Periodic convolution operator ($Af(x)={\int }_0^{2\pi }K(x-x^{\prime })f(x^{\prime })d{x}^{\prime }$)

A periodic convolution operator with continuous periodic kernel is compact on $L^2[0,2\pi ]$, since its kernel is square integrable. In the normalized Fourier series basis it is diagonal, with multipliers $c_n={\int }_0^{2\pi }K(x)e^{-inx}dx$. These multipliers include the full integral rather than its normalized Fourier coefficient.

With $e_n(s)=(2\pi {)}^{-1/2}{e}^{ins}$, periodic $L^2$ functions have the displayed convolution identity for every real $t$. Apply the polarized Parseval identity to $f$ and $\overline{g(t-\cdot )}$; a change of variable gives the latter function's pairing with $e_n$. The Cauchy-Schwarz inequality bounds the sum of absolute coefficient products by $\Vert f{\Vert }_2\Vert g{\Vert }_2$, so the resulting series converges absolutely and uniformly. This proves continuity of the convolution even though neither original Fourier series need converge pointwise.

When all multipliers are nonzero, the Moore–Penrose inverse of an operator acts by dividing each Fourier coefficient $y_n$ by $c_n$. Its data domain is precisely the square-summability condition displayed. The dense range need not be closed, and writing the formal inverse series does not establish the Picard criterion.

For a continuously differentiable periodic kernel, integration by parts and the Riemann-Lebesgue lemma give $c_n=o(1/|n|)$. Nonperiodic data $e^{\alpha x}$ with $\alpha \notin i\mathbb{Z}$ have Fourier coefficients asymptotic to a nonzero multiple of $1/n$. Their inverse coefficients $y_n/c_n$ therefore fail square summability, and there is no Hilbert-space inverse or least-squares minimizer. Periodic exponential data $\alpha =im$ instead give the single-mode solution $e^{imx}/c_m$.