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For a one-step translation-invariant recurrence, the amplification factor is the multiplier satisfyingThe Parseval identity converts the pointwise bound $|H|\le 1$ into non-growth of the discrete $2$-norm.

For a bounded function on an interval with finitely many jumps and piecewise $C^{\alpha }$ bounds, localized orthonormal wavelets with sufficient vanishing moments give squared best N-term approximation error $O(N^{-2\alpha })$. There are $O(2^j)$ smooth-region coefficients of size $O(2^{-j(\alpha +1/2)})$, and only $O(K)$ jump-crossing coefficients of size $O(2^{-j/2})$ per level. Retain all coefficients through level $J$ and the jump-crossing coefficients through $L=\lceil 2\alpha J\rceil$ when $\alpha >1/2$. This costs $O(2^J+KJ)$ and leaves squared tails $O(2^{-2\alpha J}+2^{-L})$. Choosing $2^J$ comparable to $N$ proves the rate by the Parseval identity.

The complex form of a Fourier series for period $2L$ uses coefficients $c_n=(2L{)}^{-1}{\int }_{-L}^{L}f(x)e^{-in\pi x/L}dx$ and complex exponentials $e^{in\pi x/L}$. For square-integrable functions it converges in the $L^2$ norm, and the Parseval identity is $(2L{)}^{-1}\int |f{|}^2={\sum }_n|c_n{|}^2$. Real functions satisfy $c_{-n}=\overline{c_n}$.

For a set of density $\alpha$ in a finite cyclic group, select frequencies whose normalized Fourier coefficients on a finite abelian group have magnitude at least ${\alpha }^{3/2}/2$. Parseval identity bounds their number by $4{\alpha }^{-2}$. Fourier inversion of the fourfold convolution shows it is positive on the Bohr set in phase-distance convention of radius $1/6$. This is an explicit form of the Bogolyubov lemma.

Suppose $|f|\le 1$ on a finite cyclic group, $H$ has density $\beta$, and for each $h\in H$ the multiplicative derivative has a Fourier coefficient on a finite abelian group of magnitude at least $\eta$ at $\theta (h)$. Align the correlation phases, apply Cauchy-Schwarz, group pairs by their shift and frequency differences, and apply Parseval identity. This proves the displayed lower bound on additive energy of a frequency graph without replacing exact frequency equalities by approximate ones.

Let $I=[n]$ for odd $n\ge 3$, embed it in the cyclic group $G=\mathbb{Z}/(2n+1)\mathbb{Z}$, and let $A\subseteq I$ have density of a finite subset $\alpha$. If $A$ has no nonconstant three-term arithmetic progression and $n\ge 4{\alpha }^{-2}$, its balanced indicator function $f=1_A-\alpha 1_I$ satisfiesIndeed, the normalized trilinear arithmetic progression count has the Fourier analysis on a finite abelian group formula $\Lambda (h_1,h_2,h_3)={\sum }_r{\hat{h}}_1(r){\hat{h}}_2(-2r){\hat{h}}_3(r)$. Its values on $1_A$ and $\alpha 1_I$ are $\alpha n/|G{|}^2$ and ${\alpha }^3(n^2+1)/(2|G{|}^2)$. Telescoping their difference into three terms containing $f$, the Parseval identity and Cauchy-Schwarz inequality bound its magnitude by $3\max |\hat{f}|\alpha n/|G|$. The difference is at least ${\alpha }^3{n}^2/(4|G{|}^2)$, giving the claim. The zero Fourier coefficient vanishes by the definition of $\alpha$.

Fourier stability analysis transforms a translation-invariant finite-difference scheme into scalar or matrix recurrences indexed by wavenumber. The Parseval identity then converts uniform multiplier bounds into discrete $2$-norm bounds.

The first Gaussian Sobolev space consists of functions and their weak first derivatives square-integrable for standard Gaussian measure. If $f=\sum c_n{h}_n$ in the normalized Probabilists' Hermite polynomial basis, Gaussian integration by parts gives $⟨f^{\prime },h_{n-1}⟩=\sqrt{n}{c}_n$. Thus Parseval identity identifies it with $\sum (1+n)|c_n{|}^2<\infty$, and polynomial truncations converge in its norm. It is the form domain of the Gaussian number operator, with Gaussian Dirichlet energy $\sum n|c_n{|}^2$.

The Hilbert-Schmidt norm of a Hilbert-Schmidt operator is the square root of the displayed sum. The value is independent of the orthonormal basis, by the Parseval identity. For a finite matrix, it is the Frobenius norm.

For the Fourier transform ${\hat{E}}_L(\nu )=(2\pi {)}^{-1}{\int }_{-L/2}^{L/2}E(z)e^{-i\nu z}dz$ of a stationary random field, the periodogram measure $(2\pi /L)\mathbb{E}|{\hat{E}}_L(\nu ){|}^{2}d\nu$ converges weakly to the spectral measure of a stationary random field. The expected squared modulus is $(4{\pi }^2{)}^{-1}{\int }_{-L}^L(L-|\zeta |)C(\zeta )e^{-i\nu \zeta }d\zeta$. The factor $2\pi /L$ follows from Parseval identity, and preserves the total power per unit length. It also avoids meaningless squares of Dirac delta distributions for coherent plane-wave components.

Use $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$ and let $P_j$ be the orthogonal projection onto the closed span of the orthonormal translates and dilates of a scaling function. If $\hat{f}$ is supported in $[-R,R]$ and $R<\pi 2^j$, thenIndeed, the Plancherel theorem writes the coefficient against ${\phi }_{j,k}$ as $2^{-j/2}(2\pi {)}^{-1}\int \hat{f}(\xi )\overline{\hat{\phi }(2^{-j}\xi )}{e}^{ik{2}^{-j}\xi }d\xi$. With $\xi =2^j\theta$, these are $2^{j/2}$ times the Fourier series coefficients of $\hat{f}(2^j\theta )\overline{\hat{\phi }(\theta )}$ on $[-\pi ,\pi ]$. The Parseval identity proves the formula. It shows that continuity and unit modulus at zero imply density of the refinement spaces, and conversely that density forces this unit modulus when the Fourier transform is continuous at zero.

An N-term approximation in an orthonormal basis uses at most $N$ basis functions. For a chosen index set $\Lambda$, its optimal coefficients are the inner products with those functions, and its squared error is the sum of the omitted squared coefficients by the Parseval identity.