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Local Bernstein theory, and lower bounds for Lebesgue constants

43 pointsby jjgreenlast Tuesday at 9:36 AM7 commentsview on HN

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_alternator_today at 2:09 PM

Fascinating discussion including showing how LLMs are helping push the state of the art:

> I still did not see how to prove this inequality, but I decided to try my luck giving it to ChatGPT Pro, which recognized it as an {L^1} approximation problem and gave me a duality-based proof (based ultimately on the Fourier expansion of the square wave). With some further discussion, I was able to adapt this proof to functions of global exponential type (replacing the Fourier manipulations with contour shifting arguments, in the spirit of the Paley-Wiener theorem), which roughly speaking gave me half of what I needed to establish (2).

> As a side note, this latter argument was provided to me by ChatGPT, as I was not previously aware of the Nevanlinna two-constant theorem.

This mirrors my experience with these tools for math. Great for local problems and chatting through issues. Still can’t do the whole thing in one shot but getting there.

tensegristtoday at 1:59 PM

Presumably one reason this is of interest here:

    The proof proceeds by a modification of the Duffin–Schaeffer argument, together with the two-constant theorem of Nevanlinna (and some standard estimates of harmonic measures on rectangles) to deal with the effect of the localization. (As a side note, this latter argument was provided to me by ChatGPT, as I was not previously aware of the Nevanlinna two-constant theorem.)
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adampunktoday at 2:55 PM

Think of this as an algebraic jewelers loupe. Zoom in on local structure and push aside the global. It’s fascinating.