Let $u:(x_0,\infty)\to\Bbb R$ be a monotonically increasing function which is differentiable everywhere and such that $\lim_{x\to\infty}u(x)=l\in\Bbb R$. Does it follow that $\lim_{x\to\infty}u'(x)$ exists?

Surely, if it exists, then it would equal zero. In fact $$0=\lim_{x\to\infty}[u(x+1)-u(x)]=\lim_{x\to\infty}u'(\xi_x)=\lim_{x\to\infty}u'(x)$$ (this is obtained by applying the mean value theorem: for every $x$ there exists a $\xi_x$ in between $x$ and $x+1$ such that $u'(\xi_x)=u(x+1)-u(x)$).

But I think that under the above assumptions $u'$ must have a limit at infinity. Is it true?

  • $\begingroup$ Let $a_n$ be a increasing sequence such that $a_n\to \infty$. Try to define a function $g:(x_0,\infty)\to (0,\infty)$, which oscillates, for example, between $0$ and $1$ and such that the measure of the set $\{x\in (a_n,a_{n+1}): g(x)>b_n\}$ is small, for some sequence $b_n\in (0,1)$. If this is possible then, the function $f(t)=\int_{x_0}^t g(s)ds$ satisfies your hypothesis, however its derivative does not converge. $\endgroup$ – Tomás Feb 15 '15 at 23:39
  • $\begingroup$ No $u'$ may not have a limit at infinity, in fact the limit set of $u'$ may be $[0,+\infty)$. $\endgroup$ – Did Feb 15 '15 at 23:41

The statement is false. The trick is to make a function with nonnegative values that keeps making bumps that get smaller and smaller in width, such that the integral of the function converges on $(1,\infty)$, but the function itself does not have a limit for $x\to\infty$.

We could for instance construct a function $f$ that is $0$ everywhere but makes a spike towards $1$ at every positive integer $n$, where we choose the width of the spike in such a way that the area under the spike is $\frac{1}{n^2}$. Then the function $g(t):=\int_{1}^{t} f(x)dx$ is monotonically increasing (since $f$ admits only nonnegative values) and will converge as $t\to\infty$ but $g'(t)=f(t)$ will not have a limit.


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