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Let $f:\mathbb{R}\to\mathbb{R}$ be twice differentiable with $$|f(x)|\leq M, |f''(x)|\leq M'',\forall x\in\mathbb{R}$$ Prove that $|f'(x)|\leq\sqrt{2MM''},\forall x\in\mathbb{R}$

I am thinking about using Taylor's theorem: For any $x\in\mathbb{R}$ and $a>0$, by Taylor's theorem $\exists \xi\in(x,x+a)$ s.t. $$f(x+a)=f(x)+f'(x)a+\frac{f''(\xi)}{2}a^2$$ Thus $$|f'(x)|\leq\frac{|f(x)|+|f(x+a)|}{a}+\frac{|f''(\xi)|}{2}a$$ However with this approach the best bound we can get is $$|f'(x)|\leq 2\sqrt{MM''}$$ Thus I feel that there is probably a completely different trick.

Kato yu
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3 Answers3

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A variant which is basically equivalent but with a slightly more geometric touch calculates a minimum bound for M in terms of $f'(x)$.

Assume without loss of generality that $f(a)\ge 0$ and $f'(a) > 0$.

For $x > a$ we find from integrating the lower bound of the second derivative that $$f'(x) \geq f'(a) - M''(x-a)$$ Integrating again from a to x, we get $$f(x) \geq f'(a)(x-a) - M'' \frac{(x-a)^2}{2} $$ This second degree polynomial has a maximum when $x-a = \frac{f'(a)}{M''} $ and the corresponding value is $\frac{f'(a)^2}{2M''}$, giving us $$M \geq f(x) \geq \frac{f'(a)^2}{2M''} $$ This is equivalent to the inequality that should be proved.

mlu
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Hint: You have \begin{equation*} f'(x) = \frac{f(x+a)-f(x)}{a} + \frac{a}2 f''(\xi). \end{equation*} Let us assume that $f'(x) \ge 0$. Then, \begin{align*} f'(x) & \le \frac{|f(x+a)|}{|a|}-\frac{f(x)}{a} + \frac{|a|}2 |f''(\xi)|. \\ & \le \frac{M}{|a|}-\frac{f(x)}{a} + \frac{|a|}2 M''. \end{align*} Now the trick is to choose the sign of $a$, such that $-\frac{f(x)}{a} \le 0$.

Edit: A function which attains this bound can be constructed as follows: Set $$f''(x) = \begin{cases}-1 & x \in [0,\sqrt{2})\\ -1 + (x - \sqrt{2}) & x \in [\sqrt{2}, \frac32\,\sqrt{2}+1) \\ \frac12 \sqrt{2} + (\frac32 \, \sqrt{2}+1-x) & x \in [\frac32 \, \sqrt{2} +1 , 2\sqrt{2}+1) \\ 0 & x \ge 2\sqrt2+1\end{cases}$$ Together with $f(x) = 0$ and $f'(x)=\sqrt2$ we find $f(x)$ for $x \ge 0$ and we set $f(-x ) = -f(x)$. Then, $M = M'' = 1$, but $f'(0) = \sqrt2$.

gerw
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  • How exactly does this work? – Mark Viola Dec 16 '15 at 20:45
  • Where is your problem? You choose $a = -\mathrm{sign}(f(x)) \sqrt{2 , M / M''}$. – gerw Dec 16 '15 at 21:20
  • Ah, there was a typo in the question. Also the comment should read $a = \mathrm{sign}(f(x)) , \sqrt{2 , M / M''}$. And of course, you use $-f(x)/a \le 0$. – gerw Dec 16 '15 at 21:36
  • That works now. Pleased I was able to help with the typo! - Mark – Mark Viola Dec 16 '15 at 21:39
  • @gerw do you know a function which proves that this bound is sharp? – mrprottolo Dec 17 '15 at 16:42
  • @mrprottolo: see the edit. – gerw Dec 17 '15 at 21:30
  • Sick example! Everything fits perfectly. May I ask if it is taken from some books/article or it's just a your construction? – mrprottolo Dec 18 '15 at 14:32
  • It was constructed only for you :) But the idea behind it is pretty simple: Choose $M = 1$ and $M''=1$. Now. we try to construct $f$ (within these bounds) starting from $f(0) = 0$ and $f'(0) = \sqrt{2}$ (to get the sharp estimate). Using $f''(x) = -1$, we can brake the growth. The stuff with $x > \sqrt{2}$ is only for the continuous second derivative. If you can live with a discontinuous $f''$, which is also not existant at $x = \sqrt{2}$, you can simply take $f''(x) = 0$ for $x > \sqrt{2}$. – gerw Dec 18 '15 at 18:58
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Edit: Oops, someone already said this in a comment. No harm in leaving this up - no votes, please.

Old trick: Consider

$$f(x+a)=f(x)+f'(x)a+\frac{f''(\xi_1)}{2}a^2$$

and

$$f(x-a)=f(x)-f'(x)a+\frac{f''(\xi_2)}{2}a^2.$$

Subtract:

$$|f'(x)|\le\frac Ma+\frac a2M''.$$

Let $a=\sqrt{2M/M''}$.

David C. Ullrich
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