Absolutely convergent Fourier reconstruction for C2 periodic functions

For $n\mathrm{\ne }0$, integrate by parts twice in the definition of $c_n$. The boundary terms vanish because $f$ and $f^{\mathrm{\prime }}$ have matching endpoint values and $e^{in\pi }=e^{-in\pi }$, so $\widehat{f^{\mathrm{\prime }\mathrm{\prime }}}(n)=-n^2{c}_n$, where $\widehat{f^{\mathrm{\prime }\mathrm{\prime }}}(n)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{f}^{\mathrm{\prime }\mathrm{\prime }}(x)e^{-inx}dx$. Thus $\mathrm{\mid }{c}_n\mathrm{\mid }\le (2\pi n^2{)}^{-1}{\int }_{-\pi }^{\pi }\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid }dx$. The comparison ${\sum }_{n\ge 1}{n}^{-2}\le 1+{\int }_1^{\mathrm{\infty }}{x}^{-2}dx<\mathrm{\infty }$, together with finite $c_0$, gives ${\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }<\mathrm{\infty }$.

The bound ${\sup }_x\mathrm{\mid }{\sum }_{N<\mathrm{\mid }n\mathrm{\mid }\le M}{c}_n{e}^{inx}\mathrm{\mid }\le {\sum }_{\mathrm{\mid }n\mathrm{\mid }>N}\mathrm{\mid }{c}_n\mathrm{\mid }$ makes the partial sums uniformly Cauchy. They converge uniformly to a continuous periodic function $g$; integrating the uniformly convergent series term by term shows $\widehat{g}(n)=c_n$: for integer $\mathrm{\ell }=k-n$, the normalized integral of $e^{i\mathrm{\ell }x}$ is $1$ if $\mathrm{\ell }=0$, while if $\mathrm{\ell }\mathrm{\ne }0$ its antiderivative is $e^{i\mathrm{\ell }x}\mathrm{/}(i\mathrm{\ell })$ and its endpoint values at $-\pi ,\pi$ agree, so the integral is $0$. Hence $h=f-g$ is continuous, periodic, and all its Fourier coefficients vanish.

For an integer $N\ge 0$, put $K_N(t)=(N+1{)}^{-1}\mathrm{\mid }{\sum }_{j=0}^N{e}^{ijt}{\mathrm{\mid }}^2$. Expanding the finite square gives $K_N(t)={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}(1-\mathrm{\mid }k\mathrm{\mid }\mathrm{/}(N+1))e^{ikt}$; therefore $K_N\ge 0$ and $(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{K}_N(t)dt=1$. For a continuous periodic $q$, define ${\sigma }_{N}q(x)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }q(x-t)K_N(t)dt$. The finite expansion shows that if every Fourier coefficient of $q$ is zero, then ${\sigma }_{N}q(x)=0$ for every $x$.

Given $\epsilon >0$, uniform continuity of $q$ gives $0<\delta <\pi$ such that $\mathrm{\mid }q(x-t)-q(x)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }t\mathrm{\mid }<\delta$, uniformly in $x$. For $\delta \le \mathrm{\mid }t\mathrm{\mid }\le \pi$, the geometric-sum identity gives $K_N(t)=(N+1{)}^{-1}\mathrm{\mid }\sin ((N+1)t\mathrm{/}2)\mathrm{/}\sin (t\mathrm{/}2){\mathrm{\mid }}^2\le ((N+1){\sin }^2(\delta \mathrm{/}2){)}^{-1}$. Splitting the convolution integral at $\delta$, positivity and unit mass of $K_N$ bound ${\sup }_x\mathrm{\mid }{\sigma }_{N}q(x)-q(x)\mathrm{\mid }$ by $\epsilon +2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}((N+1){\sin }^2(\delta \mathrm{/}2){)}^{-1}$. Thus ${\sigma }_{N}q\to q$ uniformly. Applying this to $h$, whose Fejér means vanish by step 3.1, gives $h=0$, so $g=f$ and the Fourier series reconstructs $f$ absolutely and uniformly.

Integration by parts once gives the $n$-th Fourier coefficient of $f^{\mathrm{\prime }}$ as $in{c}_n$. The Fejér means ${\sigma }_N{f}^{\mathrm{\prime }}$ are trigonometric polynomials of degree at most $N$ and converge uniformly to $f^{\mathrm{\prime }}$ by step 4.1. The symmetric Fourier partial sum $S_N{f}^{\mathrm{\prime }}={\sum }_{\mathrm{\mid }n\mathrm{\mid }\le N}in{c}_n{e}^{inx}$ is the orthogonal projection onto that finite span, since the exponentials are orthonormal by the elementary integral in step 2.1. Therefore $\mathrm{\parallel }{f}^{\mathrm{\prime }}-S_N{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{f}^{\mathrm{\prime }}-{\sigma }_N{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{f}^{\mathrm{\prime }}-{\sigma }_N{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, proving the stated $L^2$ convergence.