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I am trying to prove the following or find a counterexample. Suppose a function $f: \mathbb R \rightarrow \mathbb R$ is bounded, increasing, and continuously differentiable. Then $\lim_{x\rightarrow \infty}xf'(x)=0$.

So far, I have shown that $\lim\inf_{x\rightarrow \infty} xf'(x)= 0$: Suppose $\lim \inf_{x\rightarrow \infty }xf'(x)>0$. Then exists a $\delta>0$ and $\epsilon >0$ such that when $x>\delta$, $x f'(x)>\epsilon$. Then for $x>\delta$, $f'(x)>\epsilon /x$. The antiderivative of $\epsilon /x$ is $\epsilon \ln x$ which converges to $\infty $ as $x\rightarrow \infty$. This contradicts the boundedness of $f$. Therefore $\lim\inf_{x\rightarrow \infty} xf'(x)\leq 0$. Since $f$ is increasing, it must be that $\lim\inf_{x\rightarrow \infty} xf'(x) \geq 0$. Combining these conditions yields $\lim\inf_{x\rightarrow \infty} xf'(x)= 0$.

Also from $f$ increasing, we know $\lim\sup_{x\rightarrow \infty} xf'(x) \geq 0$. So it remains to show that $\lim\sup_{x\rightarrow \infty} xf'(x) = 0$.

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Counter-example: Let $g(x)=n$ for $n \leq x \leq n+\frac 1 {n2^{n}},g(x)=0$ for $x >n+\frac 2 {n2^{n}}$ and $0$ for $x <n-\frac 1 {n2^{n}}$ with a straight line graph between $n$ and $n+\frac 2 {n2^{n}}$ as well as between $n-\frac 1 {n2^{n}}$ and $n$ (for each $n$). Let $f(x)=\int_0^{x}g(t)dt$. You can easily check that this is a counter-example. (In fact $f'(n)\to \infty$).

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  • $\begingroup$ Thank you. This construction is along the lines of the intuition I had for why it may be false--an increasingly large derivative on an interval shrinking to zero as x grew large. So is f twice continuously differentiable a sufficient condition for the statement to be true? How about f Lipschitz continuous? $\endgroup$
    – Igor
    Commented Nov 4, 2021 at 21:08

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