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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.

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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
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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 $f'(t)=r(t)$.

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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.

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