If an eventual inequality holds for $f$ and $g \sim f$ then does it hold for $g$?

Question

Suppose $f$, $g$, are non-negative functions such that $f \sim g$ (meaning that $f(x)/g(x) \to 1$ as $x \to \infty$). If for all $\lambda > 1$ there exists an $\eta > 0$ such that $$f(\lambda x) - f(x) > \eta x\tag{1}$$ for all $x$ sufficiently large, is the same true of $g$?

I think it is. Here's what I've tried: let $\lambda > 0$ and let $1 > \varepsilon > 0$, to be chosen later. Get $\eta, x_0 > 0$ so that $f(\lambda x) - f(x) > \eta x$ for all $x \ge x_0$. Let $x_1$ be such that $$1 - \varepsilon < \frac{g(x)}{f(x)} < 1 + \varepsilon$$ for all $x \ge x_1$. Then for $x \ge \max\{x_0, x_1\}$ we have $$g(\lambda x) - g(x) > f(\lambda x)(1-\varepsilon) - f(x)(1+\varepsilon) = [f(\lambda x) - f(x)] - \varepsilon f(\lambda x) - \varepsilon f(x).$$ The bracketed term is bounded below by $\eta x$, and, on multiplying (1) through by $\varepsilon$ we can bound the last term below by $\varepsilon\eta x - \varepsilon f(\lambda x)$. Putting it all together, we get $$g(\lambda x) - g(x) > \eta(1-\varepsilon) x - 2\varepsilon f(\lambda x).\tag{2}$$ But this bound is too low. I understand I can get away with just a small multiple of $\eta$ as the coefficient on $x$; the trouble is, I don't know if $f(\lambda x)$ is dominated by some $\alpha x$ for $\alpha < 1$. So maybe there is a different approach?

Answer 1

Your proof doesn't work because: counterexample.

Let $(a_n)_{n=0}^\infty$ be a non-decreasing sequence of non-negative reals. Let $f,g\colon[1,\infty)\to\Bbb R$ be given by $$\begin{align}f(x)&=x+a_{\lfloor\ln x\rfloor}\\ g(x)&=f(e^{\lfloor\ln x\rfloor})\end{align} $$ Then $f$ has the desired property because $$ f(\lambda x)-f(x)\ge (\lambda-1)x.$$ On the other hand, if $1<\lambda<e$ then there are arbitrarily large $x$ with $g(\lambda x)=g(x)$. Hence $g$ does not have the property.

Remains to adjust $a_n$ to achieve $f\sim g$. Note that for $e^n\le x<e^{n+1}$ we have $$g(x)=a_n+e^n\le f(x)<a_n+e^{n+1}=g(x)+(e-1)e^n$$ Hence it suffices to ensure that $\frac{(e-1)e^n}{a_n+e^n}\to 0$, so for example $a_n=e^{2n}$ will work.

Answer 2

As Hagen von Eitzen's counterexample shows, the claim is false. However, it can be salvaged if we assume the stronger asymptotic relation: $$g(x) = f(x) + o(x).\tag{*}$$ Assuming $\lim_{x \to \infty}f(x) \ne 0$ then (*) implies $g \sim f$, since $$\lim_{x\to\infty}\frac{g(x)}{f(x)} = \frac{\lim_{x\to\infty}g(x)-f(x)}{\lim_{x\to\infty}f(x)} + 1 = 0 + 1 = 1.$$ Hagen's counterexample does not satisfy (*), since $$\frac{g(x) - f(x)}{x} = \frac{e^{\lfloor\log x\rfloor} - x}{x} = e^{-\{\log x\}} - 1\tag{^}$$ where $\{\xi\} \in [0, 1)$ denotes the fractional part of the real number $\xi$. The limit of (^) does not exist as $x \to \infty$, because $e^{-\{\log x\}}$ oscillates too much.

To prove that (*) implies (1), write $g(x) = f(x) + r(x)$ where $\lim_{x\to\infty} \frac{r(x)}{x} = 0$. Let $\lambda > 0$ and get $\eta$ and $x_0$ such that $f(\lambda x) - f(x) > \eta x$ for all $x > x_0$. Then $$g(\lambda x) - g(x) = [f(\lambda x) - f(x)] + [r(\lambda x) - r(x)] > \eta x + x\left[\lambda \frac{r(\lambda x)}{\lambda x} - \frac{r(x)}{x}\right].\tag{3}$$ Now for $\varepsilon > 0$ to be chosen later, let $x_1$ be such that $|r(x)/x| < \varepsilon$ for all $x > x_1$. Since $\lambda > 1$, we have $\lambda x > x > x_1$, so $|r(\lambda x)/\lambda x| < \varepsilon$ also. Thus we can bound below the bracketed term in (3): $$\left[\lambda \frac{r(\lambda x)}{\lambda x} - \frac{r(x)}{x}\right] > -\lambda\varepsilon - \varepsilon = -\varepsilon(\lambda + 1).$$ Hence, for $x > \max\{x_0, x_1\}$, $$g(\lambda x) - g(x) > (\eta - \varepsilon(\lambda + 1))x.$$ We want the coefficient on $x$ to be positive, so we complete the proof by taking $\varepsilon < \eta/(\lambda + 1)$.