I have been trying to prove the following:
Consider an i.i.d. sample of random variables $\left\{ X_n \right\}$. Their distribution $F$ is said to satisfy the rapidly varying tail condition, if $\forall a > 1$ the following holds: $$ \lim_{x \to \infty} \frac{1 - F(ax)}{1 - F(x)} = 0 $$ Show that if $\displaystyle \frac{\max_{i=1,\dots,n} X_i}{b_n} \overset{\mathbb{P}}{\to} 1$ (convergence in probability), for a sequence $b_n \to \infty$, then $F$ must satisfy the rapidly varying tail condition.
I consider the definition of convergence in probability, $$ \lim_{n \to \infty} \left( \left|\frac{\max_{i=1, \dots, n}(X_1, \dots, X_n}{n} - 1 \right| > \varepsilon \right) = 0 $$ and write (for the positive part):
\begin{align*} \mathbb{P}\left(\max_{i=1,\dots,n} X_i > (1 + \varepsilon) b_n\right) &= \mathbb{P}\left[ \left( \bigcap_1^n X_i \leq (1 + \varepsilon) b_n \right)^c \right] \\ &= \mathbb{P}\left( \bigcup_1^n X_i > (1 + \varepsilon) b_n \right) = n \mathbb{P}(X_1 > (1 + \varepsilon) b_n) \\ \end{align*} For the negative part, I know that \begin{align*} \mathbb{P}\left(\max_{i=1,\dots,n} X_i < (1 - \varepsilon) b_n\right) &= \mathbb{P}\left[ \bigcap_1^n X_i < (1 - \varepsilon) b_n \right] \\ &= 1 - n\mathbb{P}(X_i \geq (1 - \varepsilon)b_n) \end{align*} Taking limits in the positive and negative parts and equating them as they both have to be equal to 0 obtain $$ \lim_{n \to \infty} n \mathbb{P}(X_1 > (1 + \varepsilon) b_n) = 0, \quad \lim_{n \to \infty} n \mathbb{P}(X_1 \geq (1 - \varepsilon) b_n) = 1 $$ However, at this point I'm stuck. I can obviously deduce that $$ \lim_{n \to \infty} \frac{1 - F(az_n)}{1 - F(z_n)} = 0, \quad a := \frac{1}{1-\varepsilon} > 1, \; z_n = (1 - \varepsilon) b_n $$ however this does not exactly imply the rapidly varying tail condition, which must hold for arbitrary $x \to \infty$. Any ideas?
We need to prove that for any $x_k\to\infty$, $$ \frac{\overline F(ax_k)}{\overline F(x_k)} \to 0 \text{ as } k\to\infty. $$ Here $\overline F(x)=1-F(x)=\mathbb P(X_1 >x)$ is the right tail of distribution of $X_1$ and $a=\frac{1+\varepsilon}{1-\varepsilon}>1$.
We have that for $z_n=(1-\varepsilon)b_n\to\infty$ as $n\to\infty$ $$n\overline F(z_n)\to 1\ \text{ and }\ n\overline F(a z_n)\to 0. $$ We can assume without loss of generality that $b_n$ (and also $z_n$) is a monotone increasing sequence.
Then for any sufficiently large $k$, there exists $n_k$ s.t. $z_{n_k}\leq x_k < z_{n_k+1}$. Then we can use that $\overline F$ is decreasing and estimate $$ \frac{\overline F(ax_k)}{\overline F(x_k)} \leq \frac{\overline F(az_{n_k})}{\overline F(z_{n_k+1})}=\frac{\overline F(az_{n_k})}{\overline F(z_{n_k})}\cdot \frac{\overline F(z_{n_k})}{\overline F(z_{n_k+1})}$$ $$= \underbrace{\frac{\overline F(az_{n_k})}{\overline F(z_{n_k})}}_{\to 0}\cdot \underbrace{\frac{n_k\overline F(z_{n_k})}{(n_k+1)\overline F(z_{n_k+1})}}_{\to 1}\cdot \underbrace{\frac{n_k+1}{n_k}}_{\to 1}\to 0 $$