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For example, $f(T)\ll_T 1$ where $T$ is a positive number.

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    $\begingroup$ Normally the subscript should mean something the constant evolved in \ll depend on, but this case it looks weird. $\endgroup$ Mar 4, 2013 at 4:57
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    $\begingroup$ Maybe you could give some more context? Such as where you encountered this notation? $\endgroup$
    – Todd Trimble
    Mar 4, 2013 at 5:18
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    $\begingroup$ In analytic number theory, $f(T) \ll 1$ means that $f(T) < C$ for an unspecified constant $C$, perhaps only in the limit as $T \rightarrow \infty$. With the subscript $T$, this means that $f(T) \leq g(T)$ for some unspecified function $g(T)$ which depends only on $T$. This might make sense if the expression $f$ also depended on other variables; if $f(T)$ depends only on $T$, then your formula is a tautology. See the Wikipedia article for "Big O notation" (the bottom in particular). $\endgroup$ Mar 4, 2013 at 5:41
  • $\begingroup$ And beware that the meaning of $\ll$ differs in different sub-areas of analytic number theory. When someone questions its use, the reply is often: "That is standard notation in the area." $\endgroup$ Mar 4, 2013 at 13:41
  • $\begingroup$ The subscript means the constant depends on T, so this doesn t make sense except for that f(T) is alway a finite number:-) $\endgroup$
    – Marc Palm
    Mar 4, 2013 at 19:43

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I like Frank Thorne's comment-answer, but I would like to stress a detail he mentions in passing, but which does not seem adressed at all on Wikipedia.

What exactly $\ll$ means is not completly uniform even if on just takes the meaning(s) common in Analytic Number Theory (and ignores those informal ones used in some other context like "a lot less than" or "negligible").

One meaning is this: For maps $f: D \to \mathbb{C}$, and $g: D \to \mathbb{R}_{\ge 0}$ one says $f = O(g)$ or equivalenlty $f \ll g$ if there exists some constant $C$ such that $|f(x)| \le C g(x)$ for all $x \in D$.

In particular, this notation is not in itself asymptotic.

Another meaning is this:

For a map $f: D \to \mathbb{C}$, and a function $g: D \to \mathbb{R}_{\ge 0}$ and an accumulation point $x_0$ (possibly infinity) of $D$ one says $f = O(g)$ as $x\to x_0$ or equivalenlty $f \ll g$ as $x \to x_0$ if for each neighborhood $U$ of $x_0$ there exists some constant $C$ such that $|f(x)| \le C g(x)$ for all $x \in D \cap U$.

And not infrequently $f = O(g)$ and $f \ll g$ is used with the second meaning implictly assuming that $x_0$ is "infinity." In particular this is common if the domain of the function is discrete, as it seems to be in you case (positive integers).

A detailed discussion of this can be found for instance in the introduction of Iwaniec and Kowalski's book, who use the first meaning.

Now, after this general discussion for the subscripts. No matter what precise of the two meaning one chooses one always has some constant $C$ called the implied constant.

In practise, it can arise that one uses the above notation not just for one map at a time but rather for a family of maps at the same time, or the map depends on some parameter one choose earlier.

Say, consider $f_t (x) = t x^2/(1+x^2)$ with $D= \mathbb{R}$ and $t$ a real, then for fixed $t$ it is true that there exists a constant $C$ for example $|t|$, such that
$|f_t(x)| \le C \cdot 1$ for all $x$ (or of course also just as $x \to \infty$).

So if $t$ is fixed one has $f_t \ll 1$. However the constant depends on $t$, and you cannot find a constant that works for all $t$ simultaneously. To express this more conveniently and clearly one typically writes this as $f_t \ll_t 1$.

More generally, a subscript or also multiple subscripts of $\ll$ indicate quantities on which the implied constant will (or can) depend.

Finally, the exact expression you give is somewhat unusual as Frank Thorne already explained. I share his opinion, and just would like to over an additional interpretation. It could be that somebody uses the subscripts in a different sense then the one I mentioned namely to indicate which quantitly tends to infinity. So that this could also mean $f(T) \ll 1$ as $T \to \infty$.

I hope this explication sheds some light on the matter, as opposed to causing additional confusion. As Todd Trimble said without additional context it will be impossible to decide what exactly is meant, since the usage is not uniform as Gerald Edgar stressed.

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  • $\begingroup$ The "each neighborhood" should be "some neighborhood". I do not change it right away, since I might add something later anyway in particular if there should be some more info from OP. $\endgroup$
    – user9072
    Mar 4, 2013 at 16:06

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