# What does Hardy mean in this lemma?

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.stack.imgur.com/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 $$f\sim A\phi^a"$$?

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) : $$1/x \succ (\log(1/x))^\Delta\text{ for all }\Delta>0,$$ $$1/x\sim (1/x^2)^{1/2},\text{ and}$$ $$(1/x)^{1-\delta} \prec (1/x)\log(1/x)\prec (1/x)^{1+\delta} \text{for all \delta>0}.$$