How does one read aloud Vinogradov's notation $\ll$ and $\ll_{\epsilon }$? How does one read aloud the Vinogradov's notation $\ll$ and $\ll_{\epsilon }$ as in
$$f(x)\ll g(x)$$
and
$$c\ll_{\epsilon }\left( \prod\limits_{p\mid abc}p\right) ^{1+\epsilon}.$$
Is the first one “is very much less than”? (This is a direct translation from Portuguese as was used  informally in engineering formulae).
Added: I am not asking the meaning of this notation, rather how it is read. 
This answer to the question mentioned in Unreasonable Sin's comment points to this Wikipedia entry. According to it in Analytic number theory the symbol $\ll$ in $f(x)\ll g(x)$ is to be read as “is of smaller order than”. 
Added 2.  As a response to LVK's comment:

"I am not asking the meaning of this notation, rather how it is read."
  But the way to spell out notation depends on what it means in the
  particular text. If Vinogradov used it to mean $f=O(g)$ in one of his
  papers, then in that paper the symbol should be read differently.

Let me give a specific example. I would like to know how to read $\ll _{\epsilon }$ in the following conjecture

ABC Conjecture. Suppose $A,B,$ and $C$ are positive integers, suppose
           $$\gcd (A,B,C)=1,$$
           and suppose 
           $$A+B=C.$$
           Then
           $$C\ll _{\epsilon }\left( \prod\limits_{p\mid ABC}p\right) ^{1+\epsilon}.$$

Source: Arithmetic Algebraic Geometry, Brian Conrad, Karl Rubin, ch. 5, p.123.
The meaning as I understand it is the same as:
Suppose: 
i) $A,B$ and $C$ are positive integers,
ii) $\gcd (A,B,C)=1,$
iii) $A+B=C,$
iv) $\epsilon >0$ is a positive real number. 
Then there is a constant $K_{\epsilon}$ such that 
$$C\leq K_{\epsilon}\left( \prod\limits_{p\mid ABC}p\right)^{1+\epsilon }.$$
P.S. The symbol $\ll_{\epsilon}$ is also used in Terence Tao's post The probabilistic heuristic justification of the ABC conjecture.
 A: As a default, I second Gerry Myerson's answer "less-than-less-than", although just reading "sub-epsilon" seems a little odd. In a situation where I'm trying to reinforce the meaning, I might read $A\ll B$ as "$A$ is dominated by $B$", and $A\ll_\varepsilon B$ as "$B$ dominates $A$ depending on $\epsilon$", but this may be defeating the purpose/intent of the question. (Edit: or, better, as noted by Americo Tavares, to say "$A$ is dominated by $B$ depending on $\varepsilon$", to avoid the need to read ahead...)
A: I read $\ll$ as "less than less than". I don't think I have ever put $\ll_{\epsilon}$ in words, but if I did it would probably come out "less than less than sub epsilon". 
A: "Controlled by a constant multiple of ..., with the constant possibly depending on epsilon", or if you don't want to be quiet as you write it on blackboard, "less less sub epsilon".
A: From my experience, for $\ll$ we say "Less than less than" and as for $\ll_\epsilon$, I would say "Less than less than where the constant depends on epsilon." 
A: Extending Charles' answer in the thread I linked to, $f(x)\ll_{\epsilon} g(x)$ means that for a given $\epsilon$ there exists some $N$ and $k$ such that for all $x > N$, $f(x) < k\cdot g(x)$. In other words, $k$ and $N$ depend on the value of $\epsilon$. As to how you would read or pronounce that, I am not sure. 
A: I prefer to say " doubly less than ", which is more convenient than " less than less than " when speaking Chinese.
