Kolmogorov simultaneous phase approximation

For $r=0$, take $L=1$. Otherwise put $b_j(t)=\max (0,1-2\mathrm{\mid }{e}^{2\pi it}-z_j\mathrm{\mid }\mathrm{/}\epsilon )$. This continuous periodic function lies between zero and one and is positive only when $\mathrm{\mid }{e}^{2\pi it}-z_j\mathrm{\mid }<\epsilon \mathrm{/}2$. It equals one at an argument of $z_j$, and continuity makes its integral ${\beta }_j$ strictly positive. Set $b(t_1,\dots ,t_r)={\prod }_j{b}_j(t_j)$ and $\beta ={\prod }_j{\beta }_j>0$.

By F2, approximate each $b_j$ uniformly within $\rho \le 1$ by a trigonometric polynomial $p_j$. Then $\mathrm{\mid }{p}_j\mathrm{\mid }\le 2$, and telescoping products yields $\sup \mathrm{\mid }{\prod }_j{p}_j-{\prod }_j{b}_j\mathrm{\mid }\le r{2}^{r-1}\rho$. The constant coefficient of $p={\prod }_j{p}_j$ as a polynomial in $r$ coordinates is the product of the individual constant coefficients. Each differs from ${\beta }_j$ by at most $\rho$, by the integral definition in F1. Thus that coefficient differs from $\beta$ by at most $r{2}^{r-1}\rho$ as well. This argument needs no multivariable approximation theorem or interchange of infinite series.

For each nonzero integer vector $k\in {\mathbb{Z}}^r$, independence gives $k\cdot x\mathrm{\notin }\mathbb{Z}$. Put $u=e^{2\pi ik\cdot x}\mathrm{\ne }1$. Then $M^{-1}{\sum }_{L=1}^M{u}^L=u(1-u^M)\mathrm{/}(M(1-u))\to 0$. The zero vector gives average one. Applying these identities to the finitely many terms of $p$ proves that $M^{-1}{\sum }_{L=1}^{M}p(L{x}_1,\dots ,L{x}_r)$ tends to its constant coefficient. Step 2.1 bounds the limsup of the absolute difference between the corresponding average of $b$ and $\beta$ by $2r{2}^{r-1}\rho$. Since every positive $\rho \le 1$ is allowed, the average of $b$ tends to $\beta >0$.

Some positive integer $L$ therefore has $b(L{x}_1,\dots ,L{x}_r)>0$. Every factor is positive, so step 1.1 gives all the required strict phase inequalities (indeed with epsilon/2). Positivity of $L$ follows from using averages indexed from one, not from a symmetry argument about negative times. Only finitely many approximants are selected for each fixed rho; the proof uses no axiom of choice.