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Consider a function $f:[0,\infty)\longrightarrow [0,\infty)$. Suppose that $f(0)=0$ and that $f$ is strictly increasing. Moreover, assume that $f$ is continuously differentiable except at zero and that $f$ is differentiable (from right) at zero.

Is it true that $f$ has a bounded derivative on some neighborhood of zero?

If not, under which conditions on $f$ can we ensure this?

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The answer is no, although it seems surprisingly not easy to find an actual formula of a function that would disprove it.

But essentially the idea is that as you approach $0$ from the right, the graph of the function could look like (smooth) stairs with smaller and smaller height and width, but greater and greater maximal slope, where the oscillation with become so insignificant as you approach $0$ that the right derivative at $0$ would be equal to, say, $1$.

One way to prevent this kind of monsters and ensure a positive answer to your question would be to require $f''$ to (exist and) be of constant sign.

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  • $\begingroup$ Nice description. $\endgroup$
    – zhw.
    Commented May 22, 2018 at 16:47
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I believe the following function is a counterexample:

$$f(x) = \int_0^x \frac{[\cos^2(1/t)]^{1/t^4}}{t}\, dt.$$

The idea being that the exponent $1/t^4$ is huge when $t$ is small, and should be enough to smash $\cos^2(1/t)$ down to almost $0$ so that even when divided by $t,$ we have a nice convergent improper integral. Even more, we will have $f'(0)=0.$ Note that $f'(1/(2n\pi)) = 2n\pi, n=1,2,\dots$ so that $f'$ is unbounded in every neighborhood of $0.$ A number of details remain to be checked of course ...

It's probably more instructive to play around with tall thin triangles however. So if over the points $1/n,n=1,2,\dots$ we put triangles of base lengths less than $1/n^4$ and heights $n$ (keep the bases disjoint), we arrive at a candidate integrand. Call this function $g$ and set $f(x) = x+ \int_0^x g(t)\,dt.$ This should be an example, this time with $f'(0)=1.$

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