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Suppose we are given the following inequality, which holds for any real number $s$ and $x,y \in \mathbb{R}^n$:

\begin{equation} (1 + |x + y|)^s \leq (1 + |x|)^s(1 + |y|)^{|s|} \end{equation}

(This is sometimes called Peetre's Inequality). How can I use this to show that, for a given integer $k$ and real number $d$, and $\xi,\eta \in \mathbb{R}^n$

\begin{equation} (1 + |\xi|)^{d}(1 + |\xi - \eta|)^{-k}(1 + |\eta|)^{-k} \leq (1 + |\xi|)^{|d|-k}(1 + |\eta|)^{|d| - k} \end{equation}

This is stated in some notes I am currently studying to learn about Fourier Analysis. I cannot fill in the details, I always end up with more terms than there ought to be .. any help would be great !!

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    I guess we will apply it in an integration, so you will be able to deduce what he/she meant.2012-04-22

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Since $(1+|\eta|)^{-k}$ is a common factor, we just have to show that $(1+|\xi|)^d(1+|\xi-\eta|)^{-k}\leq (1+|\xi|)^{|d|-k}(1+|\eta|)^{|d|-k}.$ We can write $(1+|\xi|)^d(1+|\xi-\eta|)^{-k}=(1+|\xi-\eta+\eta|)^d(1+|\xi-\eta|)^{-k}$ and since $(1+|\xi-\eta+\eta|)^d\leq (1+|\xi-\eta|)^d(1+|\eta|)^{|d|}$ we get $(1+|\xi|)^d(1+|\xi-\eta|)^{-k}\leq (1+|\xi-\eta|)^d(1+|\eta|)^{|d|}(1+|\xi-\eta|)^{-k}$ and finally $(1+|\xi|)^d(1+|\xi-\eta|)^{-k}\leq (1+|\xi-\eta|)^{|d|-k}(1+|\eta|)^{|d|}.$

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    Thanks for your answer ! I am not sure, maybe I miss the last step but what I need is the first factor of the right hand side to be $(1 + |\xi|)^{|d| - k}$ instead of $(1 + |\xi - \eta|)^{|d| - k}$2012-04-22