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Consider a continuously differentiable function $f: [0, 1] \to \mathbf{R}_{\geq 0}$, such that $f(0)=0, f'(0)=0$ and $f(x)>0$ for $x>0$.

I want to prove that there exists $\bar{x}>0$ such that $f(x)$ is increasing on $[0, \bar{x}]$. (or, even better, strictly increasing)

This question is nearly identical to this Derivative of positive function is positive close to zero However, apart from continuity, I have additionally continuous derivatives and, because of that, cannot make the counterexample proposed there to work. The answer to that question is (paraphrasing): 'for a function oscillating near 0 between $x^2$ and $x^4$ we have $f(0)=0$, $f(x)>0$ for $x>0$ and the function is continuous'. However, I cannot construct such a function that would not violate continuity of the derivative. For example, consider: $$f(x)= (\sin(1/x)+1)x^2+(\cos(1/x)+1)x^4.$$ The derivative of this function oscillates between $0$ and $1$ and thus is not continuous (whilst all the other conditions are met).

In short: is the continuity of the derivative sufficient to make the original statement true, or does there exist a counterexample with continuous derivatives?

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for $f(x) = x^4 \left(\sin(\frac{1}{x})\right)^2$ you get $$f^\prime(x)= 4x^3 \left(\sin(\frac{1}{x})\right)^2 - 2x^2 \sin(\frac{1}{x})\cos(\frac{1}{x}) = 2x^2\sin(\frac{1}{x})\left(2x\sin(\frac{1}{x})- \cos\frac{1}{x} \right)$$

This can be clearly continuously extended to $x=0$ , and if $x$ becomes small behaves like $$-2x^3\sin(\frac{1}{x}) \cos(\frac{1}{x})$$ from which it is not difficult to see that $f^\prime$ will have arbitrary many points where it is $<0$.

Now this function is not positive but only $\ge0 $, but by adding, say $x^{10}$, you get a positive function (for $x>0$) with similar properties for $f^\prime$.

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  • $\begingroup$ Thank you, I've been obsessing over this longer than I care to admit! And this counterexample puts it to rest... $\endgroup$ – Tsiar Oct 4 '18 at 15:34

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