Kolmogorov analytic partial sum maximal function

Use $\mathbb{T}=\mathbb{R}\mathrm{/}\mathbb{Z}$, normalized measure $dx$ and $e_k(x)=e^{2\pi ikx}$ as in Period-one Fourier coefficients, partial sums, and convolution on the torus. An analytic trigonometric polynomial is $P={\sum }_{k=0}^d{a}_k{e}_k$ for an integer $d\ge 0$, with no negative-frequency terms. Set $A_{n}P={\sum }_{k=0}^{\min (n,d)}{a}_k{e}_k(n\ge 0),A_{-1}P=0,A_{\ast }^{d}P={\max }_{0\le n\le d}\mathrm{\mid }{A}_{n}P\mathrm{\mid }\mathrm{.}$ Trailing zero coefficients do not affect this maximum: every cutoff beyond the degree equals $P$, already attained at its degree. The zero polynomial may be represented with $d=0$ and has maximal function zero. Each partial sum and the finite maximum are continuous, hence measurable.

The coefficients agree with the integral Fourier coefficients of the cited convention: finite linearity reduces to ${\int }_0^1{e}_{k-\mathrm{\ell }}(x)dx$, which is one for $k=\mathrm{\ell }$ and is $[e^{2\pi i(k-\mathrm{\ell })x}\mathrm{/}(2\pi i(k-\mathrm{\ell })){]}_0^1=0$ otherwise. Thus for an analytic polynomial, the symmetric partial sum $S_{n}P$ equals $A_{n}P$. No infinite series or choice of representatives is involved in these finite identities.