Kolmogorov block polynomial with large partial sums

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.