i'm currently trying to understand the proof of the following statement:
For any $y$, $z$ $\in$ $R^N$ with $N\leq1$,
\begin{align} \bigl| |z|^{p-2}z - |y|^{p-2}y \bigr | \leq \begin{cases} \gamma |z - y|^{p-1} \quad \quad \quad \quad \quad ~\text{for $1\leq p \leq 2$} \\ \gamma |z - y|~(|z| + |y|)^{p-2} \quad \text{for $p \geq 2$} \end{cases} \end{align}
and $\gamma > 0$.
At the moment im looking at the proof of the case $1 < p \leq 2$ which is stated exactly like this in the paper im looking at:
For $y = 0$ the inequality holds. For $y, z \not= 0$
$ \bigl| |z|^{p-2}z - |y|^{p-2}y \bigr |^2 \leq \bigl ||z|^{p-2}|z| - |y|^{p-2}|y| \bigr |^2 - 2 |z|^{p-2} |y|^{p-2} (|z||y| - zy) $
(i get this one... basicly just out multiplying, then plus and minus $2 |z|^{p-2}|z||y|^{p-2}|y|$ and then adjusting the terms cleverly)
$\leq (\bigl | |z| - |y| \bigr |^{p-1})^2 + 2 |z|^{p-1} |y|^{p-1} (1 - \frac{zy}{|z||y|})^{p-1} (1 - \frac{zy}{|z||y|})^{2-p}$
(this one i also understand: the first term we get because $t \rightarrow |t|^{p-1}$ is hölder continious for $1 \leq p \leq 2$ and the second we get by multiplying and dividing with $|z||y|$ and then cleverly adjust the terms... and because the exponents $p-1$ and $2-p$ are $1$ when added)
but now the estimate part, which i can't understand:
$ \leq |z - y|^{2(p-1)} + 2^{2(2-p)} |z - y|^{2(p-1)}$
the only things the author wrote to these lines/result are: "because of the hölder continuity and $|1 - \frac{zy}{|z||y|}|\leq 2$
unfortunaly i still can't get it...
why is $2 |z|^{p-1} |y|^{p-1} (1 - \frac{zy}{|z||y|})^{p-1} (1 - \frac{zy}{|z||y|})^{2-p} \leq 2^{2(2-p)} |z - y|^{2(p-1)}$?
i seem to be blind and miss something :/ would be nice, if someone more experienced could help me!