Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $T<\infty$, $f:[0,T)\to \mathbb{R}$ is increasing, differentiable function and $\lim_{t\to T}f(t)=l<\infty.$ I also have $\lim_{t_k\to T}f'(t_k)=0,$ for some subsequence $\{t_k\}.$ Is it true that $\lim_{t\to T}f'(t)=0,$ and $f$ is concave beyond some point.

share|cite|improve this question
Try to modify a step function with infinitely many steps that is constant on $(1-1/(n+1),1/n)$ to get an increasing function on $[0,1)$ with a sequence $t_k\to 1$ such that $f'(t_k)=0$ and another sequence $s_k\to 1$ such that $f'(s_k)\to\infty$. – SteveH Jun 23 '11 at 21:31
up vote 4 down vote accepted

For a counterexample, just let $r(t)$ be any nonnegative continuous function on $[0,T)$ such that $\lim \inf _{t \to T} r(t) = 0$ but $\lim \sup _{t \to T} r(t) > 0$, and such that $\int_0^T {r(u)du} = l$. Now define $f$ by $f(t) = \int_0^t {r(u)du}$, $t \in [0,T)$. Then $f$ is monotone increasing on $[0,T)$, continuously differentiable, $$ \lim _{t \to T} f(t) = \lim _{t \to T} \int_0^t {r(u)du} = \int_0^T {r(u)du} = l, $$ and so we are done, since $f'(t)=r(t)$.

share|cite|improve this answer

It isn't necessarily true that $\lim_{t\rightarrow T} f'(t)=0$. Consider the function $f:[0,1)\rightarrow\mathbb{R}$, $$f(t)=1-\left(2^{\lfloor\log_2(1-t)\rfloor+1}-\sqrt{\left(2^{\lfloor\log_2(1-t)\rfloor}\right)^2-\left(2^{\lfloor\log_2(1-t)\rfloor}-(1-t)\right)^2}\right)$$ which consists of an increasing sequence of quarter-circles of decreasing radius, each connected to the last:

   Plot[1 - (2^(Floor[Log[2, (1 - t)]] + 1) - 
Sqrt[2^(2 Floor[Log[2,(1-t)]]) - (2^(Floor[Log[2,(1-t)]]) - (1-t))^2]),   
 {t, 0.000001, 1}, PlotRange -> {0, 1}, AspectRatio -> 1]

enter image description here
We can smooth $f$ out at the cusps to produce a function $F$ which is differentiable on all of $[0,1)$, and for which $F\equiv f$ except on the intervals $((1-2^{-k})-a_k,(1-2^{-k})+a_k)$, where $a_k$ is some sequence that decreases very rapidly (I think $a_k=2^{-2k}$ will suffice, but not sure).

The sequence $t_k=(1-2^{-k})-b_k$, where $b_k\geq a_k$ but is still decreasing very rapidly, has the property that $F'(t_k)=f'(t_k)\rightarrow0$, but we won't have $\lim_{t\rightarrow1}F'(t)=0$ because there will exist $t$ arbitrarily close to 1 for which $f'(t)$ is big.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.