Prove $\sup \left| f'\left( x\right) \right| ^{2}\leqslant 4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right|$ [closed]

Let $f\left( x\right)$ be a $C^{2}$ function on $\mathbb{R}$. Show that $$\sup \left| f'\left( x\right) \right| ^{2}\leqslant4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| .$$

closed as off-topic by Andrés E. Caicedo, Amzoti, Adriano, Dominic Michaelis, Hanul JeonJul 25 '13 at 5:53

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• This is not true. Take $f(x)=1$ for all $x\in\mathbb{R}$ which is $C^2$. The LHS is $1$, but RHS is $0$. – Paul Dec 12 '12 at 11:25
• There is a typo. The LHS should be $|f'(x)|^2$. Rudin's PoMA has this as an exercise. – lhf Dec 12 '12 at 11:41
• I downvoted. It is very annoying to see questions in the imperative like this. – Giuseppe Negro Dec 12 '12 at 13:23

Let $\sup|f^{(n)}(x)|=M_n$. Since $f$ is $C^2$-continuous, we can use Taylor's theorem to write:

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}2f''(t)$$

for some $x<t<x+h$. Then we can solve for $f'(x)$:

$$f'(x)=\frac1h(f(x+h)-f(x))-\frac h2f''(t)$$ $$|f'(x)|\leq\frac1h(|f(x+h)|+|f(x)|)+\frac h2|f''(t)|\leq\frac2hM_0+\frac h2M_2.$$

Since we can apply this for any $h$, choose $h=2\sqrt{M_0/M_2}$, so that

$$|f'(x)|\leq\sqrt{M_2/M_0}\cdot M_0+\sqrt{M_0/M_2}\cdot M_2=2\sqrt{M_0M_2}.$$

Since this inequality is true for any $f'(x)$, it is true for the supremum, i.e.

$$M_1\leq2\sqrt{M_0M_2}\Rightarrow\sup|f'(x)|^2\leq4\sup|f(x)|\sup|f''(x)|.$$

• +1 Nice! I'm linking an old post of mine which uses essentially the same technique, for reference: math.stackexchange.com/a/44538/8157 – Giuseppe Negro Dec 12 '12 at 13:22
• Sweet solution! I think you have typo in the second last equation. It should be $+$ instead of $-$ between the terms. – Prism Jul 24 '13 at 20:28
• @Prism You're right; I fixed it. – Mario Carneiro Jul 25 '13 at 5:12
• The question is old and closed but it is online so I think is correct to write the following: The statement holds if we assume that $\sup|f''(x)|>0$, when you chose $h$ you assume this property. The statement doesn't hold for $f(x)=x$ which has $f''(x)=0$. – Cuoredicervo May 6 at 11:56
• @Cuoredicervo It's true that I assumed $M_2\ne 0$, but it's not hard to show that the theorem still holds in the case $M_2=0$ when $M_0$ is finite. Starting from the third line, we have $|f'(x)|\le\frac2hM_0+0$, so we let $h\to\infty$ and deduce $|f'(x)|=0$. Thus $f$ is a constant function and $M_1^2=0\le 4\cdot M_0M_2=0$. Your counterexample involves the indeterminate form $\infty \cdot 0$ for the product of the suprema. With this combination you can get whatever value you like for $M_1$. – Mario Carneiro May 6 at 12:09