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For a periodic absolutely continuous function with derivative in $L^2$, integration by parts gives $\widehat{f^{\prime }}(n)=in\hat{f}(n)$. Bessel's inequality and the Cauchy-Schwarz inequality prove the displayed bound in the normalized circle norm. Adding the constant coefficient gives an absolutely and uniformly convergent Fourier series. Merely bounding individual coefficients by $O(1/|n|)$ would not establish absolute convergence.

The absolute Fourier convergence from a square-integrable derivative uses more than the pointwise estimate $|\hat{f}(n)|=O(1/|n|)$. Periodic integration by parts givesBy Bessel's inequality,Now apply the Cauchy-Schwarz inequality:Adding the finite constant coefficient proves absolute summability. Since a continuously differentiable periodic function has $f^{\prime }\in L^2$, all hypotheses of part (iii) hold and its Fourier series converges uniformly.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.

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.

Choose a singular system of a compact operator with $K{v}_j={\sigma }_j{u}_j$, $K^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. The Picard criterion, together with range orthogonality, givesFor necessity, expand a solution's component in $(\ker K{)}^{\perp }$ in the orthonormal vectors $v_j$; its coefficients must be $⟨f,u_j⟩/{\sigma }_j$. Their square sum is finite by Bessel's inequality. Conversely, the displayed square sum defines a convergent Hilbert space seriesApplying the bounded linear operator $K$ to the partial sums yields the expansion of $f$ in $\overline{\mathrm{ran}K}=(\ker K^{\ast }{)}^{\perp }$. Thus its limit solves $Ku=f$. The perpendicularity condition is essential here: the Picard criterion alone characterizes the domain of the generalized inverse, which also includes components in $\ker K^{\ast }$.

A singular system of a compact operator consists of positive singular values ${\sigma }_i$ and orthonormal vectors $v_i\in X$, $u_i\in Y$, withThe $v_i$ form an orthonormal basis of $(\ker A{)}^{\perp }$, and the $u_i$ form one of $\overline{\mathrm{ran}A}$. There are finitely many positive singular values for finite-rank $A$; otherwise ${\sigma }_i\to 0$. Obtain the system by applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$, choosing $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$, and setting $u_i=A{v}_i/{\sigma }_i$. The inner product identities verify orthonormality and the displayed relations.

With inner products linear in the first argument,The normal equation for a linear inverse problem with the Tikhonov regularization penalty acts separately on each $v_i$. Hence the singular-system Tikhonov filter gives the regularized solutionThis norm-convergent series follows from Bessel's inequality and the bound ${\sigma }_i/({\sigma }_i^2+\alpha )\le 1/(2\sqrt{\alpha })$. Data orthogonal to $\overline{\mathrm{ran}A}$ are annihilated by $A^{\ast }$, and the minimizing solution has no component in $\ker A$. The same formula with $y^{\delta }$ reconstructs noisy data. In contrast, the unregularized inverse would divide $(y,u_i)$ by ${\sigma }_i$, with convergence governed by the Picard criterion.

When the continuous Fourier transform $F$ of an integrable function vanishes outside $[-\pi ,\pi ]$, its periodic Fourier coefficients are the integer samples of the continuous representative of $f$, with the sign of the index reversed. Projecting $F$ in $L^2[-\pi ,\pi ]$ and then applying the Fourier inversion theorem gives the displayed reconstruction using the normalized sinc function, with each quotient defined as one when $t=n$. The Cauchy-Schwarz inequality controls the reconstruction error uniformly in $t$. Bessel's inequality bounds the vector of shifted sinc functions, giving absolute uniform tails when the sample sequence belongs to ${\ell }^2$.