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Rudin 5.15

  • Thread starter ehrenfest
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  • #1
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Homework Statement


Suppose [itex]a \in \mathbb{R}[/itex], f is a twice-differentiable real function on (a, \infinty) and M_0,M_1,M_2 are the least upper bound of [itex]|f(x)|,|f'(x)|,|f''(x)|[/itex], respectively on (a,\infinity). Prove that

[tex]M_1^2\leq 4 M_0 M_2[/tex]


Homework Equations





The Attempt at a Solution


That is equivalent to showing that M_2 x^2 +M_1 x +M_0=0 has a real solution.

I was trying to use Taylor's Theorem which says that if \alpha and \beta are distinct points in (a,\infinity) then there exists x between \alpha and \beta that makes the following equation true:

[tex]f(\beta) = f(\alpha) + f'(\alpha)(\beta-\alpha) + f''(x) (\beta-\alpha)^2/2[/tex]

I could take the absolute value of both sides and then use triangle inequality but I did not see how to get anywhere with that.
 
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Answers and Replies

  • #2
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Did you look at the hint given in rudin?
 
  • #3
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How was I supposed to know there was a hint on the next page? That is a serious question.
 
  • #4
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OK. The hint says:

If h>0, Taylor's theorem show that

[tex]f'(x) = \frac{1}{2h}[f(x+2h)-f(x)]-hf''(\xi)[/tex]

for some [itex]\xi \in (x,x+2h)[/itex]. Hence

[tex]|f'(x)| \leq hM_2+M_0/h[/tex]

I don't see how the last inequality is useful (even if you square it).
 
  • #5
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Square it and remember that it holds for ANY h > 0. There is a particular choice of h that will yield the result.
 

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