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Prove or Disprove: Let $f:\mathbb{R} \to \mathbb{R}$ be a bounded uniformly continuous function that whose first and second derivative exists and is continuous, in other words $f \in C^2_{unif} (\mathbb{R},\mathbb{R})$. Then $f'(x)$ is bounded.

This is a problem that I came across while working on a project. At first we felt that it wasn't true but we've been unable to find a counterexample. Any advice on how to prove it (if it is true) would be much appreciated. I apologize if this has been posted before.

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It's not true, as a counter example take a sine curve with decreasing amplitude but frequency increasing to $\infty$ (this will mean unbounded derivative). Something like:

$$\frac{1}{1+x^2}\sin(x^5)$$

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You need a slightly stronger version of continuity: Lipschitz. Recall Lipschitz continuity implies uniform continuity

Let $f:[a,b]\rightarrow\mathbb{R}$ be a function. $f'(x)$ is bounded if and only if $f$ is Lipschitz.

$\implies)$ $f'(x)$ is bounded. WTS $f$ is Lipschitz.

If the derivative is bounded, we have $|f(x)|\leq M$ for some $M$. By definition of a derivative, we have that $f'(x)=$$\lim_{h\to 0}$${|(f(x+h)-f(x)|\over h}\leq M$.

For Lipschitz continuity, we have $|f(x)-f(y)|\leq M|x-y|$. In particular when $x\neq y$, we have $f'(c)={|f(x)-f(y)|\over |(x-y)|}$ by MVT (Mean Value Theorem), and so $f'(c)={|f(x)-f(y)|\over |(x-y)|}$ $\leq M$ $\rightarrow$ $|f(x)-f(y)|\leq M|x-y|$ as desired

$\impliedby)$ $f$ if Lipschitz, WTS $f'(x)$ is bounded.

if $f$ is Lipschitz, then we can write it as ${|f(x)-f(y)|\over |(x-y)|}\leq M$ fix $y=x+h$ so ${|f(x)-f(x+h)|\over |(x-(x+h))|}={|f(x+h)-f(x)|\over |h|}\leq M$ (notice in the numerator we can switch order because of absolute value). now we just take $h\rightarrow 0$ and we have our derivative, which is bounded by $M$

while not the standard approach, this can be also be proved by contrapositive. If $f'(x)$ is unbounded then $f$ isn't lipschitz. This follows from $\implies)$ since there is now no way to choose an upper bound $M$ for $|f(x)-f(y)|\leq M|x-y|$, so we have the result.

Also, I'll have to check to be sure, but Lipschitz functions are differentiable a.e. (almost everywhere) since we can always find points $x\neq y$ where the derivative exists, the only time this might fail is if $x=y$ but this is an isolated point, and in particular, forms a singleton set of measure zero. For $f(x)=|x|$, differentiability fails at $x=y=0$. For more general functions, other points may fail, but the points at which it fail have measure zero (worth checking).

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