Additional conditions necessary to establish a property of the derivative of a bounded monotone function I am trying to find sufficient conditions for the following to be true. Suppose a function $f: \mathbb R \rightarrow \mathbb R$ is bounded, increasing and differentiable. What further conditions are necessary to establish that $\lim_{x\rightarrow \infty}x f'(x)=0$?
Boundedness and monotonicity are sufficient to establish that $\lim \inf _{x\rightarrow \infty}′()=0$: Suppose $\lim \inf _{x\rightarrow \infty}′()>0$. Then exists a $\delta>0$ and $\epsilon>0$ such that when $x>\delta$, $′()>\epsilon$. Then for >, ′()>/. The antiderivative of / is ln which converges to ∞ as →∞. This contradicts the boundedness of . Therefore $\lim \inf _{x\rightarrow \infty}′()\leq 0$. Since  is increasing, it must be that $\lim \inf _{x\rightarrow \infty}′()\geq 0$. Combining these conditions yields $\lim \inf _{x\rightarrow \infty}′()=0$
Also from  increasing, we know $\lim \sup _{x\rightarrow \infty}′()\geq 0$. So it remains to find conditions that ensure that $\lim \sup _{x\rightarrow \infty}′()= 0$.
 A: We can in fact prove :

Suppose that $f$ is bounded monotonically increasing differentiable and $\lim_{x \to \infty} (xf(x))' = a$. Then, $\lim_{x \to \infty} f(x) = a$, $\lim_{x \to \infty} xf'(x) = 0$.

Proof : From this post, the first part is proved by taking $g(x) = xf(x)$ and noting that $g'(x) \to a$ as $x \to \infty$ so $\frac{g(x)}{x} \to a$ as $x \to \infty$ i.e. $f(x) \to a$ as $x \to \infty$. Following this, we use the product rule for differentiation and note that :
$$
(xf(x))' = f(x) + xf'(x)
$$
Therefore, using limit rules it follows that $xf'(x) \to 0$ as $x \to \infty$. We didn't use monotonicity. $\blacksquare$
Note that the converse is also true : if $\lim_{x \to \infty} f(x) = a$ and $\lim_{x \to \infty} xf'(x) = 0$ then using the product rule and limit rules will tell us that $\lim_{x \to \infty}(xf(x))' = a$ readily.
Thus, we exactly need conditions which ensure that $(xf(x))'$ has some finite limit as $x \to \infty$. Note that $xf(x)$ is an increasing function, therefore we do know that $(xf(x))' \geq 0$ everywhere. The question is, can we do better?
Let's put $g(x) = xf(x)$. $g$ is a monotonically increasing function. When does $g'$ have a finite limit as $x \to \infty$? For the purposes of the analysis below, we assume by shifting $f$ by a constant ,that $a>0$.

We first see what we have. We know that $\frac{g(x)}{x} = f(x)$ is increasing, so we take the derivative and get :$$
f'(x) \geq 0 \implies xg'(x)- g(x) \geq 0 \implies g'(x) \geq \frac{g(x)}{x} \geq 0 
$$
this is what we know. This statement is incapable of providing anything about $g'(x)$ because we don't know an upper bound for it. Therefore, we will have to operate with $g'(x)$ as a separate entity.
We are entering a domain which I would call as interesting in its own right, because it contains some fascinating insights into function growth. The results I talk about often get used in probability theory because the densities of some random variables end up having properties that are reflected in the hypothesis of these questions, and the conclusions of these theorems usually leads to some kind of result around that random variable.

The first of these is called the monotone density theorem (there are many such results : this one is due to Bingham ,Goldie and Teugels). We first make a definition.

A function $h : (0,\infty) \to \mathbb R$ is called regularly varying at $\infty$ with index $\alpha \in \mathbb R$ if , for all $\lambda>1$, we have that $\lim_{x \to \infty} \frac{h(\lambda x)}{h(x)} = \lambda^{\alpha}$. A function is called slowly varying if it is regularly varying with index $0$.

A remark at this point : if, for even one $\lambda_0>1$, it is true that $\lim_{x \to \infty} \frac{h(\lambda_0 x)}{h(x)} = \lambda_0^{\alpha}$ for some $\alpha$, and if it is merely true that for every other $\lambda>1$ that $\lim_{x \to \infty} \frac{h(\lambda x)}{h(x)}$ exists (but we don't know its value), then $\lim_{x \to \infty} \frac{h(\lambda x)}{h(x)} = \lambda^{\alpha}$ is actually true for all $\lambda>1$.
With this, we can state the monotone density theorem.

Suppose that $h : (0,\infty) \to \mathbb R$ is of the form $h = x^{\alpha}l(x)$ for some slowly varying function $l(x)$  and $\alpha \neq 0$ (You can see that $h$ is regularly varying with index $\alpha$). Then, if $h$ is differentiable (note : can replace with absolute continuity) and $h'$ is an eventually monotone function, we get that $h'$ is regularly varying with index $\alpha-1$ , and $\lim_{x \to \infty} \frac{h'(x)}{\alpha x^{\alpha-1}l(x)} = 1$.

Let us apply this to our case. Let $g = h$. Then, $g = xf(x)$, where $f$ is actually slowly-varying : we have $\lim_{x \to \infty} f(x) = a$, therefore $\lim_{x \to \infty} f(\lambda x) = a$ for all $\lambda>1$ and we can use the quotient rule to see that $\lim_{x \to \infty} \frac{f(\lambda x)}{f(x)} = 1 = \lambda^0$.
Now, IF $g'$ is eventually monotone, then the result applied with $\alpha=1$ tells you that $$
\lim_{x \to \infty} \frac{g'(x)}{f(x)} = 1 \implies \lim_{x \to \infty} g'(x) = \lim_{x \to \infty} f(x) = a
$$
as desired. Therefore, we have proved that :

If $f$ is bounded increasing differentiable and $g(x) = xf(x)$ has an eventually monotone derivative (Note : this can also be expressed by saying that $g''(x) \neq 0$ for sufficiently large $x$, provided that $g$ is twice differentiable), then $\lim_{x \to \infty} xf'(x) = 0$.

But can we do better? Perhaps monotonicity is too much to ask for , and we should be providing a more flexible set of conditions? That is a more difficult question to answer : I can provide the result which is in the book of Bingham, Goldie and Teugels titled Regular variation. This will require a complicated definition.

For a function $h: (0,\infty) \to \mathbb R$, the quantities $\alpha_h$ and $\beta_h$ defined as :
$$
\alpha_h = \inf\{A:\exists C>0,\forall \gamma>0, x \text{ sufficiently large depending on $\gamma$} , \lambda \in [1,\gamma], \text{ we have that }\frac{h(\lambda x)}{h(x)} \leq C \lambda^A + o(\lambda^A) \} \\ \beta_h = \sup\{B : \exists C>0,\forall \gamma>0, x \text{ sufficiently large depending on $\gamma$} , \lambda \in [1,\gamma],\text{ we have that }\frac{h(\lambda x)}{h(x)} \geq C \lambda^B + o(\lambda^B) \}
$$
are respectively referred to as the upper and lower Matuszewska indices of $h$.

These basically capture the worst uniform behaviour that the ratio can exhibit across $x$, and help capture a finer notion of variation, than just monotonicity alone. We can write down the result in terms of these Matuszewska indices.

Prop 2.10.3, Regular Variation [O-monotone density theorem/Matuszewska density theorem] :  Suppose that $h$ is an eventually strictly increasing function with derivative $h'$ such that $\alpha_{h'}<\infty$ or $\beta_{h'}>-\infty$. Suppose furthermore that $\alpha_h<\infty$ and $\beta_h > 0$. Then, $\lim_{x \to \infty} \frac{xh'(x)}{h(x)} = 1$.

(Note : I differentiated every function of the actual statement so we can make it conform to our situation more, but it's basically the same thing).
This result doesn't assume anything except some basic growth conditions on the quantity $\frac{h(\lambda x)}{h(x)}$, except in the limit. Let us try to apply it to our situation.
We naturally want $g(x) = h(x)$ so that $\frac{xg'(x)}{g(x)} = \frac{xg'(x)}{xf(x)} = \frac{g'(x)}{f(x)} \to 1$ which would give us the conclusion. So, we just need to fit this in. Note that $g$ is already monotone : that will tell us that $\beta_{g} > 0$. Therefore, when we fulfill the rest of the conditions, we get :

Theorem : If $f(x)$ is bounded increasing differentiable and $g(x) = xf(x)$ has the following properties : either $\beta_{g'} > -\infty$ OR $\alpha_{g'}<\infty$, and $\alpha_{g} < \infty$. Then, $\lim_{x \to \infty} xf'(x) = 0$.

Of course, I will not be discussing the proof.
These are two sufficient conditions of markedly different nature that show why regular variation is a very well studied subject, and how one can link derivative growth to function growth for special classes of functions.
