Let $f\left( x\right)$ be a $C^{2}$ function on $\mathbb{R}$. Show that $$\sup \left| f'\left( x\right) \right| ^{2}\leqslant4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| .$$
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| $
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calculus
real-analysis
functions
inequality
derivatives
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1This is not true. Take $f(x)=1$ for all $x\in\mathbb{R}$ which is $C^2$. The LHS is $1$, but RHS is $0$. – 2012-12-12
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5There is a typo. The LHS should be $|f'(x)|^2$. Rudin's PoMA has this as an exercise. – 2012-12-12
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2I downvoted. It is very annoying to see questions in the imperative like this. – 2012-12-12
1 Answers
10
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
$$f'(x)=\frac1h(f(x+h)-f(x))-\frac h2f''(t)$$ $$|f'(x)|\leq\frac1h(|f(x+h)|+|f(x)|)+\frac h2|f''(t)|\leq\frac2hM_0+\frac h2M_2.$$
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.
$$M_1\leq2\sqrt{M_0M_2}\Rightarrow\sup|f'(x)|^2\leq4\sup|f(x)|\sup|f''(x)|.$$
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2+1 Nice! I'm linking an old post of mine which uses essentially the same technique, for reference: http://math.stackexchange.com/a/44538/8157 – 2012-12-12
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0Sweet solution! I think you have typo in the second last equation. It should be $+$ instead of $-$ between the terms. – 2013-07-24
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0@Prism You're right; I fixed it. – 2013-07-25