# Problems in understanding Lacey's proof of Carleson's theorem.

I am new to the Stackexchange community so do let me know if I can improve my question in any way.

Right, I have just started reading Michael Lacey's proof of Carleson's theorem (http://people.math.gatech.edu/~lacey/research/esi.pdf), and have already hit a problem.

In Proposition 1.4 (Page 4), I don't understand a few things:

1) Why has the author proceeded in this method for establishing that the desired set is closed?

My instincts would have said, take a sequence in the set of functions for which the results hold, say they converge in the norm to some function and finally prove that the said function is member of the desired set.

I think I can, however, see how the two might be equivalent: if one proves the desired set is closed the criterion that Lacey is trying to establish follows. Vice-versa if one proves what Lacey sets out to prove in the first line of the proof then the set must be closed.

2) This is just to confirm my own instincts, but I believe the result follows from an application of Chebyshev's Inequality.

I would be grateful for any help/pointers and as above please don't hesitate to let me know if I haven't complied with any other rules of the forum.

Thank you.

## 1 Answer

(1) I take it you're asking why he used this for showing that the set of functions satisfying 1.2 is closed? This is a standard technique; 1.4 is a "maximal function inequality", and maximal function inequalities are simply how people prove almost-everywhere convergence results.

How would you do it? You say "take a sequence of functions satisfying 1.2 which converge in norm and show the limit satisfies 1.2". How do you show that, exactly? You've left out a lot of details... (Hint: The fact that the set satisfying 1.2 is closed is precisely equivalent to Carleson's theorem. So if you actually do see a proof that's a simple as you're implying you're going to be famous.)

Not to, um, but if you're not familiar with the notion of maximal functions versus almost everywhere convergence it may well be that you're in over your head here. (Note I happen to be very familiar with this notion, but I'm in way over my head.)

(2) No, it would be a consequence of Chebyshev if we knew that $Cf$ was in $L^2$. We don't know that.

• Thank you so much for your answer. – Samir Dec 24 '15 at 16:35
• You have clearly noticed a few things that I should explain too. 1) My first question should really have been: 'How does what Lacey has shown imply that the set of functions satisfying (1.2) closed?' Is that any better? The subsequent idea of sequences/norms and convergence is a direction I think we can take ASSUMING the weak-(2,2) inequality. Not without it of course; I think I would 'be famous' only if I did it without using that assumption :P – Samir Dec 24 '15 at 16:43
• He gives a proof that it shows the set is closed. Sorry, but it really seems somewhat pointless to be going through this in this sort of detail - it's a very complicated technical thing, and you're already stuck on the introduction. Hmm. Find a copy of Garsia Topics in Almost Everywhere Convergence and read that for a warmup... – David C. Ullrich Dec 24 '15 at 16:46
• 2) Again, I really have missed out a lot of details in my desire to avoid TeX (all thumbs!) but by the Chebyshev inequality I only meant how Lacey sets out to prove that |\{L_{f} > \epsilon\}| we need to use Chebyshev's inequality in the final line to complete the result. That's all. – Samir Dec 24 '15 at 16:52