This is a question in Treves.

Suppose $a>1$ and $\tau \in \mathbb R $,

(i) show that for all $(\tau, \xi) \in \mathbb R^{n+1}$, $|(\tau-ia)^2 - |\xi|^2| \ge(\tau ^2+|\xi|^2+a^2)^{1/2}$

(ii) show that if $k>n/2$,

$\iint [(\tau-ia)^2-|\xi|^2]^{-1}(1+a^2+(\tau-ia)^2+|\xi|^2)^{-k}\,d {\tau}\, d {\xi}\quad$

is bounded by a constant independent of $a>1$

I've figured out the first question, and I guess one has to use (i) to prove (ii), but I really don't have an idea of integrating such a double integral. Everything seems to be messy to me. I just don't dare to try, and actually I don't know where to start. This really embarrassed me. Can anyone give me some ideas? I'll really appreciate it.


You could perhaps begin with $a>1$ which gives $$|(\tau-ia)^2+|\xi|^2 +1+a^2|> (\tau^2+|\xi|^2+1)^{1/2}$$ implying $$|(\tau-ia)^2+|\xi|^2 +1+a^2|^k> (\tau^2+|\xi|^2+1)^{k/2}$$ and using (i) $$ |(\tau-ia)^2-|\xi|^2|^{-1}|1+a^2+(\tau-ia)^2+|\xi|^2|^{-k} <(\tau^2+|\xi|^2+1)^{-(k+1)/2}. $$ The latter function has a bounded integral over $\mathbb R^{n+1}$ taking into account the condition on $k$ ...

  • $\begingroup$ Thank you again for helping me... I feel so sorry that I got stuck every time on inequalities of this simple. Could you tell me how to be sensitive to this? From what direction I can train myself? I'm so embarrassed. $\endgroup$ – user167839 Aug 4 '14 at 18:05

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