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Let $f$ be twice differentiable on an interval $I$,Let

$$M_{0}=\sup_{x\in I}{|f(x)|},M_{1}=\sup_{x\in I}{|f'(x)|},M_{2}=\sup_{x\in I}{|f''(x)|}$$

show that

(a):$$M_{1}\le 2\sqrt{M_{0}M_{2}}$$ if the length of $I$ is not less than $2\sqrt{\dfrac{M_{0}}{M_{2}}}$

(b):the numbers $2$ and $\sqrt{2}$ (in part a) cannot be replaced by smaller numbers.

My try:for part $(a)$ I can prove if the length of $I$ is not less than $4\sqrt{\dfrac{M_{0}}{M_{2}}}$

my proof:

$$f(c)-f(x)=f'(x)(c-x)+\dfrac{f''(x+\theta_{1}(c-x))}{2}(c-x)^2$$ let $c-x=h$ then we have $$f'(x)=\dfrac{f(c)-f(x)}{h}-\dfrac{f''(x+\theta_{1}h)}{2}h$$ Thus $$f'(x)\le\dfrac{2M_{0}}{h}+\dfrac{1}{2}M_{2}h$$ Now taking $h=2\sqrt{\dfrac{M_{0}}{M_{2}}}$, which in turn implies that $$M_{1}\le2\sqrt{M_{0}M_{2}}$$ and $$c=x+h=2\sqrt{\dfrac{M_{0}}{M_{2}}}+x>2\sqrt{\dfrac{M_{0}}{M_{2}}}$$ so the length of $I$ is $2c$ and not less than $4\sqrt{\dfrac{M_{0}}{M_{2}}}$

so Now How part(a)?

and for part(b) How prove it? and How take example for the best numbers$\sqrt{2}$?

By the way

This problem is from Mathematical Analysis I(Zorich )Page 233 problem 9.

Thank you evryone

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1  
This problem statement is missing crucial information compared to the book (which can be previewed on Google Books). For example, $I$ is assumed to be a closed interval or $\mathbb R$ in the book. Which version of the exercise do you want solved? Please fix the problem statement here. –  Ayman Hourieh Sep 28 '13 at 9:33
    
Hi, china math. As commented by Ayman Hourieh, the statement of your question is different from the original question in the book. For your question, given part (a), as shown in my answer below, the correct statement of part (b) should be "the number 2 in part (a) cannot be replaced by a smaller number". Could you please edit your question accordingly? –  23rd Oct 1 '13 at 17:22

2 Answers 2

up vote 4 down vote accepted
+50

(a) We should assume that $M_2>0$. Denote the length of $I$ by $|I|$ and denote $$l=2\sqrt{\frac{M_0}{M_2}}.$$ If $|I|\ge l$, then for any $x\in I$, there exist $h,k\ge 0$, such that $x+h,x-k\in I$ and $h+k=l$. As shown in your argument, using Taylor's expansion, we have: $$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2\tag{1}$$ and
$$f(x-k)=f(x)-f'(x)k+\frac{f''(\zeta)}{2}k^2\tag{2},$$ where $x\le\xi\le x+h$ and $x-k\le\zeta\le x$. Noting that $h^2+k^2\le l^2$, $(1)-(2)$ yields $$|f'(x)|\cdot l\le |f(x+h)|+|f(x-k)|+\frac{1}{2}\left(|f''(\xi)|h^2+|f''(\zeta)|k^2\right)\le 2M_0+\frac{M_2}{2}l^2,$$ and the conclusion follows.


(b) The best constant in (a) should be $2$. That is to say,

(i) there exists $f:I\to\Bbb R$, such that $|I|=l=2\sqrt{\frac{M_0}{M_2}}$ and $M_1=2\sqrt{M_0M_2}$;

(ii) for every $0<\lambda<1$, there exists $f:I\to\Bbb R$, such that $I\ge \lambda l$ and $M_1>2\sqrt{M_0M_2}$.

Example: For any $a\in [0,1)$, let $$f(x)=2x^2-1-a^2,\quad I=[a,1].$$ Then $|I|=1-a$, $M_0=1-a^2$, $M_1=4$, $M_2=4$, $l=\sqrt{1-a^2}$ and $2\sqrt{M_0M_2}=4\sqrt{1-a^2}$.

(i) When $a=0$, $|I|=l=1$, and $M_1=2\sqrt{M_0M_2}=4$.

(ii) Given $\lambda\in (0,1)$, let $a=\frac{1-\lambda^2}{1+\lambda^2}\in (0,1)$. For this $a$, $|I|=\lambda l$ and $M_1=4>2\sqrt{M_0M_2}$.

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Let's assume that $f\geq0$. Then $$ f(x+t)=f(x)+f'(x)t+f''(\xi)t^2/2\geq0. $$ In particular, the points where the equation (for $t$) vanishes are $$ t=\frac{-f'\pm\sqrt{(f')^2-4ff''}}{2f}. $$ From here we have $$ (f')^2\leq 4ff''\rightarrow f'\leq 2\sqrt{ff''}. $$ Here the length of the interval plays no role. I think that this inequality is called Landau inequality. I hope this helps.

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Thank you,But I have see this if the length of I is not less than $$2\sqrt{\dfrac{M_{0}}{M_{2}}}$$ –  china math Sep 23 '13 at 6:02
    
I mean that the lenght of the interval plays a role due to the absolut value present in $M_0$ (otherwise any length gives you the same inequality). I'm still trying to figure out why... –  guacho Sep 23 '13 at 6:19
    
Thank you,So I consider somedays,But I don't know this is why. –  china math Sep 25 '13 at 11:49

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