Sufficient condition for decreasing function $\phi:(0,\infty)\to (0,\infty)$ to obey $\int_x^\infty\phi(t)\mbox{d}t\le Cx\phi(x)$, with C = constant

Question

Question. What is a simple characterization of strictly decreasing functions $\phi:[1,\infty) \to (0,\infty)$, with $\int_1^\infty \phi(t)\mbox{d}t \lt \infty$ and $\lim_{x \to \infty}\phi(x) = 0$, such that $$ \int_x^{\infty} \phi(t)\mbox{dt} \le C x\phi(x)\,\forall x \ge 1, $$ for some positive constant $C$ which only depends on $\phi$ ?

For example, the functions $x \mapsto x^{-1-\gamma}$ (with $\gamma \gt 0$) and $x \mapsto e^{-x}$ verify the above condition.

Note. In the absense of a succinct characterization, I'm fine with a sufficient condition which covers a diverse choises of $\phi$ at once.

Answer 1

Denote $F(x) = \int_x^{\infty}\phi(t)\mbox{dt}$, then $$\begin{align} &\Longleftrightarrow F(x) \le - Cx F'(x) \\ &\Longleftrightarrow F'(x)+\frac{1}{Cx}F(x) \le 0 \\ &\Longleftrightarrow \frac{F'(x)}{F(x)}+\frac{1}{Cx} \le 0 \qquad \text{as } F(x)>0 \\ &\Longleftrightarrow \left(\ln\left(F(x)x^{\frac{1}{C}} \right)\right)' \le 0 \\ &\Longleftrightarrow \ln\left(F(x)x^{\frac{1}{C}} \right) \quad \text{is non-increasing}\\ &\Longleftrightarrow F(x)x^{\frac{1}{C}} \quad \text{is non-increasing} \tag{1} \\ &\Longleftrightarrow x^{\frac{1}{C}}\int_x^{+\infty}\phi(t)dt \quad \text{is non-increasing} \\ &\Longleftrightarrow x^{1+\frac{1}{C}}\int_1^{+\infty}\phi(xt)dt \quad \text{is non-increasing} \tag{2} \\ \end{align}$$

There are two solutions.

First solution: The necessary and sufficient condition is: there exists a $\gamma >0$ such that $F(x)x^{\gamma}$ is a non-increasing function. We determine the function $\phi(x)$ as follows:

1. Step 1: Define a function $p(x)$ and a $\gamma >0$ satisfying: $p(x)$ is non-increasing function and $x^{-\gamma}p(x)$ is convexe (the convexity is necessary and sufficient for the strictly decreasing of $\phi(x)$).

   For example: $p(x) = x^{\gamma}e^{-x}$ with $\gamma<1$. We have $x^{-\gamma}p(x) = e^{-x}$ is convexe

2. Step 2: With the $\gamma$ and $p(x)$ already defined in step 1, assume that $F(x)x^{\gamma} = p(x) \Longleftrightarrow F(x) = x^{-\gamma}p(x)$ then $$\color{red}{\phi(x) = -F'(x) = -\left(x^{-\gamma}p(x) \right)'}$$ We have $\phi(x)$ is strictly decreasing (as $\left(x^{-\gamma}p(x) \right)''>0$).

   For example, if $p(x) = x^{\gamma}e^{-x}$ then $\phi(t) = e^{-x}$

Second solution: from $(2)$, we deduce that the sufficient solution is: there exists a $\gamma >0$ such that for all $t>1$ $x \mapsto x^{1+\gamma}\phi(xt)$ is a non-increasing function

Remark: I thought that the second solution would be more interesting, but now I think the first one is much better. Indeed, the first solution is the necessary and sufficient condition and it's easy to create the function $\phi(x)$.

Answer 2

This is not a full solution but a general method, reminiscent of Gronwall's inequality, that may shed some light into the OP's question.

Setting $h(x)=\int^\infty_x\phi$, we obtain that $$-h'(x)=\phi(x)\geq \frac{1}{cx}h$$ or equivalently $$h'+\frac{1}{cx}h\leq0$$ multiplying by $x^{1/c}$ gives \begin{align} x^{1/c}\Big(h'+\frac{1}{xc}h\big)=(x^{1/c}h)'\leq0\tag{1}\label{one} \end{align} Integrating over $(t,T)$ $$t^{-1/c}T^{1/c}h(T)\leq h(t)\leq ct\phi(t)$$ That is \begin{align} \phi(t)\geq\frac{1}{c}T^{1/c}h(T)t^{-1/c-1}\tag{2}\label{two} \end{align}

If $\phi$ is such that $T^{1/c}\int^\infty_T\phi\xrightarrow{T\rightarrow\infty}\alpha$ (the limit exists since $x\mapsto x^{1/c}h(x)$ is nonnegative and monotone nonincreasing), then \begin{align} \phi(t)\geq \frac{\alpha}{c}t^{-1/c-1}\tag{3}\label{three} \end{align}

Conversely, suppose $\phi$ is non negative, continuous, integrable in $(1,\infty)$, such that $x\mapsto x^{1/c}\int^\infty_x\phi$ is monotone non increasing, then $$\big(x^{1/c}\int^\infty_x\phi\big)'=x^{1/c}\Big(-\phi(x)+\frac{1}{cx}\int^\infty_x\phi\Big)\leq0$$ Hence $\int^\infty_x\phi \leq cx\phi(x)$, and by the first part of the posting, it satisfies \eqref{three}.

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Continuity is to avoid issues of differentiability of $x\mapsto \int^\infty_x\phi$ and avoid passing to almost surely type of arguments.