= Solution
The discrete <Bochner–Herglotz theorem> says that a sequence $r:\mathbb Z\to\mathbb C$ is a <positive-definite function>, meaning
$$
\sum_{j,k=0}^{N-1}c_j\overline{c_k}\,r(j-k)\geq0\quad\hbox{for every }N\text{ and every }c_0,\ldots,c_{N-1},
$$
if and only if there is a unique finite positive <Borel measure> $\sigma$ on $\mathbb T=\mathbb R/\mathbb Z$ such that
$$
\boxed{r(n)=\int_{\mathbb T}e^{2\pi int}\,d\sigma(t)\quad(n\in\mathbb Z).}
$$
The mass is $\sigma(\mathbb T)=r(0)$. The necessity of positive definiteness follows by integrating $|\sum_j c_je^{2\pi ijt}|^2$.
For sufficiency, positive definiteness makes the relevant <Gram matrices> Hermitian positive semidefinite, so $r(-n)=\overline{r(n)}$ and $|r(n)|\leq r(0)$. Define
$$
p_N(t)=\frac1N\sum_{j,k=0}^{N-1}r(j-k)e^{-2\pi i(j-k)t}
=\sum_{|n|<N}\left(1-\frac{|n|}{N}\right)r(n)e^{-2\pi int}.
$$
Taking $c_j=e^{-2\pi ijt}$ in the positive-definiteness condition gives $p_N(t)\geq0$, while $\int_{\mathbb T}p_N(t)\,dt=r(0)$. Moreover,
$$
\int_{\mathbb T}e^{2\pi imt}p_N(t)\,dt
=\begin{cases}(1-|m|/N)r(m),&|m|<N,\\0,&|m|\geq N.
\end{cases}
$$
Thus the integrals against every <trigonometric polynomial> converge. These polynomials are uniformly dense in continuous functions on the circle: smooth periodic approximation followed by <Fourier inversion> for smooth functions gives this density. The uniform mass bound $r(0)$ then makes the integrals against every continuous $\varphi$ converge as well. Their limit is a positive linear functional $L$ with $|L(\varphi)|\leq r(0)\|\varphi\|_\infty$. The standard <Riesz-Markov-Kakutani representation theorem> gives a finite positive <Borel measure> $\sigma$ with $L(\varphi)=\int\varphi\,d\sigma$. The preceding coefficient computation proves its required <Fourier coefficients>. Finally, two such measures agree on all <trigonometric polynomials>, hence on all continuous functions by uniform density, and therefore are identical. This proves the <Bochner–Herglotz theorem>, including existence and uniqueness; if $r(0)=0$, the estimate $|r(n)|\leq r(0)$ gives the zero measure directly.
Now let $U$ be the <Koopman operator> $Uf=f\circ T$ on the complex <L2 space> of the probability <measure-preserving system>. We use $\langle f,g\rangle=\int f\overline g\,d\mu$, linear in the first entry. Since $U$ is an <isometry>, the sequence $r_f(n)=\langle U^nf,f\rangle$ for $n\geq0$, extended by $r_f(-n)=\overline{r_f(n)}$, is positive-definite: its matrix is the <Gram matrix> of $f,Uf,\ldots$. The <Bochner–Herglotz theorem> gives its <spectral measure of a Koopman observable> $\sigma_f$.
We first prove exactly how an <atom of a measure> detects an <eigenfunction>. Fix $\zeta=e^{2\pi it_0}$ and set
$$
A_N^\zeta f=\frac1N\sum_{j=0}^{N-1}\zeta^{-j}U^jf.
$$
The <mean ergodic theorem>, applied to the <isometry> $V=\zeta^{-1}U$, says these averages converge in <L2 space> to the <orthogonal projection> $P_\zeta f$ onto $\ker(U-\zeta I)$. Its usual Hilbert-space proof works without invertibility: the orthogonal complement of $\overline{\operatorname{ran}(I-V)}$ is $\ker(I-V^*)=\ker(I-V)$, and the averages of $(I-V)h$ telescope to $(h-V^Nh)/N\to0$. The equality of the two kernels follows from $V$ being an <isometry>, by expanding $\|Vh-h\|^2$ when $V^*h=h$.
Expanding the averaged norm through the <spectral measure of a Koopman observable> gives
$$
\|A_N^\zeta f\|_2^2=\int_{\mathbb T}\left|\frac1N\sum_{j=0}^{N-1}e^{2\pi ij(t-t_0)}\right|^2\,d\sigma_f(t).
$$
The integrand is at most one and converges to $\mathbf1_{\{t_0\}}$. By the <dominated convergence theorem>,
$$
\boxed{\|P_\zeta f\|_2^2=\sigma_f(\{t_0\}).}
$$
Thus a nonzero atom corresponds to a nonzero <Koopman operator> <eigenfunction>, with eigenvalue $\zeta$, in the cyclic spectral data of $f$.
We also need the <Wiener Fourier coefficient lemma>, which here has a short proof. For any finite positive circle measure $\sigma$ with coefficients $r(n)$,
$$
\frac1N\sum_{n=0}^{N-1}|r(n)|^2
=\iint\left(\frac1N\sum_{n=0}^{N-1}e^{2\pi in(t-s)}\right)\,d\sigma(t)\,d\sigma(s).
$$
The bracket is bounded in absolute value by one and tends to the indicator of the diagonal $t=s$. Another application of the <dominated convergence theorem> gives
$$
\boxed{\lim_{N\to\infty}\frac1N\sum_{n=0}^{N-1}|r(n)|^2=\sum_{t\in\mathbb T}\sigma(\{t\})^2.}
$$
In particular the limit is zero for an <atomless measure>.
Suppose there are no nonconstant <Koopman operator> <eigenfunctions>. For every $f$ in the <mean-zero L2 space>, all its projections $P_\zeta f$ vanish: a nonzero projection would be a nonconstant <eigenfunction>, since it has mean zero. Hence $\sigma_f$ is atomless, and the <Wiener Fourier coefficient lemma> gives
$$
\frac1N\sum_{n<N}|\langle U^nf,f\rangle|^2\longrightarrow0.
$$
The <Cauchy-Schwarz inequality> also gives vanishing averaged absolute autocorrelations. For two mean-zero observables $f,g$, the sesquilinear polarization identity
$$
\langle U^nf,g\rangle=\frac14\sum_{j=0}^3i^j\langle U^n(f+i^jg),f+i^jg\rangle
$$
then gives vanishing averaged absolute cross-correlations. Taking $f=\mathbf1_A-\mu(A)$ and $g=\mathbf1_B-\mu(B)$ proves
$$
\frac1N\sum_{n<N}|\mu(T^{-n}A\cap B)-\mu(A)\mu(B)|\longrightarrow0,
$$
which is <weak mixing>.
Conversely, this set formulation of <weak mixing> extends to all pairs of mean-zero <L2 space> observables by finite-simple-function approximation and the uniform estimate $|\langle U^nf,g\rangle|\leq\|f\|_2\|g\|_2$. If $Uh=\zeta h$ and $h\ne0$, the <isometry> property gives $|\zeta|=1$. If $\zeta\ne1$, invariance of the integral gives $\int h=0$. If $\zeta=1$ and $h$ is nonconstant, replace it by $h-\int h$, a nonzero mean-zero <eigenfunction>. But in either case
$$
|\langle U^nh,h\rangle|=\|h\|_2^2>0
$$
for every $n$, contradicting <weak mixing>. We have proved \b[weak mixing is equivalent to absence of nonconstant <eigenfunctions>], including for noninvertible transformations.
Finally, the <doubling map> $Tx=2x\pmod1$ preserves normalized <Lebesgue measure>, as substitution on its two inverse branches shows. Let $f\in L^2(\mathbb T)$ be an <eigenfunction>, $f(2x)=\zeta f(x)$, with <Fourier coefficients> $c_k$. The <Fourier basis> calculation gives
$$
\widehat{Uf}(k)=\begin{cases}c_{k/2},&k\text{ even},\\0,&k\text{ odd}.
\end{cases}
$$
Since $\zeta\ne0$, all odd coefficients vanish. Every nonzero integer is $2^rm$ with $m$ odd, and repeated use of $\zeta c_{2^rm}=c_{2^{r-1}m}$ makes all nonzero coefficients vanish. Completeness of the <Fourier basis> then says $f$ is constant. Therefore the <doubling map> has no nonconstant <eigenfunctions>, and
$$
\boxed{x\mapsto2x\pmod1\text{ is weakly mixing for Lebesgue measure}.}
$$
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