1
$\begingroup$

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)

  • 0
    @GEdgar I think it is$a$constant only depending on 1 and $x$... :)2012-11-05

1 Answers 1

1

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

$\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|}

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.

  • 0
    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..2012-11-05