While reading a paper Hengang Li and Weiping Yan - Asymptotic stability and instability of explicit self-similar waves for a class of nonlinear shallow water equations, I experienced that my calculation results kept differing from the author's calculation results.
The authors of the paper seem to believe that the following equation holds: $$\int_\mathbb{R} xf'(x) \Lambda^{2s}f dx= -\int_\mathbb{R} (\Lambda^s f)^2 dx-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]' dx$$ where $s>3$ is a constant and $\Lambda^s f:=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}$, $\hat{f}(\xi):=\int_\mathbb{R}f(x) e^{-2\pi i\xi x}dx$ (Fourier transform) and $\check{f}(x):=\int_\mathbb{R} f(\xi)e^{2\pi i \xi x}dx$ (Inverse Fourier transform). $f$ is in Schwartz space $S(\mathbb{R})$.
However, my result is different with above. Following all lemmas and theorem are just my opinions.
Lemma 1. $\Lambda^s f=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}$.
Proof) This holds because of change of variable $\xi \rightarrow -\xi$. $$\Lambda^s f(x)=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\Bigl{(}\int_\mathbb{R}f(y)e^{-2\pi i\xi y}dy\Bigr{)}d\xi$$ $$=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{-2\pi i\xi x}\Bigl{(}\int_\mathbb{R}f(y)e^{2\pi\xi y}dy\Bigr{)} d\xi=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}.\qquad\blacksquare$$
Lemma 2. $\int_\mathbb{R}\Lambda^sg\Lambda^sfdx=\int_\mathbb{R}g\Lambda^{2s}(f)dx$.
Proof) This results from weak Parseval's theorem i.e. $$\forall f,g\in S(\mathbb{R})\ \ \ \ \int_\mathbb{R}\hat{f}gdx=\int_\mathbb{R}f\hat{g}dx.$$ If we use weak Parseval's theorem, \begin{gather*} \int_\mathbb{R}\Lambda^s{f}\Lambda^s{g}dx=\int_\mathbb{R}[(1+\xi^2)^{\frac{s}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}\check{g}]^{\wedge}dx=\int_\mathbb{R}\hat{f}(1+\xi^2)^s\check{g}d\xi=\int_\mathbb{R}[(1+\xi^2)^s\hat{f}]^{\vee}(\check{g})^{\wedge}dx \\ =\int_\mathbb{R}g\Lambda^{2s}f dx.\qquad \blacksquare \end{gather*}
Lemma 3. $\frac{d}{dx}(\Lambda^sf)(x)=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}$. In other words, the linear operators $\Lambda^s$ and $\frac{d}{dx}$ commute.
Proof) In this proof, we use the well-known fact about Fourier transforms $$2\pi i\xi \hat{f}(\xi)=(\frac{d}{dx}f)^{\wedge}(\xi),\ \ \ \frac{d}{d\xi}\hat{f}(\xi)=(-2\pi i) (xf(x))^\wedge(\xi).$$
If we use the above fact, $$\Lambda^s\Bigl{(}\frac{df(x)}{dx}\Bigr{)}=[(1+\xi^2)^{\frac{s}{2}}(f')^\wedge]^{\vee}=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}.$$ Then, $$\frac{d}{dx}(\Lambda^sf)(x)=\frac{d}{dx}\int_{\mathbb{R}}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\hat{f}(\xi)d\xi=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}=\Lambda^s\Bigl{(}\frac{df(x)}{dx}\Bigr{)}.\qquad\blacksquare$$
Theorem $\int_\mathbb{R}xf'\Lambda^{2s}fdx=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx$ holds.
If we use integration by parts and Lemma 3 and Lemma 2 \begin{gather*} \int_\mathbb{R}xf'(x)\Lambda^{2s} f dx=\Bigl{[}xf(x)\Lambda^{2s} f\Bigr{]}_{-\infty}^\infty-\int_\mathbb{R}xf(x)(\Lambda^{2s}f)'dx-\int_\mathbb{R}f(x)\Lambda^{2s}fdx \\ =-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx-\int_\mathbb{R}(\Lambda^sf)^2dx. \end{gather*}
Meanwhile, if we use $(h')^{\vee}=-2\pi ix\check{h}$ \begin{gather*} x\Lambda^sf=x[(1+\xi^2)^{\frac{s}{2}}\hat{f}(\xi)]^{\vee}=\frac{i}{2\pi}[\frac{d}{d\xi}((1+\xi^2)^{\frac{s}{2}}\hat{f})]^{\vee} \\ =\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}+(1+\xi^2)^{\frac{s}{2}}\hat{f}'(\xi)]^{\vee} \\ =\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+[(1+\xi^2)^{\frac{s}{2}}(xf)^{\wedge}]^{\vee} \\ =\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+\Lambda^s(xf). \end{gather*} In other words, $$\Lambda^s(xf)=x\Lambda^sf-\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}.$$ Therefore, $$-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx=-\int_{\mathbb{R}}x\Lambda^sf\Lambda^s(f')dx +\int_\mathbb{R}\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx.$$ If we focus on the second term, $$\frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx =\frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}(f')^{\vee}]^{\wedge}dx.$$ If we use weak Parseval theorem and $(h')^{\vee}=-2\pi ix\check{h}$, \begin{gather*} =\int_\mathbb{R}s\xi^2(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi =\int_\mathbb{R}s(1+\xi^2)^s\hat{f}\check{f}d\xi-\int_\mathbb{R}s(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi \\ =\int_{\mathbb{R}}s(\Lambda^sf)^2dx-\int_{\mathbb{R}}s(\Lambda^{s-1}f)^2dx. \end{gather*}
Therefore, \begin{gather*} \int_\mathbb{R}xf'(x)\Lambda^{2s}fdx=-\int_\mathbb{R}(\Lambda^sf)^2dx-\int_\mathbb{R}x\Lambda^sf\Lambda^s{f'}dx+ \frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx =-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx.\qquad \blacksquare \end{gather*}
I wonder if the calculations of the paper authors are correct or if my calculations are correct. If anyone has any ideas for the calculations, any help would be greatly appreciated.
