What does f(t) Subscripted to g(t) Epsilon Mean?

[Swamp Thing]

What does this mean: ##f(t) \ll_\epsilon (g(t))^\epsilon## ?

Ref: Beginning of the introduction here.

 

[jedishrfu]

Could you quote that part of the paper?

I couldn't access the researchgate link as it may be a paywall.

Usually the double less than symbols means "much less than"

https://en.wikipedia.org/wiki/Less-than_sign

Double less-than sign​

The double less-than sign, <<, may be used for an approximation of the much-less-than sign (≪) or of the opening guillemet («). ASCII does not encode either of these signs, though they are both included in Unicode.

In the wikipedia Lindelof hypothesis article, they reference a result by Ingham:

https://en.wikipedia.org/wiki/Lindelöf_hypothesis

Other consequences
Denoting by pn the n-th prime number, a result by Albert Ingham shows that the Lindelöf hypothesis implies that, for any ε > 0,

if n is sufficiently large. However, this result is much worse than that of the large prime gap conjecture.

 

[Swamp Thing]

I couldn't access the researchgate link as it may be a paywall.

I don't have a subscription either. If you scroll down, you can read it (embedded in the same page).

It says,

Let s=σ+it be a complex variable and ζ(s)the Riemann zeta-function. The classical Lindelof hypothesis asserts that ##\zeta(\sigma+it) \ll_\epsilon (1+|t|)^\epsilon## for every positive ##\epsilon##.

<< means "much less than" ... that's familiar, but what about the subscript?

 

[jedishrfu]

The subscript ##\epsilon## is used to denote a very small quantity often used in epsilon-delta limit proofs but I'm sure when its used as a subscript to the much less than.

Maybe @fresh_42 or @Mark44 will know.

 

[Mark44]

Let s=σ+it be a complex variable and ζ(s)the Riemann zeta-function. The classical Lindelof hypothesis asserts that ##\zeta(\sigma+it) \ll_\epsilon (1+|t|)^\epsilon## for every positive ##\epsilon##.

I've never seem that notation before and I don't know what it, ##\ll_\epsilon##, means. The Wikipedia page, https://en.wikipedia.org/wiki/Lindelöf_hypothesis, says something different.
##\zeta(\frac 1 2 + it) = O(t^\epsilon)## with a similar equation that follows.

 

[Petek]

The expression ##f(x) \ll g(x)## uses what is called Vinogradov notation. It means that there exists a constant ##C > 0## such that

##|f(x)| \leq Cg(x)##

where it's usually understood that ##x \rightarrow \infty##.

The Vinogradov notation ##f(x) \ll g(x)## is the same as the so-called Big O notation. That is, ##f(x) = O(g(x))## means the same as ##f(x) \ll g(x)##. The subscript ##\epsilon## means that the implied constant ##C## depends on ##\epsilon##. See this link and the Wikipedia link there for more information.

 

[Swamp Thing]

Thanks. Another quick question: When we read "for ##\epsilon>0## ", does it imply ##\epsilon<1## even if that's not stated? I mean specifically when talking about big-O to the power of ##\epsilon## ?.

 

[fresh_42]

I haven't seen it either, I think. My interpretation would be: ##f(t)## is small compared to ##g(t)^\varepsilon ## for suitable ##\varepsilon ,## e.g. probably "any ##\varepsilon >0.##

Anyway, it is used for the first time in whatever you read. This is the context you need to answer your question.

 

[malawi_glenn]

I think the same as fresh, that is is a "conditional" much less than. In this case, for a suitable ##\varepsilon##.

I think I have seen it been used in "mathematical methods for physics" books, but I can not think of a title right of the bat.

 

[Mark44]

Another quick question: When we read "for ##\epsilon>0## ", does it imply ##\epsilon<1## even if that's not stated?

Yes, in most cases. The inequality ##\epsilon>0## is almost universally meant to convey the idea the ##\epsilon## is a small, positive number.

I mean specifically when talking about big-O to the power of ##\epsilon## ?.

This doesn't make any sense. The "big O" business is applied to some function; e.g., O(f(x)).

 

[Swamp Thing]

This doesn't make any sense. The "big O" business is applied to some function; e.g., O(f(x)).

Sorry, that was sloppy language. I meant big-O applied to some f(x) that involves something raised to ##\epsilon##