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Choose an orthonormal basis $(e_n)$ of the Dirichlet energy space and independent variables $({\xi }_n)$ with the standard normal distribution. The zero-boundary Gaussian free field is the formal seriesIts precise energy-indexed interpretation is the isonormal Gaussian processFor every fixed $f$, this sum converges in $L^2$ of the probability space by Parseval identity, and it is a centered Gaussian random variable. For $f,g\in H_0^1(D)$,This covariance characterizes its law independently of the chosen basis. The series does not define an $H_0^1$-valued random element: ${\sum }_n{\xi }_n^2=\infty$ almost surely. It is realized as a random distribution, and its pairing with ordinary test functions is constructed in part (b).

Let $P_j$ be the orthogonal projection onto $V_j$. The dilation property and the orthonormal basis property give the orthonormal basis ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ of $V_j$. The periodized nonnegative functionhas period one and ${\int }_0^{1}F=\Vert \phi {\Vert }_2^2=1$, by Tonelli theorem. For a function $g\in L^2(\mathbb{R})$ with compact support contained in a bounded interval $I$, Cauchy-Schwarz inequality and the Parseval identity giveas $j\to -\infty$. The last step uses absolute continuity of the integral of a locally $L^1$ function, since the interval's length tends to zero. If $f\in {\bigcap }_j{V}_j$, thenFunctions with compact support are dense in $L^2$, so $f=0$. ConsequentlyThis proof needs neither the density axiom nor any compact support assumption on the scaling function.

Write $c_m=⟨f,g_m⟩$. A linear N-term approximation fixes the indices independently of $f$, normally the first $N$ in a prescribed ordering:A best N-term approximation chooses the indices using $f$: retain $N$ coefficients of largest absolute value, resolving ties arbitrarily, and set $f_N^n={\sum }_{m\in {\Lambda }_N(f)}{c}_m{g}_m$. The Parseval identity shows why this choice is optimal:For any fixed index set, the orthogonal projection coefficients minimize the error; optimizing the set then means discarding the smallest squared coefficients. Thus “linear” requires a specified ordering, while “nonlinear” refers to the data-dependent selection.

Use $N$ distinct periodic grid points $x_m=m/N$, $0\le m<N$, and $h=1/N$. Values at the two ends of the interval represent the same point, so the endpoint must not be stored twice. With indices interpreted modulo $N$, setThis replaces the Dirichlet matrix by the periodic discrete Laplacian. The normalized Fourier modes $N^{-1/2}{e}^{2\pi ijm/N}$ form an orthonormal basis and have amplification factorsEach satisfies $0<g_j\le 1$ for every $\mu >0$. The Parseval identity therefore gives $\Vert U^{n+1}{\Vert }_h\le \Vert U^n{\Vert }_h$ in the discrete L2 norm, uniformly in the grid and time step. Thus all positive Courant numbers remain stable. The constant mode has $g_0=1$, expressing preservation of the spatial mean; all nonconstant modes decay. The formula with modulo indices also handles small grids, where two neighbours can coincide.

Remove the quadratic phase by settingThe phase around the parallelogram isso the hypothesis says exactly that $\Vert g{\Vert }_{U^2}^4\ge c$, with harmless negative choices of the increments. The Fourier identity for the Gowers U2 norm and the Parseval identity now givewhere the last inequality uses $\Vert f{\Vert }_{\infty }\le 1$. Thus the Quadratic phase detection by the Gowers U2 norm supplies an $r\in {\mathbb{Z}}_N$ such that

Let $I=[N]$ and let $f=1_A-\delta 1_I$ be the balanced indicator function, extended by zero to the integers. For finitely supported functions define the linear configuration countSince $A$ has no nonconstant three-term arithmetic progression, $\Lambda (1_A,1_A,1_A)=|A|=\delta N$. On the other hand, $\Lambda (1_I,1_I,1_I)\ge N^2/4$. Hence, once $N$ is sufficiently large in terms of $\delta$,for an absolute $c_0>0$.

Use the unnormalized Fourier transform $\hat{h}(\theta )={\sum }_{x}h(x)e(-\theta x)$ on $\mathbb{R}/\mathbb{Z}$. Telescoping $1_A=\delta 1_I+f$ writes the preceding difference asLet $M=\Vert \hat{f}{\Vert }_{\infty }$. The Fourier-integral formula for $\Lambda$, followed by the Cauchy-Schwarz inequality and the Parseval identity, bounds these three terms respectively byThey are all at most $M\delta N$, sofor an absolute $c_1>0$. Choose $\theta$ with $|\hat{f}(\theta )|=M$.

By the Dirichlet approximation theorem, some integer $1\le d\le \sqrt{N}$ satisfies $\Vert d\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le N^{-1/2}$. Choose a sufficiently small absolute multiple $\eta$ of ${\delta }^2$. Partition each residue-class progression of common difference $d$ into progressions $P_j$ whose lengths lie between $\eta \sqrt{N}$ and $2\eta \sqrt{N}$; for sufficiently large $N$, short final pieces can be joined to the preceding piece. On each $P_j$, the linear phase $e(-\theta x)$ varies by at most $4\pi \eta$. Choosing $\eta$ small enough compared with $c_1{\delta }^2$ therefore yieldsThe sums over all cells add to ${\sum }_{I}f=0$, so their positive parts total half their absolute values. At least one $P_j$ consequently satisfiesThis is the Roth density-increment step, and it giveswith $c=c_1/4$ and $\eta >0$ depending only on $\delta$.

Put $N=p^n$, $\alpha =|A|/N$, and use normalized Fourier analysis on a finite abelian group. If $F=1_A\ast 1_A$ denotes unnormalized convolution and $h=N^{-1}F$, thenby the Parseval identity.

Sample $k=O(m{\epsilon }^{-2})$ characters independently, choosing $\gamma$ with probability $|\hat{h}(\gamma )|/\alpha$, and attach the phase of $\hat{h}(\gamma )$ to the sampled character. The Marcinkiewicz–Zygmund inequality, followed by averaging over $x$, shows that some sampled Fourier sum $g$ satisfiesThis is the sampling argument recorded in the finite-field character approximation principle.

Let $V$ be the intersection of the kernels of the sampled characters. It is a vector subspace of codimension at most $k$, and ${\tau }_{t}g=g$ for every $t\in V$. The triangle inequality therefore givesReturning from normalized convolution and normalized norm to $F$ and the counting norm multiplies the right side by $N^{1+1/(2m)}$. Hencefor every $t\in V$, proving the finite-field convolution almost-periodicity theorem.

For the Cauchy problem, insert the Fourier mode $u_m=e^{im\vartheta }$. The spatial symbol isIts real part is nonpositive for every $\vartheta$, so each mode has modulus $e^{t\mathrm{Re}{\lambda }_h}\le 1$. By the discrete Fourier transform and Parseval identity, the scheme is stable in the discrete $L^2$ norm.

For the discrete Fourier mode $u_m^n=G^n{e}^{im\theta }$, the centred second difference has multiplier $-4{\sin }^2(\theta /2)$. Hence the amplification factor isFor every $\mu >0$, $|G(\theta )|\le 1$. The discrete Parseval identity then shows directly that the discrete $L^2$ norm cannot increase at any time step, proving unconditional stability for the Cauchy problem.

The finite-field Bogolyubov lemma states that if $A\subseteq G={\mathbb{F}}_p^n$ has density $\alpha$, then $2A-2A$ contains a subspace of codimension at most $2{\alpha }^{-2}$.

Use normalized Fourier analysis on a finite abelian group and put $f=1_A$. DefineBy Parseval identity,so $|S|\le 2{\alpha }^{-2}$. LetThen $V$ is a subspace of codimension at most $|S|$.

The normalized representation function of $2A-2A$ iswhere $\tilde{f}(x)=f(-x)$. For $x\in V$, all terms indexed by $S$ are nonnegative real numbers, whileThe trivial character alone contributes ${\alpha }^4$, so $r(x)>0$. Hence $x\in 2A-2A$, proving the Finite-field Bogolyubov lemma.

Put $M(a)={\max }_{\gamma \in \hat{G}}|\widehat{{\partial }_{a}f}(\gamma ){|}^2$. By the Derivative identity for the Gowers U3 norm, the Fourier formula for the $U^2$ norm, and Parseval identity,because $\Vert f{\Vert }_{\infty }\le 1$. Letand for every $a\in B$ choose $\varphi (a)\in \hat{G}$ attaining $M(a)$. Thenfor every $a\in B$, and deleting the complement of $B$ from the preceding average leaves total mass at least $c/2$.

Let ${\Gamma }_{\varphi }=\{(a,\varphi (a)):a\in B\}\subseteq G\times \hat{G}$. The box-norm inequality applied to the selected Fourier coefficients and the cocycle identity for multiplicative derivatives givesThe left side is at least $(c/2{)}^8$. An additive quadruple in ${\Gamma }_{\varphi }$ is exactly a tuple $(a,b,c,d)\in B^4$ satisfyingTherefore there are at least $(c/2{)}^8|G{|}^3$ such quadruples, which is the Frequency graph extracted from a large Gowers U3 norm.

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

Consider first a scalar constant-coefficient Cauchy problem for a partial differential equation on the whole line. A translation-invariant spatial discretization is a convolution operator, so the discrete Fourier transform turns it into multiplication by its Fourier symbol. This reduction is exact because Fourier modes are the simultaneous generalized eigenfunctions of every translation-invariant stencil.

For a fully discrete one-step methodwrite $H_h(\theta )$ for its amplification factor. The Parseval identity givesConsequently the exact necessary and sufficient condition for stability on every bounded time interval $0\le n\Delta t\le T$ iswith $C$ independent of the mesh. Sufficiency follows directly from Parseval; necessity follows by choosing transformed initial data concentrated where the multiplier is largest. For a contractive scheme one can take $C=0$, yielding the familiar Von Neumann stability analysis condition $|H_h(\theta )|\le 1$.

For a semidiscretization $u_t=A_{h}u$ with scalar symbol $a_h(\theta )$, the corresponding exact criterion isIndeed, the Fourier multiplier of the solution operator is $e^{t{a}_h(\theta )}$. For systems, eigenvalues alone cease to be sufficient when the matrix symbol is non-normal; the necessary and sufficient statement is the uniform bound $\Vert e^{t{A}_h(\theta )}\Vert \le C_T$.

On the whole lattice, a finite stencil defines a Laurent operator, whose operator norm is the essential supremum of its symbol. Restricting the same stencil to a half-line gives a Toeplitz operator. The Cauchy symbol condition remains necessary for a stable initial-boundary scheme because data localized far from the boundary behave like the whole-line problem for a finite time. It is not sufficient: the boundary closure can support growing modes or amplify incoming modes even when every interior Fourier mode is stable.

Boundary stability therefore requires a uniform estimate for the forced half-line recurrence. After a Laplace transform in time and a Fourier transform in tangential variables, one solves a normal-direction recurrence. The Uniform Kreiss--Lopatinskii condition requires the boundary equations to determine its decaying roots with a uniformly bounded inverse. In practical terms, one imposes one independent boundary condition for each incoming characteristic or numerical mode and none for outgoing modes. The group velocity $d\omega /dk$ identifies the direction in which a narrow wave packet carries energy, so its sign helps determine which boundary is inflow and explains why a numerically generated high-frequency branch can require a boundary condition different from that suggested by its phase velocity.

For a two-step method, the Fourier substitution produces an amplification polynomial of a multilevel finite difference scheme. Every root must lie in the closed unit disk, and unit-modulus roots must be simple, uniformly in the wavenumber. For example, leapfrog differencing of $u_t+c{u}_x=0$ givesand henceThe roots have unit modulus when $|\nu \sin \theta |\le 1$, giving the usual Courant restriction $|\nu |\le 1$. At the endpoint, a wavenumber with $|\nu \sin \theta |=1$ produces a repeated unit root and violates the uniform root condition; strict $|\nu |<1$ avoids this marginal linear growth. The second root is the familiar oscillatory computational mode, illustrating why a multilevel scheme requires its full amplification polynomial rather than a single multiplier.

Iteration givesIf $|H(\theta )|\le 1$ everywhere, the Parseval identity givesConversely, if $|H({\theta }_0)|>1$, continuity supplies a neighbourhood on which $|H|\ge 1+ϵ$. Choose nonzero square-integrable Fourier data supported there. Its norm then grows at least as $(1+ϵ{)}^n$, contradicting boundedness. HenceThis is the one-step Von Neumann stability analysis criterion.

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus

The function is odd, so only sine coefficients occur. For $n\ge 1$,Integration by parts givesandThe terms of order $1/n$ cancel, leavingThus

The Parseval identity givesDirect integration yieldsThereforeas recorded in the Fourier series of x cubed minus pi squared x.

The Karhunen–Loève expansion has the following Hilbert-space form. Let $X$ be a square-integrable Random element of a Hilbert space $H$, with mean $\mu$ and covariance operator $C$. There are nonnegative eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$ and orthonormal eigenvectors ${\varphi }_k$ spanning the closure of the range of $C$ such thatin $L^2(\Omega ;H)$. The functional principal component scoressatisfyIf $X$ is a Gaussian random element, the scores are independent normal random variables.

For the proof, $C$ is positive, self-adjoint, and trace-class, withIt is therefore compact, so the spectral theorem for compact Hermitian operators supplies the eigenpairs. Their score covariance isFor the residual $R_m=X-\mu -{\sum }_{k=1}^m{\xi }_k{\varphi }_k$, Parseval identity and the trace formula giveThe centered variable's projection onto $\ker C$ has zero second moment and is therefore zero almost surely, which completes the mean-square expansion. When $C$ has a continuous covariance kernel, Mercer's theorem additionally expands that kernel as $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

Separation $y=T(t)X(x)$ and the fixed-end conditions givePutThe initial displacement and zero initial velocity then give the separated solution of the damped string equation

For the triangular initial displacement,Thus the even coefficients vanish and the odd ones alternate in sign.

LetThenOrthogonality of the sine and cosine modes, equivalently the Parseval identity, gives

The sign of its derivative is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yieldsThis is the energy dissipation identity for a linearly damped string.

Let $B={\mathbb{F}}_2^n\setminus (A+A)$ have density $\beta$. Since $1_A\ast 1_A$ vanishes on $B$,The zero-frequency contribution is ${\alpha }^2\beta$, soPut $\Gamma ={\mathrm{Spec}}_{\alpha /2}(1_B)$. Outside $\Gamma$, Parseval identity bounds the contribution byThereforebecause $|\widehat{1_B}(t)|\le \beta$. Chang's theorem places $\Gamma$ in a subspace $W$ of dimensionSet $V=W^{\perp }$. Then $V$ has this codimension, $V^{\perp }=W$, and adding the zero-frequency term $|\widehat{1_A}(0){|}^2={\alpha }^2$ gives