I need to prove the following inequality: for any $Z=(z_1,\dots,z_l)\in\mathbb{C}^l$ for any $p\geq 2$ and $l\geq 2$ \begin{equation} \left|\left|\sum_{j=1}^l z_j\right|^p-\sum_{j=1}^l\left|z_j\right|^p\right|\leq C_{p,l}\sum_{i\neq j}\left|z_i\right|\left|z_j\right|^{P-1}. \end{equation}
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$\begingroup$ Why do you "need" to do anything? -- Is this homework? $\endgroup$– Stefan Kohl ♦Commented Jan 19, 2014 at 18:59
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$\begingroup$ No it is not homework. I was trying to understand a paper of Cazenave were he proved a scattering property for focusing nonlinear Schrödinger equation. I found this inequality not trivial. $\endgroup$– Felice IandoliCommented Jan 19, 2014 at 19:40
1 Answer
Here is a somewhat silly way to do it. You need to prove that:
$$\frac{\left|\sum_{j=1}^l z_j\right|^p-\sum_{j=1}^l\left|z_j\right|^p}{\sum_{i\neq j}\left|z_i\right|\left|z_j\right|^{P-1}}$$
is bounded function on the unit sphere $S^{l-1}$. $S^{l-1}$ is compact, and this function is continuous away from the places where $z_i=\pm 1$, where the denominator is $0$. So it is sufficient to prove boundedness in a neighborhood of those points. Say $z_1$ is very close to $1$. Then the denominator is very close to $p\sum_{j=2}^l z_j$ and the numerator is very close to $\sum_{j=2}^l |z_j|$. So the ratio is bounded.