= Solution
Use the orthonormal <Fourier series> basis $e_n(x)=(2\pi)^{-1/2}e^{inx}$, indexed by $n\in\mathbb Z$, and the <Hilbert space> <inner product> $\langle f,g\rangle=\int_0^{2\pi}\overline{f(x)}g(x)dx$. The integral operator is a <periodic convolution operator>. Changing variables $s=x-x'$ and using periodicity gives
$$
(Ae_n)(x)=\frac{e^{inx}}{\sqrt{2\pi}}\int_0^{2\pi}K(s)e^{-ins}ds=c_n e_n(x).
$$
There is no extra factor $2\pi$: the paper's $c_n$ already includes the full integral, rather than its normalized Fourier-series coefficient. The adjoint kernel is $\overline{K(x'-x)}$, so $A^*e_n=\overline{c_n}e_n$.
A <singular system of a periodic convolution operator> is therefore
$$
\boxed{\sigma_n=|c_n|,\qquad v_n=e_n,\qquad u_n=\frac{c_n}{|c_n|}e_n\quad(n\in\mathbb Z).}
$$
Indeed $Av_n=\sigma_nu_n$ and $A^*u_n=(c_n/|c_n|)\overline{c_n}e_n=\sigma_nv_n$. The unit factor determined by the <complex argument> of $c_n$ in $u_n$ is necessary when the complex <Fourier coefficients> are not positive real numbers. Both families are orthonormal and complete, since every $c_n\ne0$. One may enumerate $\mathbb Z$ by $\mathbb N$, or order the positive <singular values> by decreasing magnitude.
The continuous kernel on a finite square makes $A$ a <Hilbert-Schmidt operator>, hence compact. Since $K$ is continuously differentiable and periodic, integration by parts gives, for $n\ne0$,
$$
c_n=\frac1{in}\int_0^{2\pi}K'(s)e^{-ins}ds=o(1/|n|)
$$
by the <Riemann-Lebesgue lemma>. In particular the <singular values> tend to zero. Nonzero multipliers imply both $\ker A=0$ and $\ker A^*=0$: the range is dense, but the inverse is unbounded and the range is not closed.
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