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The difference between the two cumulative sums of Fourier partial sums isHence the de la Vallée Poussin sum satisfiesUsing the uniform-norm contraction of Fejér summation on each term and the triangle inequality in the supremum norm,This proves the requested operator norm bound. When $n=0$ the identity reduces to $v_{0,m}={\sigma }_m$ and the bound is one.

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:

Write $S_{N}f(t)={\sum }_{|n|\le N}\hat{f}(n)e^{int}$ for the symmetric Fourier partial sum, with normalized Fourier coefficients on $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$. We will prove the stronger conclusionIf $E$ is empty, take $f=0$. Otherwise the proof of the Kahane-Katznelson divergence theorem has three steps: a bounded trigonometric polynomial whose partial sum is large on short arcs, a finite-arc batching of an arbitrary null set, and a uniformly convergent sequence of separated frequency blocks. In particular, no assumption that $E$ is closed or compact is needed.

First consider a nonempty finite family of arcs with centers ${\theta }_{\nu }$, half-lengths $0<d_{\nu }\le 1$, and $D={\sum }_{\nu }{d}_{\nu }$. On the closed unit disk defineEach summand has positive real part there, and $\Phi (0)=1$. Thus the principal complex logarithm $A=\log \Phi$ is holomorphic on a neighborhood of the closed disk, $A(0)=0$, and $|\mathrm{Im}A|<\pi /2$. To bound its real part on an arc, put $h=t-{\theta }_{\nu }$ with $|h|\le d_{\nu }$ and $q=1+d_{\nu }-\cos h$. ThenAfter multiplication by the weight $d_{\nu }/D$, this contribution is at least $1/(4D)$; all the other contributions have positive real part. Throughout the arc union we therefore haveThis is the positive-real logarithmic amplifier for short arcs. Since $A$ is holomorphic beyond the closed unit disk, its Taylor series converges uniformly there. Choose a polynomial $Q(z)={\sum }_{\nu =1}^d{a}_{\nu }{z}^{\nu }$ with ${\sup }_{|z|\le 1}|Q(z)-A(z)|<1$. There is no constant term because $A(0)=0$. Set $C=\pi /2+1$ andwhere the integer $M>d$ may be chosen arbitrarily large. We have $\Vert P{\Vert }_{\infty }\le 1$, and its frequencies lie in $[M-d,M+d]$. At the cut $N=M$, the positive-frequency half of $Q$ is excluded and the negative-frequency half is included. Henceon the arc union. In particular, for any $K>0$, a total half-length $D\le \frac{1}{4}{e}^{-1-2CK}$ gives $|S_{M}P|\ge K$ there. This proves the needed compact-set Fourier amplification lemma, with the additional freedom to place its entire frequency block above any prescribed frequency.

Next we construct the null-set limsup cover by finite arc families. SetSince $E$ has Lebesgue measure zero, for every row $\ell \ge 1$ choose a countable arc cover $\{I_{\ell ,m}:m\ge 1\}$ of $E$ with positive half-lengths $d_{\ell ,m}$ and ${\sum }_m{d}_{\ell ,m}<2^{-\ell }{D}_1$. We can take closed arcs by slightly enlarging an open cover; their half-lengths are less than one. The whole two-index array has total half-length less than $D_1$. Absolute convergence of this sum lets us choose strictly increasing integers $N_j$ such that the sum outside the finite square $1\le \ell ,m\le N_j$ is less than $D_{j+1}$. Put $N_0=0$ and let ${\mathcal{F}}_j$ contain those arcs for whichEach family is finite. Its total half-length is less than $D_j$: for $j=1$ use the total sum, and for $j>1$ use the tail outside the preceding square. Moreover, every $t\in E$ belongs to arcs with unbounded row index, so belongs to the unions of infinitely many of the families ${\mathcal{F}}_j$. This batching is essential: a dense null set need not have a finite cover of small total length, whereas the infinitely-often finite covers above suffice.

Apply the amplification construction to ${\mathcal{F}}_j$ with height $K_j$, obtaining $P_j$ with $\Vert P_j{\Vert }_{\infty }\le 1$ and $|S_{M_j}{P}_j(t)|\ge K_j$ on that family's union. Choose the modulation integers successively so that the frequency intervals $[M_j-d_j,M_j+d_j]$ are positive and each lies strictly above the preceding one. Empty families may be represented by $P_j=0$ and an unused frequency interval. Define the frequency-separated Fourier block seriesThe bound ${\sum }_j{a}_j\Vert P_j{\Vert }_{\infty }\le 1$ gives absolute uniform convergence, so $f$ is continuous. Uniform convergence also justifies computing every Fourier coefficient term by term. At frequency $M_j$, all earlier blocks have been included in full and all later blocks are absent:For $t$ in the union of ${\mathcal{F}}_j$, the reverse triangle inequality givesEvery $t\in E$ lies in infinitely many such unions, and $M_j\to \infty$. Therefore the boxed unboundedness holds at every point of $E$, completing the proof for arbitrary null sets.

Use $\hat{f}(\lambda )=\int f(t)e^{-i\lambda t}dt$ and interpret the regularity assumption to include $f\in L^2(\mathbb{R})$ with the continuous representative furnished by Fourier inversion. Write $F=\hat{f}$. The bandwidth assumption implies $F\in L^1$ as well as $L^2$, by the Cauchy-Schwarz inequality on its finite support, sois continuous and defined at every $t$. The integer samples are precisely the Fourier coefficients of $F$ with the sign convention appropriate to the basis $e^{-in\lambda }$:The completeness and Parseval identity for the exponential orthonormal basis on $[-\pi ,\pi ]$ give $F={\sum }_{n\in \mathbb{Z}}f(n)e^{-in\lambda }$ in $L^2$ andThe last equality is the Plancherel theorem. Apply the inverse integral to the finite symmetric Fourier partial sums. Sincewe obtain the Nyquist–Shannon sampling theorem in this normalization:where the quotient is one at $t=n$. This is a sampling expansion by periodic Fourier projection, not merely an almost-everywhere inversion. Indeed, the inverse integral of an error $G\in L^2[-\pi ,\pi ]$ is bounded, uniformly in $t$, by $\Vert G{\Vert }_2/\sqrt{2\pi }$. Hence the symmetric finite sampling sums converge uniformly on all of $\mathbb{R}$. Moreover, the Parseval identity applied to $e^{it\lambda }$ givesThe Cauchy-Schwarz inequality therefore makes the sampling series absolutely convergent at every $t$, with its absolute tail bounded uniformly by $({\sum }_{|n|>N}|f(n){|}^2{)}^{1/2}$. Thus the integer samples recover the function everywhere.

For the nonuniqueness beyond this bandwidth, fix $0<{ϵ}_0<\min (ϵ,\pi )$ and a nonzero smooth function $\Phi$ compactly supported in $(-{ϵ}_0,{ϵ}_0)$. Let $h$ be its inverse Fourier transform and putThe function $h$, and hence $g$, is a Schwartz function. The two shifted spectral supports are disjoint, so $\hat{g}\ne 0$, but its support is contained strictly inside $(-\pi -ϵ,\pi +ϵ)$. Every integer sample of $g$ is zero. Therefore $g$ and the zero function have identical samples but are distinct, providing a strictly supercritical sampling alias counterexample for every positive bandwidth enlargement.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

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.

Let $m$ be a positive integer and let the trigonometric polynomial $t_n$ have frequencies only in $[-n,n]$. For every $j\ge n$, its Fourier partial sum is the polynomial itself: $s_j(t_n)=t_n$. Every term in the defining average of the de la Vallée Poussin sum therefore equals $t_n$, givingThis is exact reproduction of the degree-at-most-$n$ trigonometric polynomials, irrespective of the positive averaging length $m$.

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 $1+2n/m$.

Apply part (i) to the continuous difference $h=f-g$. Every Fourier coefficient of $h$ vanishes, so every Fourier partial sum is zero. The orthogonal projection convergence from part (i) therefore gives $\Vert h{\Vert }_2=0$.

If $h(t_0)\ne 0$, continuity gives an interval on which $|h|$ is bounded below by a positive number. That interval would contribute positively to $\Vert h{\Vert }_2^2$, a contradiction. HenceThe continuity hypothesis upgrades equality almost everywhere to pointwise equality.

Use normalized Fourier coefficients and the inner productThe functions $e_n(t)=e^{int}$ are orthonormal. Hence the Fourier partial sum $S_{N}f={\sum }_{|n|\le N}\hat{f}(n)e_n$ is the orthogonal projection onto the trigonometric polynomials of degree at most $N$. For any such polynomial $P$, orthogonality givesso $\Vert f-S_{N}f{\Vert }_2\le \Vert f-P{\Vert }_2$.

Given $\epsilon >0$, the permitted density result supplies a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. Once $N$ includes its degree,Therefore the Fourier partial sums converge to $f$ in the normalized $L^2$ norm, giving exactly the stated mean-square limit.

Part (i) and the Cauchy-Schwarz inequality showDirect integration of this finite Fourier partial sum givesMoreover Bessel's inequality puts both coefficient sequences in ${\ell }^2$, so their product series is absolutely convergent by the Cauchy-Schwarz inequality. Consequently the cross form of Parseval's identity isTaking $g=f$ also gives equality of the squared function norm and the squared coefficient norm.

We prove the Kahane-Katznelson divergence theorem through an explicit small-norm block construction. Let $\mu$ be normalized Lebesgue measure on the circle.

First establish the compact-set Fourier amplification lemma. If a compact set $K$ satisfieswe can make a trigonometric polynomial $q$ with $\Vert q{\Vert }_{\infty }\le \epsilon$, supported in any sufficiently high interval of positive frequencies, whose partial prefix has magnitude greater than $M$ on $K$.

To construct it, choose a smooth nonnegative function $u$ equal to one near $K$, with values at most one and meanOuter regularity and a smooth cutoff give this choice; the stipulated bound on $\mu (K)$ leaves room between the two exponentials. The Schwarz integral on the unit diskhas positive real part in the disk, $H(0)=\delta$, and boundary real part $u$. Its holomorphic logarithmsatisfies $W(0)=0$ and $|\mathrm{Im}W|<\pi /2$. On $K$, the boundary value has $\mathrm{Re}W\ge \log (1/\delta )>4\pi M/\epsilon$. Smoothness of $u$ makes $H$ continuous at the boundary, and its positive boundary real part near $K$ makes $W$ continuous there.

Choose a radius just below one, then truncate the Taylor series of $W$ at that radius. This gives an analytic polynomial $R(e^{it})$ with zero constant term, degree $d$, andThe radial function is analytic beyond the closed unit disk, so the Taylor truncation is uniform on the whole circle. For $L>d$, putIts frequencies lie between $L-d$ and $L+d$, all positive. Its prefix through frequency $L$ includes exactly the negative-frequency half of $\mathrm{Im}R$ shifted into this interval:Consequently $|S_{L}q(t)|>\epsilon (4\pi M/\epsilon -1)/(2\pi )>M$ on $K$, whereas $\Vert q{\Vert }_{\infty }<\epsilon$. Increasing $L$ places the entire block above any previously used frequency.

We next use compact batching of a small open set to handle an arbitrary null set, without assuming that it is compact or a countable union of compact null sets. SetFor each $j$, choose an open $U_j\supseteq E$ with $\mu (U_j)<\exp (-8\pi j/{\epsilon }_{j,0})$. Decompose $U_j$ into countably many closed subarcs with pairwise disjoint interiors: subdivide each open component into closed pieces accumulating only at its excluded endpoints. Group these subarcs into finite successive batches $K_{j,k}$. After batch $k$, include enough pieces that the remaining total length is less than $\exp (-8\pi j/{\epsilon }_{j,k+1})$. Require each batch endpoint in the enumeration to increase. ThenEach batch is compact; endpoints shared by pieces have zero measure and do not affect the estimates.

Enumerate the pairs $(j,k)$ in diagonal order. Apply the block lemma with target $M=j$ to each $K_{j,k}$, and shift its spectrum above all preceding blocks. Denote the resulting polynomial by $q_{j,k}$ and setSince ${\sum }_{j,k}{\epsilon }_{j,k}=1/2$, this series is uniformly convergent and defines a continuous complex-valued function.

Fix $t\in E$. For every $j$ there is a $k$ with $t\in K_{j,k}$. The difference between the Fourier partial sum just before that block and the sum at its midpoint has magnitude greater than $j$: previous blocks cancel in the difference, and future blocks have not yet entered. As $j\to \infty$, these cutoffs tend to infinity. At least one of the two partial sums therefore has magnitude greater than $j/2$. The Fourier partial sums are unbounded, hence not Cauchy, at $t$. We have provedThis establishes divergence on every prescribed null set, including dense nonclosed null sets; it does not assert that the divergence set is exactly $E$.

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:

Use the unnormalized Fourier transform in the question and the Plancherel theorem, with inverse factor $1/(2\pi )$. For a general $L^2$ scaling function, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.

Changing variables in the Fourier transform givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely the same calculation and injectivity of the Fourier transform recover refinement, with convergence understood in $L^2$.

To justify both the orthogonality assertion and this convergence precisely, put $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. It belongs to $L^1[-\pi ,\pi ]$ by monotone integration. The Plancherel theorem givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t){|}^2$ into even and odd translates,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\varphi$.

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

Substituting the Fourier coefficients into the Fourier partial sum and summing the finite geometric series giveswhere

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

Write $h_n=a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L)$. For the partial sum $S_N=a_0/2+{\sum }_{n=1}^N{h}_n$, ordinary multiplication givesOn $[-L,L]$, the integrals of every nonconstant sine or cosine vanish. Fourier orthogonality further gives $\int \cos (m\pi x/L)\cos (n\pi x/L)dx=L{\delta }_{mn}$ and the same identity for two sines, while every sine-cosine integral is zero. ThusFor square-integrable $f$, its Fourier partial sums converge to $f$ in L2 space, so their squared norms converge to the squared norm of $f$. Passing to the limit proves Parseval's identity:Working with finite partial sums before taking the norm limit avoids an unjustified interchange of an infinite double series and integration.