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Here I'm interesting in prove (or disprove and counterexamples) the following Claim :

Claim: For $p: [1,\infty)\longrightarrow\mathbb R$ and $f: [1,\infty)\longrightarrow\mathbb R$ where $p$ is positive and monotonic increasing, $f$ is non-negative and monotonic decreasing. If $\int_1^\infty f<\infty$, $\int_1^\infty\frac{1}{p}=\infty$, then $\lim\limits_{x\longrightarrow \infty}p(x)f(x)=0$.

This claim is based on the following proposition:

Proposition: For $f: [1,\infty)\longrightarrow\mathbb R$ to be nonnegative and decreasing with $\int_1^\infty f<\infty$, show that $\lim\limits_{x\longrightarrow\infty}xf(x)=0$.

For the above proposition is the special case for the claim when $p(x)=1/x$. The proof of the proposition can proceed in many ways. The following is my proof.

Proof of the proposition:

Assume not, there exists $\varepsilon_0>0$, $\forall n\in\mathbb N_+$, there exists $x_n\in \mathbb R$ such that $x_n>n$ and $f(x_n)\geq \frac{\varepsilon_0}{x_n}$. Then we integrate $f(x)$ over $[x_n/2,x_n]$, by the monotonic decreasing property of $f$, we may know that $f(x)\geq f(x_n)\geq\frac{\varepsilon_0}{x_n}\geq \frac{\varepsilon_0}{2x}dx$ for all $x\in [x_n/2,x_n]$. Hence: $$\int_{x_n/2}^{x_n} f\geq \int_{x_n/2}^{x_n} \frac{\varepsilon_0}{2x}dx=\frac{\varepsilon_0 \log 2}{2}$$ contradicting with the condition that $\int_1^\infty f<\infty$

My Question:

I have tried to prove the claim in the similar way to the proof of the proposition. Then I have $$\int_{x_n/2}^{x_n} f\geq \int_{x_n/2}^{x_n} \frac{\varepsilon_0}{p(2x)}= \int_{x_n}^{2x_n} \frac{\varepsilon_0}{2p(t)}dt$$ However I find it difficult to use Cauchy criterion or the monotone convergence theorem to argue that $\int_{x_n}^{2x_n} \frac{\varepsilon_0}{2p(t)}dt\geq \eta$ for some $\eta$. Or I can proceed the proof only if $p$ has an explicit form. Any thoughts and inspirations are welcome. Thank you!

CCQ
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2 Answers2

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The claim is does not hold in general.

Let $f(x)=(3-x)\mathbb{1}_{(1,2]}(x)+\frac{1}{(x-1)^2}\mathbb{1}_{(2,\infty)}(x)$ and $\frac{1}{p(x)}=g(x)=\frac{1}{(x-1)^2}$, for $x>1$. It is easy to check that $\int^\infty_1 f(x)\,dx=\frac32+\int^\infty_1\frac{1}{x^2}\,dx<\infty $ and that $\int^\infty_0g(x)\,dx =\int^\infty_0\frac{1}{x^2}\,dx=\infty$. Yet, $$\frac{f(x)}{g(x)}\xrightarrow{x\rightarrow\infty}1$$ Also, $f$ and $g$ are is monotone non increasing.

Edit: In my initial posting I overlooked the assumption that $g$ is assumed bounded ($g$ is real valued on $[1,\infty)$). In any event, the claim in the OP does not hold in general under this additional assumption on $g$.

In general, if $f\geq0$, $f\in L_1(1,\infty)$ (that is, $\int^\infty_1f(x)\,dx<\infty$), $g>0$ and $\int_{[1,\infty)}g=\infty$, then $$\ell:=\liminf_{x\rightarrow\infty}\frac{f(x)}{g(x)}=0$$

Otherwise, there is $X>1$ such that $\frac{\ell}{2}g(x)<f(x)$ for all $x>X$. Thus, if $f\in L_1(1,\infty)$, so is $g$.

The question is then whether under the assumptions of the OP, $u:=\limsup_{x\rightarrow\infty}\frac{f(x)}{g(x)}=0$. The answer is no in general. Here I present counterexamples based on Theo Bendit's answer to a posting of Adam Rubinson here.

Lemma: For any positive, convex, decreasing, summable sequence $(a_n:n\in\mathbb{Z}_+)$ ($a_n>0$, $a_n\leq \frac{a_{n-1}+a_{n+1}}{2}$, and $a_{n+1}<a_n$ for all $n$) there is a convex, decreasing, non-summable sequence $(b_n:n\in\mathbb{Z}_+)$ so that $\frac{a_n}{b_n}\not\rightarrow0$

Using this result, once can construct counter examples to the statement in the OP starting with convex, nonincreasing integrable functions as follows: Suppose the that $(a_n:n\in\mathbb{Z}_+)$ is as in the Lemma. Define the function \begin{align} f(x)&=\sum^\infty_{n=1}a_n\mathbb{1}_{[n,n+1)}(x)\\ g(x)&=\sum^\infty_{n=1}b_n\mathbb{1}_{[n,n+1)}(x) \end{align} Clearly $f,g>0$ are bounded, monotone nondecreasing, $f\in L_1(1,\infty)$ and $\int_{[1,\infty)}g=\infty$. Yet, $\frac{f(x)}{g(x)}\not\rightarrow0$.

Theo's construction:

  1. Set $n_0 = 0$ and $b_{n_0} = b_0 = a_0$. Once integers $n_0,\ldots, n_{k-1} \geq 0$ are defined, so that

  2. $n_{j+1} > n_j$,

  3. $(n_{j+1} - n_j)(a_{n_{j+1}} + a_{n_j}) \geq 2$,

for all $ 0\leq j\leq k - 1$.

  1. Choose: $$n_{k+1} \ge n_k + \frac{2}{a_{n_k}}.$$

  2. Define $(b_n:n\mathbb{Z}_+)$ as a linear interpolation of points $a_{n_k}$: That is, for $n \in [n_k, n_{k+1})$, set $$b_n = \frac{n - n_k}{n_{k+1} - n_k}a_{n_{k+1}} + \frac{n_{k+1} - n}{n_{k+1} - n_k}a_{n_k}. $$ Notice that $\frac{a_{n_k}}{b_{n_k}}=1$ and so, $$\limsup_n\frac{a_n}{b_n}=1$$

Details as to the proof of the Lemma are in Theo's answer.

Mittens
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You do not necessarily have that $\displaystyle \int_{x_n}^{2x_n} \frac{\varepsilon_0}{2p(t)} \, dt \geqslant \eta > 0$ for all $n$ large enough (take for example $p(t) = t\ln(t)$) but the divergence of the integral of $1/p$ implies that, $$ \sum_{n \geqslant 0} \int_{x_n}^{2x_n} \frac{\varepsilon_0}{2p(t)} \, dt = \int_{x_0}^{+\infty} \frac{\varepsilon_0}{2p(t)} \, dt = +\infty. $$ And, $$ \sum_{n \geqslant 0} \int_{x_n/2}^{x_n} f(t) \, dt = \int_{x_0/2}^{+\infty} f(t) \, dt < +\infty. $$ Therefore, assuming the inequality you computed to be true for all $n$ (I didn't verify everything), we reach a contradiction.

Cactus
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