Unbounded second derivative of continuous bounded function Does a continuous function $f$ exist where:
$f$ is continuous,
$f$ is known to be bounded with a codomain of $\left[0,1\right]$
$f''$ (second derivative) is unbounded with a codomain of $(-\infty,+\infty)$
 A: Presumably the answer is yes: a function can be bounded but wiggle very vigorously.  
So let's try to make one.  Maybe we should start with $\cos(x)$, that's between $-1$ and $1$, easily fixed by adding $1$ and dividing by $2$. Unfortunately all derivatives are bounded.  So give it a little more vigor by say looking at $f$ where 
$$f(x)=\frac{\cos(x^2) +1}{2}$$
We have $f'(x)=-x\sin(x^2)$ and 
$$f''(x)=-(2x^2\cos(x^2) +\sin(x^2))$$.
Now show that we can make $f''(x)$ arbitrarily large positive or negative.  Then the Intermediate Value Theorem will show that the range of $f''(x)$ is the one you were looking for.
To make $f''(x)$ very large positive, just choose $x$ large such that $\cos(x^2)=-1$.  To make  $f''(x)$ very large negative, choose $x$ large such that $\cos(x^2)=1$.  Each is easy to arrange.  
A: $$f(x)=\begin{cases}
\frac{1}{2}+\frac{1}{2}\sqrt{1-x^2} & \text{ if } 0\le x\le 1 \\ 
\frac{1}{2}-\frac{1}{2}\sqrt{1-(2-x)^2} & \text{ if } 1< x\le2 
\end{cases}$$
This is bounded—$0\le f(x)\le 1$—and its second derivative is unbounded in both directions (though I think $|f''(x)|\ge 1$, which may not be quite what you wanted).
