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](https://doi.org/10.1016/j.cnsns.2019.104928), 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.



Question about calculation in Schwartz space
The key computation is the commutator $$ \Lambda^s (xf) - x \Lambda^s f. $$ You can check this "classically" in the case $s = 4$ to find $$ (1 - \Delta)^2 (xf) = x (1-\Delta)^2 f - 4 (1-\Delta) f'$$
which differs from was found in the paper (their computation dropped the final commutator term).


In fact, your computations imply the following result
$$ \Lambda^s(xf) = x\Lambda^s f - s \Lambda^{s-2} f' $$
from which you'd get (returning to the author's original term)
\begin{gather\*}
\int \Lambda^s f \Lambda^s(xf') = \int \Lambda^s f\Lambda^s(xf)' - \Lambda^s f \Lambda^s f = - \int \Lambda^s f' \Lambda^s (xf) - \int (\Lambda^s f)^2 \\
= - \frac12 \int x [(\Lambda^s f)^2]' + s \int (\Lambda^{s-1} f')^2 - \int (\Lambda^s f)^2 = s \int (\Lambda^{s-1} f')^2 - \frac12 \int (\Lambda^s f)^2.
\end{gather\*}


(The fact that the extra term can be written as something with a sign means that this cannot just disappear due to some strange cancellations.)


This may have an impact on the main results of the paper; I haven't looked too carefully, but it appears that in deriving the Bernoulli type differential inequality after equation (3.21), the authors relied on this (and similar terms) to be signed to drop them from consideration. The corrected expression adds a term with the wrong sign. There may be a way to absorb this, but it is not immediately obvious.


