, proving the Dirichlet kernel harmonic lower bound. If every continuous $f$ had bounded partial sums at ${\theta }_0$, the Banach-Steinhaus theorem would bound all the operator norms, a contradiction. Consequently there exists a continuous real function $f$ with ${\sup }_N|S_{N}f({\theta }_0)|=\infty$, and its Fourier series diverges at the prescribed point. The function depends on the chosen point; the argument requires no claim of simultaneous divergence everywhere.

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 ${\sigma }_{n-1}$.

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

The Fourier coefficients satisfy $a_0=a_n=0$ andThus the square wave has Fourier seriesIt equals $f$ away from its jump discontinuities, and converges to $0$ at the jumps. The assigned endpoint values do not affect its Fourier coefficients.

For the Poisson equation, use separation of variables $\varphi ={\sum }_{n\text{ odd}}{b}_n{R}_n(r)\sin (n\theta )$. The radial equation is $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. Its bounded solution with zero Dirichlet boundary condition at $r=1$ isIndeed the negative power is excluded at the centre, and the boundary fixes the coefficient of $r^n$. No resonance at $n=2$ occurs because only odd modes are forced. ThereforeThis is a Poisson equation on a disk with angular forcing. The series converges uniformly on the closed disk: for $n\ge 3$ each term is bounded by a constant times $n^{-3}$. Its first spatial derivatives also converge uniformly. For instance the radial derivative and $r^{-1}$ times the angular derivative of each term are bounded by constants times $n^{-2}$; the $n=1$ harmonic term is smooth in Cartesian coordinates. It defines a function with zero boundary data and the required weak formulation. Applying the Laplacian to finite partial sums gives the Fourier partial sums of $f$, which converge in $L^2$; passing to the limit against smooth compactly supported test functions verifies the Poisson equation. It holds classically off the jump rays and the centre, where the forcing is locally constant. It cannot be interpreted as a global classical equation with continuous second derivatives for discontinuous forcing. Uniqueness follows from the zero-boundary Dirichlet energy identity for the difference of two solutions.

Finally orthogonality gives ${\int }_0^{2\pi }f(\theta )\sin (n\theta )d\theta =\pi b_n$. The radial integral isUniform convergence of $\varphi$ and boundedness of $f$ justify integration term by term, soThe negative sign also agrees with $\int \varphi \Delta \varphi =-\int |\nabla \varphi {|}^2$. The original PDF's radial denominator is $n^2-4$; the converted TeX's $n^2-1$ is a transcription error.

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized measure $dt/(2\pi )$. The Fourier partial sum at zero is the real linear functionalwhere the Dirichlet kernel isThe quotient follows by summing the finite geometric progression, with its removable value $D_N(0)=2N+1$. It is a real continuous periodic function. Ifthen $|L_{N}f|\le {\Lambda }_N\Vert f{\Vert }_{\infty }$. This bound is its exact operator norm even on the real continuous functions. Indeed, for $\delta >0$ setIt is real, continuous and periodic with norm at most one, andby dominated convergence. At zeros of the kernel the integrand is zero, and elsewhere its limit is $|D_N|$.

For $0<t\le \pi$, $\sin (t/2)\le t/2$. Using evenness and putting $v=(N+1/2)t$ givesFor the last inequality, split into the first $N$ complete intervals $[(j-1)\pi ,j\pi ]$: their sine integrals are two and $1/v\ge 1/(j\pi )$. The harmonic sum tends to infinity. This proves the Dirichlet kernel harmonic lower bound.

Choose $N$ with ${\Lambda }_N>K$, then choose $\delta$ sufficiently small that $L_N{f}_{N,\delta }>K$. ThusThis constructs continuous near-maximizers, rather than using the discontinuous sign of the kernel as the desired function.

A trigonometric polynomial $t_n$ of degree at most $n$ has zero Fourier coefficients outside $[-n,n]$. Its Fourier partial sum $s_j(t_n)$ therefore equals $t_n$ for every $j\ge n$. Averaging the sums with $j=n,\dots ,n+m-1$ givesThus the de la Vallée Poussin sum reproduces the entire degree-at-most-$n$ approximation space, including when the actual degree is smaller.