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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 $(-\infty,+\infty)$

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Second line answers the question... –  soandos May 22 '11 at 1:48
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The answer is yes. Have you experimented with examples? To start with, it may help to think of bounded functions with unbounded first derivative. You may want to try sketching their graphs before explicitly defining them. –  Jonas Meyer May 22 '11 at 1:56

2 Answers 2

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.

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$$f(x)=\begin{cases} \frac{1}{2}+\frac{1}{2}\sqrt{1-x^2} & \text{ if } 0\le x\le 1 \\ \frac{1}{2}-\frac{1}{2}\sqrt{1-(2-x)^2} & \text{ if } 1< x\le2 \end{cases}$$ 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).

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