0
$\begingroup$

Suppose $2<p<\infty$ and $0<\theta<1$. Let $n\geq 1$ be an integer. Assume that $$ \frac{1}{p}=\frac{1-\theta}{2^n}+\frac{\theta}{2^{n+1}}. $$

How to prove the following inequality $$ (2^n-1)^{1-\theta}(2^{n+1}-1)^{\theta}\leq 2p-1\ ? $$

$\endgroup$
1
  • 1
    $\begingroup$ Sure you want both exponents to be $1-\theta$? $\endgroup$
    – martini
    Jul 16, 2014 at 14:40

1 Answer 1

2
$\begingroup$

First, if you clear denominators, you get $$2^{n+1}=2p(1-\theta)+p\theta=2p-p\theta.$$

Hence, $$2^{n+1}-1=2p(1-\theta)+p\theta-1=2p-p\theta-1\le2p-1.$$

Similarly, we can get $$ 2^n-1 \le 2^{n+1}-1\le 2p-1.$$

But if we have three real numbers $a,b,c$ all greater than or equal to 1 such that $a\le c$ and $b\le c,$ we know that $a^\theta b^{1-\theta}\le c^\theta c^{1-\theta}=c,$ so we are done. This is an interesting fact about convexity, in my opinion.

Let me know if you'd like me to clarify any of the steps.

$\endgroup$
1
  • 1
    $\begingroup$ Small typo: A -1 is missing in the second equation.Thank you! $\endgroup$
    – Zouba
    Jul 16, 2014 at 15:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .