**Setting**


Although this detail is not relevant to my question, let me set the problem that my question arise.


We are considering an initial value problem
\begin{align\*}
\begin{cases}
u\in L^\infty(I,H^{1}\_0)\cap W^{1,\infty}(I,H^{-1})\\
iu\_t+\Delta u+\lambda |u|^\alpha u =0 \\
u(0)=\varphi
\end{cases}
\end{align\*}
on an interval $I$ containing $0$.


In order to prove the uniqueness of this problem, we use the Duhamel formula. In other words, we need to estimate
$$u\_1-u\_2=i\lambda\int\_{0}^{t}\mathcal{T}(t-s)\left[|u\_1(s)|^\alpha u\_1(s)-|u\_2(s)|^\alpha u\_2(s) \right]ds$$


**Question**


The author says that one can easily verify the following inequality.


$$||u\_2|^\alpha u\_2-|u\_1|^\alpha u\_1|\leq C (|u\_1|^\alpha+|u\_2|^\alpha)|u\_2-u\_1|$$


And this inequality was the core part that allow us to prove the uniqueness of the weak solution. However, I am not the one who can easily prove the inequality. I have tried few days but still cannot figure it out.


I post this question hoping I can have answer from someone. Thanks in advance.



An inequality for uniqueness proof of NLS
I assume that $u\_1,u\_2$ are complex and $\alpha\geqslant 0$ is fixed. Denote $u\_1=a$, $u\_2=a+b$, $f(z)=|z|^\alpha z$, then $$f(a+b)-f(a)=\int\_0^1 \frac{d}{dt}f(a+bt)dt\\=b\int\_0^1 |a+bt|^\alpha+\alpha (a+bt)|a+bt|^{\alpha-1}\frac{d}{dt}|ab^{-1}+t|dt.$$
Since $a+bt=:c$ is a point on the segment $[a,b]$ we get $|c|\leqslant \max(|a|,|a+b|)$, using this and $|\frac{d}{dt}|ab^{-1}+t||\leqslant 1$ we get the necessary bound $$|f(a+b)-f(a)|\leqslant C(\alpha)|b|\max(|a|^\alpha,|a+b|^\alpha).$$


