# Properties of a derivative on a compact interval

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.

• hmm, potential counterexample? desmos.com/calculator/1yqeep5px7 presumably this isn't differentiable in the limit at 0 – Calvin Khor May 27 at 13:37
• Interesting idea! However, indeed the derivative doesn't exist at 0. – user2258552 May 27 at 20:18
• I believe it works if you make the spikes sharper, like desmos.com/calculator/tlotdsrofw . I've made a quick calculation and it seems right, but you've got simpler answers now – Calvin Khor May 27 at 21:14

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 $$[0,1]$$).

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 $$F'(0)=1/2.$$