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.

This is a rather open-ended question; I'll interpret it as looking for conditions on $f$ that don't refer directly to $f''$ (or even $f'$) and preferably minimize any reference whatsoever to Calculus concepts.  (After all, if I could use Calculus freely in my answer, then I'd just say that $f''(x) \leq B$ for all $x$, and what would be the point of that?)
Obviously you need $f$ to be twice differentiable, say on some interval $I$.  Assuming this, $f'' \leq B$ on $I$ if and only if for all $a < b < c$ in $I$,
$$ \frac {\frac {f(c) - f(b)} {c - b} - \frac {f(b) - f(a)} {b - a}} {c - a} \leq \frac {B} {2} .$$
The only way that I can see to use that $f < A$ is to replace $f(a)$ and/or $f(c)$ (but not $f(b)$) with $A$.  However, this doesn't make things much simpler, especially since $a$ and $c$ remain in the inequality, and it loses the reverse implication.
As long as you don't use $A$, it's not actually necessary that $a < b < c$ (as long as they are distinct to avoid division by zero), although that's easiest to motivate the result; the final numerator on the left-hand side is the difference between the slopes of secant lines, which approximates a difference between slopes of tangent lines, which is related to the second derivative.
Compare the corresponding result for first derivatives: assuming that $f$ is differentiable on $I$, $f' \leq B$ on $I$ iff for all $a < b$ in $I$,
$$ \frac {f(b) - f(a)} {b - a} \leq B .$$
Note that the expression on the left-hand side here occurs within the left-hand side of the second-derivative version.  (The proof of the second-derivative version is also similar to this, based on the Mean Value Theorem, but with more steps.)  If you continue the pattern, you can find a condition for when the $n$th derivative of $f$ is bounded above by $B$ on an interval $I$, assuming that $f$ is $n$-times differentiable on $I$.  (The right-hand side in the general case is $B/n!$.)
If you simplify the complex fraction on the left-hand side above and expand the numerator and denominator, then they can be expressed as determinants:
$$ \frac {\det \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ f(a) & f(b) & f(c) \end{bmatrix}} {\det \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}} \leq \frac {B} {2} .$$
This also generalizes to higher orders; the denominator is a determinant of order $n + 1$, containing powers of the points chosen, while the numerator is almost the same, except that the bottom row (where the $n$th powers should be) is replaced by the values of the function.  (The right-hand side, as I remarked above, is $B/n!$.)
Some of the other commenters have mentioned Lipschitz conditions.  This is not directly relevant, since you asked about $f''$ itself rather than its absolute value.  However, you can slap absolute value bars around everything, and then it all remains true.  In particular, $\lvert{f'}\rvert \leq B$ on $I$ (assuming that $f$ is differentiable on $I$) iff for all $a < b$ in $I$,
$$ \left|\frac {f(b) - f(a)} {b - a}\right| \leq B ,$$
and this means precisely that $f$ is Lipschitz continuous on $I$ with Lipschitz constant $B$.  Similarly, $\lvert{f''}\rvert \leq B$ on $I$ (assuming that $f$ is twice differentiable on $I$) iff for all $a < b < c$ in $I$,
$$ \left|\frac {\frac {f(c) - f(b)} {c - b} - \frac {f(b) - f(a)} {b - a}} {c - a}\right| \leq \frac {B} {2} ,$$
which is a sort of higher-order Lipschitz condition.  (With these absolute values, you also can not only drop the requirement that $a, b, c$ come in order but even drop the requirement that they are distinct, since you can multiply by the absolute value of the denominator to avoid division by zero.  Without the absolute value, you need at least $a \leq b \leq c$ to justify this.)
