bounded function has bounded second derivative Suppose function $f < A$ in interval $(-\infty, \infty)$, what other conditions are required so that $f'' \leq B, B > 0$? $A, B$ are constants.
I appreciate any suggestions.

Building off of @YvesDaoust's example, I thought to see how general the example of differentiable but not Lipschitz functions are for bounded functions without bounded derivative.
According to Wikipedia:

An everywhere differentiable function g : R → R is Lipschitz
  continuous (with K = sup |g′(x)|) if and only if it has bounded first
  derivative; one direction follows from the mean value theorem. In
  particular, any continuously differentiable function is locally
  Lipschitz, as continuous functions are locally bounded so its gradient
  is locally bounded as well.

Hence it seems like a necessary and sufficient condition for a bounded twice-differentiable function to have bounded second derivatives is that its first-derivative be Lipschitz continuous.
Also with regards to @James's concern, we can even find a counterexample on a compact interval (i.e. all of [0,1]) as follows:

The function $f(x) = x^{\frac{3}{2}} \sin(\frac{1}{x})$ where x ≠ 0
  and f(0) = 0, restricted on [0, 1], gives an example of a function
  that is differentiable on a compact set while not locally Lipschitz
  because its derivative function is not bounded.

