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Let $f \in C^{2}(\mathbb{R})$. Then $$ ||f'||_{\infty} \leq 4 ||f||_{\infty} ||f''||_{\infty}$$

I tried to prove this using the generalized mean value theorem but couldn't. Also writing $f'$ in its limit definition and trying to use the taylor expansion of $f(x+h)$ was also an idea but didn't bring me to the end.

Thanks for any hints and proofs!

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    Using the Taylor expansion of $f(x+h)$ is the right idea here. Choose the value of $h$ that gets you the best possible upper bound for $f'$. – 2017-02-02
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    Are you assuming $f$, $f'$, and $f''$ are bounded? – 2017-02-02

2 Answers 2

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Here's a hint provided by Rudin's $\textit{Principles of Mathematical Analysis}$.

Hint: If $h>0$, Taylor's theorem shows that \begin{align} f'(x) = \frac{1}{2h}[f(x+2h)-f(x)]-hf''(\xi) \end{align} for some $\xi \in (x, x+2h)$. Hence \begin{align} |f'(x)| \leq h\|f''\|_\infty + \frac{\|f\|_\infty}{h}. \end{align} then set \begin{align} h = \sqrt{\frac{\|f\|_\infty}{\|f''\|_\infty}}. \end{align}

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    @JeanMarie You forgot the square root. – 2017-02-02
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    @JeanMarie what do you want me to elaborate more on when you say "I don't understand the global reasoning and deduction you do..."? – 2017-02-02
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    @JeanMarie where's your square root? – 2017-02-02
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    @JeanMarie It's a pointwise bound, what do you mean "its 2-norm". – 2017-02-02
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    I mean that sup $|f'(x)|^2$ is not $\|f'\|_{\infty}.$ There is a square that shouldn't be there. – 2017-02-02
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    @JeanMarie Unfortunately, the problem is actually wrong as stated by OP. – 2017-02-02
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    Rudin problem 5.15 – 2017-02-02
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    Sorry, I have not this book... But do you mean 1) That the inequality as stated by the OP has counterexamples and/or 2) That the inequality given in the book of Rudin is different ? – 2017-02-02
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    Besides, I erase my questions, note for erasing the trace of my successive misunderstandings, but because they pollute the site – 2017-02-02
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    The estimate is sharp according to the problem. – 2017-02-02
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    See the answer I have given. – 2017-02-02
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    Jackie, please, you should 1) Somewhat revise (at least the end of) your text because it proves something else that what is asked, knowing that what is asked is **erroneous**... 2) (if you agree) erase all the commentaries as I have done for most of them, because they will puzzle anybody who tries to understand what is going on. – 2017-02-02
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    Besides, I haven't understood what you mean when you say "the estimate is sharp according to the problem". – 2017-02-02
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    It would be desirable that you state in the body of your answer that the inequality as written in the question doesn't generally hold, not only in the comments. – 2017-02-02
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We assume, as remarked by @Jonas Meyer, that $f$ is such that $f$ and $f'$ are bounded on $\mathbb{R}.$

In fact, inequality

$$\tag{1}||f'||_{\infty} \leq 4 ||f||_{\infty} ||f''||_{\infty}$$

cannot hold for all $f \in C^2$ by lack of homogeneity.

Let us use "reductio ad absurdum". Let us assume there exist a non-zero function that fullfills (1) with $>0$ values on each side. (such a function exist, e.g., $f:x \mapsto 1/(1+x^2)).$

Then replacing $f$ by $\alpha f$ with $\alpha>0$, and cancelling $\alpha$, we have:

$$||f'||_{\infty} \leq 4 \alpha ||f||_{\infty} ||f''||_{\infty}$$

Taking $\alpha$ sufficiently small results in a contradiction.

The correct inequality should be:

$$\tag{2}||f'^2||_{\infty} \leq 4 ||f||_{\infty} ||f''||_{\infty}$$

as can be easily established by terminating the "proof" of Jacky Wong.

A direct consequence of (2) is that the boundedness of $f$ and $f''$ implies the boundedness of $f'^2$.