Let $g:[0, \infty) \rightarrow \mathbb{R}$ and $h:[0, \infty) \rightarrow \mathbb{R}$ be nonnegative functions let us put $\phi(y)=g(y) h(y)$. Assume that the following condition is satisfied:

$h$ is nondecreasing and $\phi$ is strictly increasing with $\phi([0, \infty))=[0,+\infty)$.

Now on some work relating to the convergence of random variables I've been using the above condition along with the following condition which I'll refer to as A1:

There exists a constant $b>0$ such that $$ \sum_{i=1}^{n} \frac{1}{h^{2}(i)} \leq b \frac{n}{h^{2}(n)}. $$

And I've been wondering if any of the two following conditions were weaker than A1:

A2: There exists $p \ge 2$, a constant $b>0$ such that $$ \sum_{i=1}^{n} \frac{1}{h^{p}(i)} \leq b \frac{n}{h^{p}(n)}. $$

A3: There exists $p \le 2$, a constant $b>0$ such that $$ \sum_{i=1}^{n} \frac{1}{h^{p}(i)} \leq b \frac{n}{h^{p}(n)}. $$

At the moment I'm reading on "slowly varying functions". I believe some of the tools there might help me reach a conclusion. If anyone knows any references or have any sort of comments that would be helpful then I would gladly appreciate.


1 Answer 1


$\def\N{\mathbb{N}}\def\paren#1{\left(#1\right)}$Here it is assumed that A1 to A3 are all to hold for any $n \in \N_+$. Since these conditions on $h$ are easily seen to be irrelevant to that on $φ$, and $h(t) > 0$ for $t > 0$, then define $a_n = \dfrac{1}{(h(n))^2}$ for $n \in \N_+$ and A1 to A3 are equivalent to:\begin{align*} &A_1\colon && \exists c > 0,\ \forall n \in \N_+,\ \frac{1}{n} \sum_{k = 1}^n a_k \leqslant ca_n,\\ &A_2\colon && \exists c > 0,\ q \geqslant 1,\ \forall n \in \N_+,\ \paren{ \frac{1}{n} \sum_{k = 1}^n a_k^q }^{\frac{1}{q}} \leqslant ca_n,\\ &A_3\colon && \exists c > 0,\ q \leqslant 1,\ \forall n \in \N_+,\ \paren{ \frac{1}{n} \sum_{k = 1}^n a_k^q }^{\frac{1}{q}} \leqslant ca_n, \end{align*} respectively.

Now, the inequality of power means shows that for any $x_1, \cdots, x_n > 0$,$$ f(t) = \begin{cases} \min(x_1, \cdots, x_n); & t = -∞\\ \paren{ \dfrac{1}{n} \sum\limits_{k = 1}^n x_k^t }^{\frac{1}{t}}; & t ≠ 0\\ \paren{ \prod\limits_{k = 1}^n x_k }^{\frac{1}{n}}; & t = 0\\ \max(x_1, \cdots, x_n); & t = +∞ \end{cases} $$ is strictly increasing with respect to $t$ on $[-∞, +∞]$ unless all $x_k$'s are equal, thus in general$$ A_2 \implies A_1 \implies A_3, $$ i.e. A3 is the weakest among the three.

  • 1
    $\begingroup$ thanks, very neat, will reward the bounty once I'm able to which is in a few hours. $\endgroup$ Commented Nov 19, 2021 at 8:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .