What does Hardy mean in this lemma?

Question

I am concerned with the paper "Oscillating Dirichlet's Integrals" by G.H. Hardy (Quarterly Journal of Pure and Applied Mathematics, Vol. XLIV, pg 1-40). I don't understand Lemma 1, but I suppose this may be a problem of English comprehension?

First, some notation. Hardy defines (for positive functions) $f\prec g$ (and $g\succ f $) to mean that $g/f \to \infty $ (as $x\to 0$), $f \asymp g$ to mean that $f/g \in (\delta, \Delta)$ for some constants $0<\delta <\Delta$, $f\sim Ag$ to mean that $f/g\to A$ (for an unspecified constant $A$, that may change from line to line) and $f\sim g$ to mean that $f/g\to 1$ (the usual asymptotic notation).

Then he further states that he uses the symbols $\delta,\Delta$ to mean two different things (which cannot both occur at the same time): he writes $$ \log(1/x)^\Delta \prec (1/x)^\delta$$ to mean that the statement holds for any positive numbers $\Delta\gg1$ sufficiently large and $\delta \ll 1$ sufficiently small. At the same time, he would write $$ (1/x)^\delta \prec f \prec (1/x)^\Delta$$ to mean that there exists $\delta,\Delta>0$ such that this statement is true.

Now the lemma (Lemma 1). He has already made some assumptions on the functions so that it is always true for any $f,g$ under consideration that one of $f\prec g, f \succ g, f \sim Ag$ is true.

Lemma 1. If $f\succ 1, \phi \succ 1$, then either $f\succ \phi^\Delta$ or there is a number $a$ ($a\ge 0$) such that $f=\phi^a f_1$, where $\phi^{-\delta} \prec f_1 \prec \phi^{\delta}$. A similar result holds when $f\prec 1, \phi \prec 1$.

(Proof) For if it is not true that $f\succ \phi^\Delta$, we can find numbers $\alpha$ such that $$ f \prec \phi^\alpha$$ and we can divide the positive real numbers $\alpha,$ including zero, with at most one exception, into two classes such that for one class $f\succ \phi^\alpha$, and for the other $f\prec \phi^\alpha$. There is at most one number, viz. $a$, the number which divides the two classes, for which $f\sim A\phi^\alpha$. If $f\sim A \phi^a$, $a$ belongs to neither class. If, however $a$ belongs to one class or the other, and we put $f = \phi^a f_1$, it is clear that $\phi^{-\delta}\prec f_1 \prec \phi^{\delta}$. Thus the result of the lemma is true in either case.

Some notes -

1. "viz." means "namely"
2. It is written $\phi^{-\delta} f_1\prec \phi^\delta$ in the original lemma statement in the paper...surely this is wrong, and I have corrected it in the above.
3. I'm quite sure that I transcribed the $\alpha s$ and $a$s correctly, but I cannot say for sure due to the quality of my digital copy, which I include a clip of here if there is any doubt - https://i.sstatic.net/KWaUE.png.

Questions (at long last)

1. In the lemma statement, which interpretation of $\delta,\Delta$ is being used? I guess it is $f\succ g^\Delta$ for all $\Delta \gg 1$? and $\phi^{-\delta} \prec f_1 \prec \phi^{\delta}$ for $\delta \ll 1$? The other interpretation seems impossible when he also claims later that there are $\alpha $ such that $f \succ \phi^a$.
2. When he claims there is at most one $a$, is he implicitly saying also that the existence of $a$ is trivial and deserves no mention? I guess he would define $a$ as a supremum/infimum?
3. How does he claim that there is at most one $a$ such that $f\sim A \phi^a$, and then write "If $f\sim A\phi^a"$?

Answer 1

I believe the following is a faithful "translation" (my own complete rewording of the proof)-

Lemma 1. If $f\succ 1, \phi \succ 1$, then either $f\succ \phi^\Delta$ for all $\Delta>0$, or there is a number $a$ ($a\ge 0$) such that $f=\phi^a f_1$, where $\phi^{-\delta} \prec f_1 \prec \phi^{\delta}$ for all $\delta>0$. A similar result holds when $f\prec 1, \phi \prec 1$.

(Proof) Suppose the first assertion was not true. If there existed $\alpha$ such that $f\sim A\phi^\alpha$, the second assertion is clearly true. So assume that $f\not\sim A\phi^a$ for all $a$ as well. Having ruled out these possibilities, by the trichotomy between $\sim,\prec,\succ$, there is some $\alpha\ge 0$ such that $$ f \prec \phi^\alpha.$$ For all $\alpha_1>\alpha$, we also have $f \prec \phi^{\alpha_1}$.In particular, The set $A=\{ \alpha \ge 0 : f \prec \phi^\alpha\}$ is a half-infinite interval, and together with $A^c$, forms a partition of $\mathbb R$. Take $\alpha_0:=\inf A=\sup A^c$. Since we have ruled out the possibility of $f\sim A\phi^{\alpha_0}$, either $f\succ \phi^{\alpha_0}$ or $f\prec \phi^{\alpha_0}$. In either case, adjusting the power by a small $\delta$ gives the required result, QED.

Examples for the lemma (meeting the assumption in the paper for the functions, which roughly restricts the paper to a bunch of exponentials or logs or polynomials) : $$ 1/x \succ (\log(1/x))^\Delta\text{ for all }\Delta>0,$$ $$1/x\sim (1/x^2)^{1/2},\text{ and}$$ $$(1/x)^{1-\delta} \prec (1/x)\log(1/x)\prec (1/x)^{1+\delta} \text{for all $\delta>0$}.$$