Properties of a derivative on a compact interval

Question

Suppose a function $F$ is differentiable on an interval $(a,b) \supset [0,1]$. Denote its derivative by $f$, and suppose that $f > 0$ on $[0,1]$.

Question 1: Is it true that $f$ can be bounded away from $0$ on $[0,1]$, i.e. that there exists some $c > 0$ such that $f(x) > c$ for all $x \in [0,1]$? If $f$ is continuous, this is clearly true, as a continuous function attains its minimum on a compact set, and this minimum is $> 0$ by assumption. If $f$ were an arbitrary function (not a derivative), this is clearly false; for instance, consider the function $f(x) = 1$ when $x = 0$ and $f(x) = x$ elsewhere. But this function has a jump discontinuity, and therefore is not the derivative of any function.

Question 2: Is it true that $f$ is bounded on $[0,1]$? Note that if we remove the $f > 0$ requirement, this is not true (for instance, consider $F(x) = x^2 \sin(1/x^2)$, with $F(0) = 0$; $F$ is differentiable, so $f$ exists, but $f$ is not bounded).

This question may be relevant, but it doesn't directly answer the above.

Answer 1

No. For instance, let $f$ be piecewise linear and positive on $[0,1)$ such that on an infinite sequence of intervals approaching $1$, $f$ alternates between jumping down to values approaching $0$, jumping up to values approaching $\infty$, and jumping back down to $1$ and remaining constant with value $1$. Define $F(x)=\int_0^xf(t)\,dt$. If we choose the intervals where $f$ takes values other than $1$ to be sufficiently small and sparse (so as $t\to 1$, $f(t)=1$ for a quickly increasingly large proportion of the time), the integrals of $f$ over these intervals will have a negligible effect on the limiting behavior of $F(x)$ as $x\to 1$. So, $F$ will extend continuously to $1$ with $F'(x)=1$. We can then extend $F$ to be differentiable on an open interval containing $[0,1]$ (e.g., by making its derivative $1$ outside of $[0,1]$).

Answer 2

A counterexample for the the first question is

$$\tag 1 F(x)=\int_0^x\sin^2 (1/t)\,dt + x^2.$$

Proof: On $(0,1],$ the FTC shows $F'(x)= \sin^2 (1/x) + 2x >0.$ We also have $F'(0)=1/2.$ This claim is harder to prove but suppose it holds. We then have $F'>0$ on $[0,1].$ But as $n\to \infty,$ $F'(1/(n\pi))$ $= 2/(n\pi)$ $\to 0.$ Thus there is no positive lower bound $c$ for $F'.$

To show $F'(0)=1/2,$ let $I(x)$ be the integral in $(1).$ Letting $t=1/y$ gives

$$\frac{I(x)}{x} = \frac{1}{x}\int_{1/x}^\infty \frac{\sin^2 y}{y^2}\,dy=\frac{1}{2x}\int_{1/x}^\infty \frac{1-\cos(2y)}{y^2}\,dy$$ $$ = \frac{1}{2} - \frac{1}{2x}\int_{1/x}^\infty \frac{\cos(2y)}{y^2}\,dy.$$

To show the second term in the last expression $\to 0$ as $x\to 0^+,$ integrate by parts. (I'll leave it here for now; ask if you have questions.) Thus $I(x)/x\to 1/2,$ which shows $F'(0)=1/2.$