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Comment by vincentchu | Hacker News Reader

A bit rusty on the math, but taking a quick gander at the Wikipedia page (https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Newman_const...):

"In brief, the Riemann hypothesis is equivalent to the conjecture that Λ ≤ 0."

This result seems to imply that the De Bruijn-Newman constant (Λ) is non-negative, i.e., >=0. Thus, were one able to prove that Λ=0, then one would have also shown the Riemann Conjecture to be true? Similarly, if one were to show Λ to be not equal to 0, then the Riemann Conjecture would be false.

Again, just a quick reading. Sure others who are better/smarter can chime in!

Update: Tao expands on this point in the bottom of his blog post.

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Yes so if you want to prove the Riemann hypothesis one possibility would be to similarly show some sort of contradiction if Λ > 0. This is sort of the opposite approach that Rogers and Tao have appeared to use. Of course maybe that side is much much harder (it is equivalent to the Riemann hypothesis after all).

I have essentially no real knowledge about this topic, but I suppose some progress has been made by tightening the bounds for this constant?

Can somebody with more info chime in--- is this a huge, groundbreaking amount of progress?

Reading https://en.wikipedia.org/wiki/De_Bruijn–Newman_constant, it moves the bound from −1.1×10−12 (probably an approximation) to zero.

So, not huge in absolute terms, but on the other hand, huge, as it removes all remaining negative numbers.

I would presume that entirely different methods would be needed to prove that the constant can't be positive. There are many types analytical proofs in math that break down to "negative", "zero" and "positive" cases and often a couple of those are relatively easy whereas others are extremely difficult. The Calabi conjecture is an example.

So really it's hard to say. You might even say this proof made things harder. It guarantees that you have to prove the constant is 0 if you want to prove the conjecture. Before you might have hoped prove negativity.

Right, near the bottom of the paper he writes “...this result does not make the Riemann hypothesis any easier to prove, in fact it confirms the delicate nature of that hypothesis”

Hmm. So the Riemann hypothesis requires Λ to be <= 0. Now they’ve showed that Λ >= 0. Since the intersection of the two ranges is non-measurable, probabilistically the Riemann hypothesis must be false! :^)

{0} is measurable (It's a borel set) with the standard lebesgue measure and has measure 0. Of course, with some other measure it might not be measurable, but that would be a weird measure.

You could identify 0 with some other point to get a topology where {0} is not a Borel set anymore.

A somewhat more useful example would be to identify x with x+1 for all x (or equivalently use the topology on R induced by the map t -> e^2πit from R to S^1), which results in a topology where only periodic functions are continuous.

For a complete measure you get the additional requirement that 0 isn't allowed to be contained in any null-set.

Been thinking for a while how such a measure would be defined, but my measure theory abilities are a bit rusty (heck, unused for 10 years already, time flies). I'd bet it's impossible (in \mbb{R})... and sadly I'll now keep thinking about it until I remember how to prove for sure it or fall asleep, thanks :D

The Lebesgue measure is not defined for all subsets of R. It is defined on the Lebesgue sigma algebra which is larger than the Borel sigma algebra.

You could instead choose a sigma algebra that does not contain {0}. Trivially, you could choose the standard Lebesgue measure restricted to {{}, R}.

Oh well, I was thinking of a measure where 0 was in some Borel set of it, otherwise it loses a bit the fun of it

In this case, 0 is contained in R which is a Borel set and is part of the sigma algebra on which the measure is defined.

I mean, one with a more rich sigma algebra.

You’re right. I should have used the term null set.

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