If a function is bounded by A and its second derivative is bounded by B, how to prove that its first derivative is bounded? Is it true that its first derivative is bounded by $kA+B/k$ for any positive real number k?
 A: Let $f:\mathbb R\rightarrow \mathbb R$ be a twice differentiable function, such that $|f(x)|\leq A$ and $|f''(x)|\leq B$ for all $x\in \mathbb R$, where $A$ and $B$ are positive constants.
We now prove $|f'(x)|\leq kA+B/k$ for all $k>0$ and $x\in \mathbb R$.
Without loss of generality suppose $f'(0)>kA+B/k$.
Notice that since $f''(x)\geq -B$ we have $f'(x)> kA+B/k -Bx$ for $x\geq0$.
Letting $w=(kA+B/k)/B$ we have:
$f(w)-f(0)=\int\limits_{0}^{(kA+B/k)/B}f'(x)dx> \int\limits_{0}^ {(kA+B/k)/B}[kA+B/k-xB] dx=(kA+B/k)^2/2B$.
Now notice $(kA+B/k)^2=k^2A^2+2AB+B^2/k^2\geq 4AB$ by AM.GM
So $\frac{(kA+B/k)^2}{2B}\geq \frac{4AB}{2B}=2A$.
We conclude $f(w)-f(0)>2A$, a contradiction, since we have $f(0)\geq -A$, which implies $f(w)>A$.
A: As a matter of fact, I myself had come up with a proof that it is bounded, but i dont know if it is rigorous.
If we take the limit of (fx - fa)/x-a when x approach infinity for any positive number a, it will get zero due to boundedness of f. Hence according to mean value theorem, where is some number t for which the first derivative is zero. By taking a as t, there is t' greater than t with the first derivative of t' equal to zero. Therefore by induction the first derivative does not approach infinity and due to Continuity, it is bounded.
