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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 sum in question is $⟨Pu,Pv⟩$. Applying Cauchy-Schwarz and then Bessel's inequality givesIf either original vector is zero, equality holds. If both are nonzero, equality forces both projection norms to equal their original norms and also forces dependence of the projected vectors. Thus apart from the zero-vector cases, equality holds exactly when $u,v\in W$ and $u,v$ are linearly dependent.

For the final unnumbered request, apply Cauchy-Schwarz in ${\mathbb{R}}^k$ to $p_i=a^{i/2}$ and $q_i=a^{-i/2}$. Their inner product is $k$, soEquality requires the positive vectors $p,q$ to be proportional, equivalently $a^i$ constant for $i=1,\dots ,k$. For $k\ge 2$ this is $a=1$. If $k=1$, equality holds for every $a>0$.

A real inner product is a symmetric bilinear map $⟨\cdot ,\cdot ⟩:V\times V\to \mathbb{R}$ with $⟨v,v⟩>0$ for $v\ne 0$. It defines $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Set $W=\mathrm{span}\{e_1,\dots ,e_k\}$ and $Pv={\sum }_i⟨v,e_i⟩e_i$. Orthonormality gives $v-Pv\perp W$, hence the Pythagorean identity givesThis proves Bessel's inequality. Equality holds exactly when $v\in W$.

Let $p={\sum }_{j=1}^m⟨u,v_j⟩v_j$. Orthonormality gives $⟨u-p,v_k⟩=0$ for every $k$, so $u-p$ is orthogonal to $p$. The Pythagorean theorem for inner product spaces givesSince a squared norm is nonnegative,This is Bessel's inequality, with equality exactly when $u$ belongs to the span of the orthonormal set.

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.

Apply continuity of the inner product to the convergent expansion of $x$:The series is absolutely convergent because the Cauchy-Schwarz inequality for scalar sequences and Bessel's inequality giveThis is the polarized Parseval identity.

For $a_n=⟨x,e_n⟩$, the finite orthogonal projection $P_{N}x={\sum }_{|n|\le N}{a}_n{e}_n$ satisfiesThis is Bessel's inequality, so the coefficients are square summable. The preceding result gives a limit $s={\sum }_n{a}_n{e}_n$. For each fixed $k$, continuity of the inner product gives $⟨x-s,e_k⟩=0$. The completeness assumption on the orthonormal system forces $x-s=0$. HenceTaking squared norms also gives Parseval's identity, $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$.

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.

The residual $x-P_{n}x$ is orthogonal to $P_{n}x\in U_n$. Thus the Pythagorean theorem gives $\Vert x{\Vert }^2=\Vert x-P_{n}x{\Vert }^2+\Vert P_{n}x{\Vert }^2$. Substitute part (ii) and rearrange:Nonnegativity of the left-hand side also gives the finite form of Bessel's inequality.

For an orthonormal family in a Hilbert space, the squared norm of every finite partial sum is the sum of squared coefficients. Square summability therefore makes the net of finite partial sums Cauchy, proving unconditional convergence by completeness. If the orthonormal family is complete, Bessel's inequality makes the coefficients of each vector square summable, and the residual after summing is orthogonal to the whole family, hence zero. This recovers the orthonormal expansion and Parseval identity.

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 ${\ell }^2$.