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Let $f : \mathbb R \to\mathbb R$ be a differentiable function such that $f(0) = 0$, and $f''(0)$ exists and is positive. Prove that there exists $x > 0$ such that $f(2x) > 2f(x)$.

By Taylor's theorem with Lagrange's reminder I can write $$f(x)=xf'(0)+x^2f''(0)/2+x^3f'''(y)/6$$ with $0<y<x$, so if I write $g(x)=f(2x)-2f(x)$ I have $$g(x)=x^2f''(0)+x^3f'''(y).$$ If $f'''$is bounded in $(0,h)$ for some $h>0$, then I have $g(x)/x^2$ tends to $f''(0)>0$ as $x$ tends to 0, and by the limit definition $g(x)>0$ for some $x$ small enough, and we're done. However we don't have conditions for $f'''(x)$ to be bounded near 0, so I don't know how to get $g(x)>0$ near 0.

Arctic Char
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Pier Francesco Peperoni
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3 Answers3

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Let $g(x)=f(2x)-2f(x)$, so $g(0)=0$ and $g'(x)=2f'(2x)-2f'(x)$ so $g'(0)=0$. Finally, $g(x)=\frac{g''(0)}{2}x^2+o(x^2)$ which in turn gives $g(0)>0$ for $x \in (0,\varepsilon)$ since it behaves like a degree two polynomial with positive degree two coefficient close to zero. That gives you the answer and you don't need to assume any more differentiability for $g$.

astro
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  • When you write $g(x) = \frac{g''(0)}{2} x^2 + o(x^3)$, do you need to assume that $g$ is three times differentiable? – Arctic Char Aug 09 '20 at 16:43
  • If you knew it, it would be straighforward that that holds. If you don't know it, you can make an $\varepsilon - \delta $ argument regarding $|g(x)-\frac{g''(0)}{2}x^2|$ to show that they are close so an analogous property to the former follows. Moreover, when your hypothesis says ''given $f$ differentiable'', how differentiable is that $f$? You may assume $f$ is smooth under that general assumption. – astro Aug 09 '20 at 16:48
  • I am asking since you claimed that you don't need more differentiability assumptions on $g$. From the question I suppose $f$ is differentiable, and $f'$ is differentiable at $x=0$. – Arctic Char Aug 09 '20 at 16:55
  • Is not a problem, take $f(x)=x^2 + \sqrt{x^5}$, it fulfills all the conditions and is not three times differentiable. Anyway, you should be a little more careful when you say differential so loosely, it could mean very different things in different contexts (for example, my function here is not ''differentiable'' in any open set containing zero, so there you go). – astro Aug 09 '20 at 16:59
  • Yes. Using your $f(x) = x^2 + x^{2.5}$, the corresponding $g$ does not satisfy $g(x) = \frac{g''(0)}{2} x^2 + o(x^3)$. Let's what I want to point out. – Arctic Char Aug 09 '20 at 17:09
  • The first part of this answers transforms the problem in the other I posted before, where I didn't need to use Taylor's expansion formula, otherwise I couldn't say that the reminder is o(x³) without assuming some conditions on the third derivative, which was the matter of this question. – Pier Francesco Peperoni Aug 09 '20 at 17:11
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    But again @ArcticChar , $f$ is not under hypothesis since it is not differentiable at zero. – astro Aug 09 '20 at 17:11
  • Exactly @PierFrancescoPeperoni , that is why I am saying\asking what OP means when he says that $f$ is ''differentiable''. – astro Aug 09 '20 at 17:14
  • @PierFrancescoPeperoni So in some sense this answer did not answer this question – Arctic Char Aug 09 '20 at 17:25
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    This answer gave an hint, from which I solved the problem, since it can be shown, without Taylor's expansion, that if g(0)=g'(0), and g"(0)>0, with g differentiable, then g(x)>0 for some x (which is my previous post). It is wrong in the second part (the Taylor expansion). – Pier Francesco Peperoni Aug 09 '20 at 17:48
  • I forgot g(0)=g'(0)=0 – Pier Francesco Peperoni Aug 09 '20 at 17:54
  • First of all I think you should give a precise meaning of '''differentiable'', does it mean infinitely differentiable? smooth? Looking at your request it looks like it is but again, is not really absolutely clear. On the other hand, you're both correct when you said $o(x^3)$ is wrong, I should have written $o(x^{2+\alpha})$ for some $\alpha >0$. In the case of $f(x)=\left{ \begin{array}{c} x^2 + \sqrt{x^5} \mbox{ if x \geq 0}\ x^2 + \sqrt{(-x)^5} \mbox{ if x<0} \end{array}\right.$ you have a full example holding for all positive $\alpha < 1/2$. – astro Aug 09 '20 at 18:29
  • A better aproach would be to use that, if you knew $ f \in C^2$, continuity of $f''$ and Lagrange's remainder theorem with the equality $g(x)=r_2(x)$ for the remainder yields the property in a better shape than any of the above. – astro Aug 09 '20 at 18:34
  • By Taylor's theorem with Peano's form of remainder your answer needs the hypotheses that $g''(0)$ exists (equivalently $f''(0)$ exists). Need not worry at all. +1 – Paramanand Singh Aug 09 '20 at 20:23
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    @astro you can't assume $f$ is smooth. –  Aug 10 '20 at 02:44
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    By differentiable I just meant that the first derivative exists everywhere, with no statement about the higher order derivatives. – Pier Francesco Peperoni Aug 10 '20 at 08:35
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At $x=0$, $f(2x)$ and $2f(x)$ are both $0$ and equal.

At $x=0$, their derivative are both $2f'(0)$ and equal.

At $x=0$, their derivative is increasing as $f"(0)>0$.

The double derivative of the former is $4f''(0)$, is larger than that of the latter, $2f''(0)$.

Hence at a point "close" to 0, the derivative of the former increases faster than the latter, and as they both start from the same point, the former has a value larger than the latter for a point "close" to 0.

To be more "rigorous", you may follow the same procedure by defining a function $g(x)=f(2x)-2f(x)$ and show that it is positive at some point(but I would prefer the more logical approach as above :) )

user600016
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I'll assume the following: $f$ is differentiable in a neighborhood of $0,$ and $f''(0)>0.$ We do not need the existence of $f''$ anywhere else and we certainly don't need the existence of $f'''$ anywhere.

From Taylor we then have

$$f(x) = 0+ f'(0)x + f''(0)x^2/2 +o(x^2).$$

Thus

$$f(2x)= 2f'(0)x + 2f''(0)x^2 + o(x^2),$$

whereas

$$2f(x) = 2f'(0)x +f''(0)x^2 + o(x^2).$$

Subtracting, we get

$$f(2x) - f(2x) = f''(0)x^2 + + o(x^2).$$

Because $f''(0)>0,$ the last expression will be positive for small nonzero $x.$

zhw.
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  • Don't you need f'''(x) to be bounded around 0 to say that the reminder in Taylor's expansion is o(x²)? – Pier Francesco Peperoni Aug 10 '20 at 08:27
  • No you don't. Lemma: Suppose $g(0)=g'(0)=g''(0)=0.$ Then $g(x)=o(x^2).$ Proof: MVT appled twice.// So if $f$ is differentiable near $0$ and $f''0)$ exists, apply the lemma to $g(x) = f(x)-(f(0)+f'(0)x+f''(0)x^2/2).$ – zhw. Aug 10 '20 at 14:06