The L1 endpoint is excluded

Assume the Axiom of Choice (The Axiom of Choice). The local theorem Kolmogorov lone fourier series diverges almost everywhere supplies an integrable function with unbounded symmetric Fourier partial sums almost everywhere. Thus a convergence theorem for every $L^1(\mathbb{T})$ function would be false.

In particular there is no finite $A\ge 0$ such that $m\{Cg>\lambda \}\le A\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{/}\lambda$ for all $g\in L^1$ and $\lambda >0$. Such a weak $(1,1)$ estimate would imply almost-everywhere convergence for every such $g$ by A weak maximal bound implies almost-everywhere Fourier convergence at $p=1$. Applying that conclusion to the supplied witness contradicts its almost-everywhere unboundedness, because a convergent complex sequence is bounded and the union of the two exceptional null sets remains null.

A strong $(1,1)$ estimate $\mathrm{\parallel }Cg{\mathrm{\parallel }}_1\le B\mathrm{\parallel }g{\mathrm{\parallel }}_1$ with finite $B$ would imply the forbidden weak estimate by Chebyshev-Markov inequality for the integral applied to $Cg$, so no such strong estimate exists either. All operators here use symmetric partial sums on the normalized period-one torus. AC supplies the witness and implies the countable choice needed by the maximal convergence principle. Failure at one prescribed point for a continuous function is a different assertion from this failure on a full-measure set for an integrable function.