The L1 endpoint is excluded

Assume countable choice. The assertion in Kolmogorov almost-everywhere divergence — recorded theorem ‡ prevents extending Carleson–Hunt maximal bound and almost-everywhere convergence — recorded theorem ‡ to all of $L^1(\mathbb{T})$. This comparison relies on the recorded Kolmogorov existence theorem, whose proof is not supplied here.

Indeed, a weak $(1,1)$ estimate $m\{Cg>\lambda \}\le A\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{/}\lambda$ for all $g\in L^1$ and $\lambda >0$ would imply almost-everywhere convergence for every such $g$ by A weak maximal bound implies almost-everywhere Fourier convergence. That implication is incompatible with the recorded witness. A strong $(1,1)$ estimate would imply this weak estimate by Chebyshev-Markov inequality for the integral applied to $Cg$.

This is an interpretation of external literature, not an independently proved weak-endpoint theorem. Failure at one prescribed point for a continuous function is a different phenomenon from failure on a full-measure set for an integrable function.