Kolmogorov’s Block Construction and Almost-Everywhere Divergence

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.

At step $j$, the rational linear span of $1,t_1,\dots ,t_{j-1}$ is countable: enumerate rational tuples by listing integer numerators and positive denominators of bounded absolute size, then increase the bound; evaluate each tuple. F3 makes this span null. The open cell $((j-1)\mathrm{/}n,j\mathrm{/}n)$ has measure $1\mathrm{/}n>0$, so contains a point $t_j$ outside the span. Finite induction chooses these points and proves their independence. Let $V$ be their full rational linear span with one; the same enumeration makes $V$ countable and null. AC supplies the countable-choice measure hypotheses in F3; selecting this fixed finite tuple adds no arbitrary-index choice.

Fix $x\in [0,1)\setminus V$ and write $y_j=x-t_j$. If $q_0+{\sum }_j{q}_j{y}_j=0$ for rational coefficients, put $q={\sum }_j{q}_j$. If $q=0$, independence of $1,t_1,\dots ,t_n$ forces all coefficients zero. If $q\mathrm{\ne }0$, the equation puts $x=({\sum }_j{q}_j{t}_j-q_0)\mathrm{/}q$ in $V$, impossible. Thus $1,y_1,\dots ,y_n$ are independent, and none of the $y_j$ is an integer.

Set $z_j=i{e}^{-\pi i{y}_j}\mathrm{sgn}(\sin (\pi y_j))$, which has modulus one. Apply F1 with error $1\mathrm{/}2$ to obtain $L\ge 1$. Multiplying the approximation by $e^{\pi i{y}_j}\mathrm{sgn}(\sin (\pi y_j))$ shows $\mathrm{\mid }{e}^{(2L+1)\pi i{y}_j}\mathrm{sgn}(\sin (\pi y_j))-i\mathrm{\mid }<1\mathrm{/}2$. Its imaginary part exceeds $1\mathrm{/}2$, hence F2 gives $D_L(y_j)>1\mathrm{/}(2\mathrm{\mid }\sin (\pi y_j)\mathrm{\mid })\ge 1\mathrm{/}(2\pi \mathrm{\mid }{y}_j\mathrm{\mid })$. All terms have the same positive sign; there is no cancellation in their average.

Let $k\in \{1,\dots ,n\}$ be the cell index with $(k-1)\mathrm{/}n\le x<k\mathrm{/}n$. Cell placement implies $\mathrm{\mid }x-t_j\mathrm{\mid }\le (\mathrm{\mid }j-k\mathrm{\mid }+1)\mathrm{/}n$. At least one side from $k$ to an endpoint contains $m+1\ge n\mathrm{/}2$ indices with distances $\mathrm{\ell }=0,\dots ,m$. Therefore $n^{-1}{\sum }_j\mathrm{\mid }x-t_j{\mathrm{\mid }}^{-1}\ge {\sum }_{\mathrm{\ell }=0}^m(\mathrm{\ell }+1{)}^{-1}\ge \log (m+2)\ge \log (n\mathrm{/}2)\ge (\log n)\mathrm{/}2$. The harmonic bound follows by integrating $1\mathrm{/}t$ over each $[\mathrm{\ell }+1,\mathrm{\ell }+2]$, where it is at most $1\mathrm{/}(\mathrm{\ell }+1)$. Step 3.1 now gives $B_L(x)>(\log n)\mathrm{/}(4\pi )$. This holds off the single countable null set $V$, proving the assertion with the stated uniform constant. Grid endpoints are covered by the half-open choice of k; singular points $t_j$ were excluded in step 2.1.

Choose an atomic configuration from F2 with $c\log n>2H+2$, and write $b_k=n^{-1}{\sum }_j{e}_{-k}(t_j)$. Then $\mathrm{\mid }{b}_k\mathrm{\mid }\le 1$ and $B_L={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le L}{b}_k{e}_k$, by expanding its finite kernel sum. The increasing measurable sets $E_K=\{x:{\max }_{1\le L\le K}\mathrm{\mid }{B}_L(x)\mathrm{\mid }>2H+2\}$ have union of measure one, so some finite $K\ge 1$ has $m(E_K)>1-\eta$ by F4. Strict inequality in the chosen logarithmic threshold ensures that a finite cutoff attains the threshold used here, even if the original supremum is not attained.

Choose $M\ge K$ so large that $K(K+1)\mathrm{/}(M+1)<1$. Set $g(x)=n^{-1}{\sum }_j{F}_M(x-t_j)$. F3 gives $g\ge 0$ and $\int g=1$, hence $\mathrm{\parallel }g{\mathrm{\parallel }}_1=1$. Expanding the square in F3 yields $F_M=(M+1{)}^{-1}{\sum }_{r,s=0}^M{e}_{r-s}$. For each $\mathrm{\mid }k\mathrm{\mid }\le M$ exactly $M+1-\mathrm{\mid }k\mathrm{\mid }$ pairs have $r-s=k$, and no pairs give larger $\mathrm{\mid }k\mathrm{\mid }$. Translating by $t_j$ multiplies this coefficient by $e_{-k}(t_j)$. Thus g has coefficients $b_k(1-\mathrm{\mid }k\mathrm{\mid }\mathrm{/}(M+1))$ for $\mathrm{\mid }k\mathrm{\mid }\le M$. Consequently, for every $1\le L\le K$ and every x, $\mathrm{\mid }{S}_{L}g(x)-B_L(x)\mathrm{\mid }\le {\sum }_{\mathrm{\mid }k\mathrm{\mid }\le L}\mathrm{\mid }k\mathrm{\mid }\mathrm{/}(M+1)=L(L+1)\mathrm{/}(M+1)<1$. For every $x\in E_K$ there is therefore an L with $\mathrm{\mid }{S}_{L}g(x)\mathrm{\mid }>2H+1>H$. This proves the nonnegative symmetric-frequency interface by finite coefficient computation alone.

Put $P=e_{M}g$, an analytic polynomial of degree at most $2M$. Since $\mathrm{\mid }{e}_M\mathrm{\mid }=1$, $\mathrm{\parallel }P{\mathrm{\parallel }}_1=1$. For $0\le L\le M$, direct reindexing gives $e_M{S}_{L}g=A_{M+L}P-A_{M-L-1}P$. At each point of $E_K$, the L from step 2.1 is at most K and hence at most M; the difference has modulus greater than $2H+1$. The triangle inequality implies one of its two terms has modulus greater than H. If the second index is -1, that term is zero by F1, and the first supplies the bound. Thus $A_{\ast }^{2M}P>H$ throughout $E_K$, proving the analytic interface with the same measure bound. AC is propagated from F2; the subsequent K and M may be chosen as least integers, and no Recorded block theorem is used.

Multiplying the finite sum gives $Q={\sum }_{k=0}^{d}a{c}_k{e}_{m+k}$. Orthogonality of the characters, verified in F1, identifies the coefficient at $m+k$ with $a{c}_k$ and all other coefficients with zero. Thus the claimed support containment holds, even when some coefficients vanish. At $N<m$ the cutoff contains no supported frequency, so $A_{N}Q=0$.

At $N=m+r$, $0\le r\le d$, exactly the terms with $k\le r$ occur, so $A_{m+r}Q={\sum }_{k=0}^{r}a{c}_k{e}_{m+k}=a{e}_m{A}_{r}P$. Taking absolute values and the finite maximum gives ${\max }_{0\le r\le d}\mathrm{\mid }{A}_{m+r}Q\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }{A}_{\ast }^{d}P$. By linearity of a finite coefficient sum, adding any polynomial supported strictly beyond a cutoff leaves that cutoff unchanged; the same conclusion applies to any finite list of later separated blocks.

Since $\mathrm{\mid }{e}_m(x)\mathrm{\mid }=1$ for every x, $\mathrm{\mid }Q(x)\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }P(x)\mathrm{\mid }$ at every point, and every separately supplied amplitude bound on E transfers unchanged. If $a=0$, Q is zero regardless of E; for E empty the pointwise bound is vacuous (use supremum zero for nonnegative functions on the empty set). In particular $a=1$, $P=1$ gives $\mathrm{\mid }Q\mathrm{\mid }=1$ for every m, so frequency shifts cannot make it smaller than one on a nonempty set. These are finite algebraic identities and use no choice axiom.

Apply F1 for each $j\ge 1$ with $H=j{2}^j$ and $\eta =2^{-j}$. AC selects one polynomial from each nonempty set of possible polynomials. Its degree can be taken as its greatest nonzero coefficient index, which exists because its norm is one. The stated exceptional sets are measurable because the maxima are finite continuous maxima. The recursion for $m_j$ is explicit; it gives $m_{j+1}>m_j+d_j$ and $m_j\ge j\ge 1$. F2 therefore places each $Q_j$ in the claimed disjoint positive interval and gives $\mathrm{\parallel }{Q}_j{\mathrm{\parallel }}_1=a_j=2^{-j}$.

For partial sums $f_J={\sum }_{j\le J}{Q}_j$ and $K>J$, the triangle inequality gives $\mathrm{\parallel }{f}_K-f_J{\mathrm{\parallel }}_1\le {\sum }_{j=J+1}^K{2}^{-j}\le 2^{-J}$. They are Cauchy, so F3 supplies an $L^1$ limit f. Passing K to infinity in the norm inequality gives $\mathrm{\parallel }f-f_J{\mathrm{\parallel }}_1\le 2^{-J}$, and $\mathrm{\parallel }f{\mathrm{\parallel }}_1\le {\sum }_{j\ge 1}{2}^{-j}=1$. AC includes the countable choice required by F3.

Apply F4 to $G_J={\sum }_{j\le J}\mathrm{\mid }{Q}_j\mathrm{\mid }$. Its increasing limit G is measurable and has integral $\int G={\lim }_J{\sum }_{j\le J}{2}^{-j}=1$. For every positive integer M, $Mm\{G=\mathrm{\infty }\}\le \int G=1$, so $G<\mathrm{\infty }$ off a null set. There the numerical series converges absolutely; set its sum h to zero on that measurable exceptional set. The resulting function is measurable, $\mathrm{\mid }h\mathrm{\mid }\le G$ a.e., and $\mathrm{\mid }h-f_J\mathrm{\mid }\le {\sum }_{j>J}\mathrm{\mid }{Q}_j\mathrm{\mid }$ a.e. A further application of F4 to the nonnegative tail gives $\mathrm{\parallel }h-f_J{\mathrm{\parallel }}_1\le 2^{-J}$. Consequently $\mathrm{\parallel }h-f{\mathrm{\parallel }}_1\le \mathrm{\parallel }h-f_J{\mathrm{\parallel }}_1+\mathrm{\parallel }{f}_J-f{\mathrm{\parallel }}_1\to 0$, so h represents f. No smallness of completed blocks is attributed to their frequency shifts.

For any integer k and integrable functions u,v, the definition gives $\mathrm{\mid }\widehat{u}(k)-\widehat{v}(k)\mathrm{\mid }\le \int \mathrm{\mid }u-v\mathrm{\mid }$, since $\mathrm{\mid }{e}_{-k}\mathrm{\mid }=1$. Hence the coefficients of F1's partial block sums converge to those of f. For each fixed k, the polynomial coefficients stabilize once the block containing k has been included; if no block contains k, all are zero. Thus f has exactly the prescribed block coefficients and no negative coefficients. In particular every fixed symmetric partial sum is determined by finitely many of these coefficients, with no pointwise infinite-series interchange.

By F4 and F1, $m({\mathrm{lim\hspace{0.17em}sup}}_j{E}_j)=0$. Remove this set and F1's null set on which $G(x)={\sum }_j\mathrm{\mid }{Q}_j(x)\mathrm{\mid }$ may be infinite. Fix any remaining x. There is $j_0(x)$ such that for all $j\ge j_0(x)$ the point lies in the strict good set of F2, so some $0\le r_j\le d_j$ has $\mathrm{\mid }{A}_{r_j}{P}_j(x)\mathrm{\mid }>j{2}^j$. These witnesses may be taken as least indices; no choice of a measurable family of cutoffs is needed for a pointwise supremum.

Set $N_j=m_j+r_j$. At this cutoff every earlier block is completed and every later block is absent. Steps 1.1 and F3 give the exact identity $S_{N_j}f(x)={\sum }_{i<j}{Q}_i(x)+a_j{e}_{m_j}(x)A_{r_j}{P}_j(x)$. Consequently $\mathrm{\mid }{S}_{N_j}f(x)\mathrm{\mid }>j-{\sum }_{i<j}\mathrm{\mid }{Q}_i(x)\mathrm{\mid }\ge j-G(x)\to \mathrm{\infty }$. Also $N_j\ge m_j\ge j$, so these are unbounded cutoffs. This proves the assertion. AC is inherited from F1's countable block choice; Borel–Cantelli requires no independence.

Take f to be the $L^1$ limit supplied by F1, represented by the absolutely convergent block sum where defined and zero on its measurable null exceptional set. This is a complex integrable function with norm at most one, so all its Fourier coefficients are well defined.

F2 applies to exactly this sequence and limit. It gives ${\sup }_N\mathrm{\mid }{S}_{N}f(x)\mathrm{\mid }=\mathrm{\infty }$ outside a measurable null set. A convergent complex sequence is bounded: beyond some index it lies within one of its limit, and its finite initial segment has a finite maximum. Therefore these partial sums in particular fail to converge almost everywhere. The choice assumption is the one used for the countable norm-one block selection and the complex completeness supplier in F1.