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For a fixed value of $a<\frac{1}{4}$ prove that

$\ |x|^{2a} < c_1(a) \frac{1+|x|}{1+|x|^{1-2a}}$

holds for all $x\in\mathbb R$

Similarly show that for fixed $a<\frac{1}{2}$

$\ |x|^{2a} < c_2(a) \frac{1+x^2}{1+|x|^{2-2a}}$

holds for all $x\in\mathbb R$

"i have encountered this inequality in existence result of fully nonlinear evolutionary Navier Stokes Equations..Exactly at : Navier Stokes Equations Theory and Numerical Analysis by Roger Temam pages;277,286. Thanks for your interest indeed." ($c_1(a)$ and $c_2(a)$ are constants depending on a)

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Did Nov 5 '12 at 16:41
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Do you mean $x\in\mathbb R$ rather than $a\in\Re$? –  Andres Caicedo Nov 5 '12 at 16:41
    
Andres: $\mathfrak{R}$ is a (not un)common notation for the real numbers. The OP almost certainly meant $x$ instead of $a$, though. –  Cameron Buie Nov 5 '12 at 17:28
    
I don't want to be rude of course .. i have used only mathematical term..i have encountered this inequality in existence result of fully nonlinear evolutionary Navier Stokes Equations..Exactly at : Navier Stokes Equations Theory and Numerical Analysis by Roger Temam pages;277,286. Thanks for your interest indeed. –  noname Nov 5 '12 at 17:28
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It seems the proposer and the solver know what is $c_1(a)$, but I don't. –  GEdgar Nov 5 '12 at 17:54
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1 Answer 1

up vote 0 down vote accepted

$$\ |x|^{2a} < c_1(a) \frac{1+|x|}{1+|x|^{1-2a}} \Leftrightarrow \ |x|^{2a} (1+|x|^{1-2a}) < c_1(a) ( 1+|x|)$$

$$\Leftrightarrow \frac{\ |x|^{2a} +|x|}{1+ |x|} < c_1(a) $$

Since $\lim_{x \to \pm \infty} \frac{\ |x|^{2a} +|x|}{1+ |x|} =1$, the inequality

$$\frac{\ |x|^{2a} +|x|}{1+ |x|} <2$$

holds for all $|x| > x_0$, for some $x_0$.

The function $\frac{\ |x|^{2a} +|x|}{1+ |x|}$ is continuous on $[-x_0,x_0]$, thus has an absolute max M on this interval.

Thus we get

$$\frac{\ |x|^{2a} +|x|}{1+ |x|} <2 \,;\, \forall |x| > x_0$$ $$ \frac{\ |x|^{2a} +|x|}{1+ |x|} <M \,;\, \forall |x| \leq x_0$$

This shows that $c_1(a) =\max \{2,M \}$ works.

I think the second one is wrong as stated. Note that

$$\lim_{x \to \infty} \frac{ |x|^{2}+|x|^{4-2a}}{1+|x|^2} = \infty$$

If instead the inequality is

$$\ |x|^{2a} < c_2(a) \frac{1+x^2}{1+|x|^{2-2a}}$$

then it can be proven exactly as above.

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Thanks a lot.. you are right for the second inequality. when editing for x being real intead of a i have deleted it by mistake.. Thank you very much. it won't be possible for me to think absolute maximum in that interval. Thanks again.. –  noname Nov 5 '12 at 17:50
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