While reading a paper, 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 hold because of change of variable $\xi \rightarrow -\xi$. $$\Lambda^s f(x)=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{-2\pi i\xi y}dy\Big{)}d\xi$$ $$=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{-2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{2\pi\xi y}dy\Big{)} d\xi=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}\blacksquare$$

Lemma 2. $\int_\mathbb{R}\Lambda^sg\Lambda^sfdx=\int_\mathbb{R}g\Lambda^{2s}(f)dx$.

proof) This is results derived 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, $$\int_\mathbb{R}\Lambda^s{f}\Lambda^s{g}dx=\int_\mathbb{R}[(1+\xi^2)^{\frac{s}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}\check{g}]^{\wedge}dx=\int_\mathbb{R}\hat{f}(1+\xi^2)^s\check{g}d\xi=\int_\mathbb{R}[(1+\xi^2)^s\hat{f}]^{\vee}(\check{g})^{\wedge}dx$$ $$=\int_\mathbb{R}g\Lambda^{2s}f dx \blacksquare$$

Lemma 3. $\frac{d}{dx}(\Lambda^sf)(x)=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}$. In other words, Linear operator $\Lambda^s$ and $\frac{d}{dx}$ are commutative.

proof) In this proof, we use well-known facts about Fourier transform $$2\pi i\xi \hat{f}(\xi)=(\frac{d}{dx}f)^{\wedge}(\xi),\ \ \ \frac{d}{d\xi}\hat{f}(\xi)=(-2\pi i) (xf(x))^\wedge(\xi).$$

If we use above facts, $$\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}=[(1+\xi^2)^{\frac{s}{2}}(f')^\wedge]^{\vee}=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}$$ Then, $$\frac{d}{dx}(\Lambda^sf)(x)=\frac{d}{dx}\int_{\mathbb{R}}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\hat{f}(\xi)d\xi=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}\blacksquare$$

$\bf Theorem$ $\int_\mathbb{R}xf'\Lambda^{2s}fdx=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx$ holds

If we use integration by parts and Lemma 3 and Lemma 2 $$\int_\mathbb{R}xf'(x)\Lambda^{2s} f dx=\Big{[}xf(x)\Lambda^{2s} f\Big{]}_{-\infty}^\infty-\int_\mathbb{R}xf(x)(\Lambda^{2s}f)'dx-\int_\mathbb{R}f(x)\Lambda^{2s}fdx$$ $$=-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx-\int_\mathbb{R}(\Lambda^sf)^2dx$$

Meanwhile, if we use $(h')^{\vee}=-2\pi ix\check{h}$ $$x\Lambda^sf=x[(1+\xi^2)^{\frac{s}{2}}\hat{f}(\xi)]^{\vee}=\frac{i}{2\pi}[\frac{d}{d\xi}((1+\xi^2)^{\frac{s}{2}}\hat{f})]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}+(1+\xi^2)^{\frac{s}{2}}\hat{f}'(\xi)]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+[(1+\xi^2)^{\frac{s}{2}}(xf)^{\wedge}]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+\Lambda^s(xf)$$ 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 $(h')^{\vee}=-2\pi ix\check{h}$, $$=\int_\mathbb{R}s\xi^2(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi =\int_\mathbb{R}s(1+\xi^2)^s\hat{f}\check{f}d\xi-\int_\mathbb{R}s(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi$$ $$=\int_{\mathbb{R}}s(\Lambda^sf)^2dx-\int_{\mathbb{R}}s(\Lambda^{s-1}f)^2dx$$

Therefore, $$\int_\mathbb{R}xf'(x)\Lambda^{2s}fdx=-\int_\mathbb{R}(\Lambda^sf)^2dx-\int_\mathbb{R}x\Lambda^sf\Lambda^s{f'}dx+ \frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx$$ $$=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx \blacksquare$$

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.

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.

While reading a paper, 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 hold because of change of variable $\xi \rightarrow -\xi$. $$\Lambda^s f(x)=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{-2\pi i\xi y}dy\Big{)}d\xi$$ $$=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{-2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{2\pi\xi y}dy\Big{)} d\xi=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}\blacksquare$$

Lemma 2. $\int_\mathbb{R}\Lambda^sg\Lambda^sfdx=\int_\mathbb{R}g\Lambda^{2s}(f)dx$.

proof) This is results derived 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, $$\int_\mathbb{R}\Lambda^s{f}\Lambda^s{g}dx=\int_\mathbb{R}[(1+\xi^2)^{\frac{s}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}\check{g}]^{\wedge}dx=\int_\mathbb{R}\hat{f}(1+\xi^2)^s\check{g}d\xi=\int_\mathbb{R}[(1+\xi^2)^s\hat{f}]^{\vee}(\check{g})^{\wedge}dx$$ $$=\int_\mathbb{R}g\Lambda^{2s}f dx \blacksquare$$

Lemma 3. $\frac{d}{dx}(\Lambda^sf)(x)=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}$. In other words, Linear operator $\Lambda^s$ and $\frac{d}{dx}$ are commutative.

proof) In this proof, we use well-known facts about Fourier transform $$2\pi i\xi \hat{f}(\xi)=(\frac{d}{dx}f)^{\wedge}(\xi),\ \ \ \frac{d}{d\xi}\hat{f}(\xi)=(-2\pi i) (xf(x))^\wedge(\xi).$$

If we use above facts, $$\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}=[(1+\xi^2)^{\frac{s}{2}}(f')^\wedge]^{\vee}=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}$$ Then, $$\frac{d}{dx}(\Lambda^sf)(x)=\frac{d}{dx}\int_{\mathbb{R}}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\hat{f}(\xi)d\xi=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}\blacksquare$$

$\bf Theorem$ $\int_\mathbb{R}xf'\Lambda^{2s}fdx=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+s\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx$ holds

If we use integration by parts and Lemma 3 $$\int_\mathbb{R}xf'(x)\Lambda^{2s} f dx=\Big{[}xf(x)\Lambda^{2s} f\Big{]}_{-\infty}^\infty-\int_\mathbb{R}xf(x)(\Lambda^{2s}f)'dx=-\int_\mathbb{R}xf(x)\Lambda^{2s}(f')dx$$

If we use Lemma 2 above is same with $$=-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx$$ Meanwhile, if we use $(h')^{\vee}=-2\pi ix\check{h}$ $$x\Lambda^sf=x[(1+\xi^2)^{\frac{s}{2}}\hat{f}(\xi)]^{\vee}=\frac{i}{2\pi}[\frac{d}{d\xi}((1+\xi^2)^{\frac{s}{2}}\hat{f})]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}+(1+\xi^2)^{\frac{s}{2}}\hat{f}'(\xi)]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+[(1+\xi^2)^{\frac{s}{2}}(xf)^{\wedge}]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+\Lambda^s(xf)$$ 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 $(h')^{\vee}=-2\pi ix\check{h}$, $$=\int_\mathbb{R}s\xi^2(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi =\int_\mathbb{R}s(1+\xi^2)^s\hat{f}\check{f}d\xi-\int_\mathbb{R}s(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi$$ $$=\int_{\mathbb{R}}s(\Lambda^sf)^2dx-\int_{\mathbb{R}}s(\Lambda^{s-1}f)^2dx$$

Therefore, $$\int_\mathbb{R}xf'(x)\Lambda^{2s}fdx=-\int_\mathbb{R}x\Lambda^sf\Lambda^s{f'}dx+ \frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx$$ $$=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+s\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx \blacksquare$$

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.

While reading a paper, 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 hold because of change of variable $\xi \rightarrow -\xi$. $$\Lambda^s f(x)=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{-2\pi i\xi y}dy\Big{)}d\xi$$ $$=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{-2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{2\pi\xi y}dy\Big{)} d\xi=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}\blacksquare$$

Lemma 2. $\int_\mathbb{R}\Lambda^sg\Lambda^sfdx=\int_\mathbb{R}g\Lambda^{2s}(f)dx$.

proof) This is results derived 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, $$\int_\mathbb{R}\Lambda^s{f}\Lambda^s{g}dx=\int_\mathbb{R}[(1+\xi^2)^{\frac{s}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}\check{g}]^{\wedge}dx=\int_\mathbb{R}\hat{f}(1+\xi^2)^s\check{g}d\xi=\int_\mathbb{R}[(1+\xi^2)^s\hat{f}]^{\vee}(\check{g})^{\wedge}dx$$ $$=\int_\mathbb{R}g\Lambda^{2s}f dx \blacksquare$$

Lemma 3. $\frac{d}{dx}(\Lambda^sf)(x)=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}$. In other words, Linear operator $\Lambda^s$ and $\frac{d}{dx}$ are commutative.

proof) In this proof, we use well-known facts about Fourier transform $$2\pi i\xi \hat{f}(\xi)=(\frac{d}{dx}f)^{\wedge}(\xi),\ \ \ \frac{d}{d\xi}\hat{f}(\xi)=(-2\pi i) (xf(x))^\wedge(\xi).$$

If we use above facts, $$\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}=[(1+\xi^2)^{\frac{s}{2}}(f')^\wedge]^{\vee}=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}$$ Then, $$\frac{d}{dx}(\Lambda^sf)(x)=\frac{d}{dx}\int_{\mathbb{R}}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\hat{f}(\xi)d\xi=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}\blacksquare$$

$\bf Theorem$ $\int_\mathbb{R}xf'\Lambda^{2s}fdx=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx$ holds

If we use integration by parts and Lemma 3 and Lemma 2 $$\int_\mathbb{R}xf'(x)\Lambda^{2s} f dx=\Big{[}xf(x)\Lambda^{2s} f\Big{]}_{-\infty}^\infty-\int_\mathbb{R}xf(x)(\Lambda^{2s}f)'dx-\int_\mathbb{R}f(x)\Lambda^{2s}fdx$$ $$=-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx-\int_\mathbb{R}(\Lambda^sf)^2dx$$

Meanwhile, if we use $(h')^{\vee}=-2\pi ix\check{h}$ $$x\Lambda^sf=x[(1+\xi^2)^{\frac{s}{2}}\hat{f}(\xi)]^{\vee}=\frac{i}{2\pi}[\frac{d}{d\xi}((1+\xi^2)^{\frac{s}{2}}\hat{f})]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}+(1+\xi^2)^{\frac{s}{2}}\hat{f}'(\xi)]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+[(1+\xi^2)^{\frac{s}{2}}(xf)^{\wedge}]^{\vee}$$ $$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+\Lambda^s(xf)$$ 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 $(h')^{\vee}=-2\pi ix\check{h}$, $$=\int_\mathbb{R}s\xi^2(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi =\int_\mathbb{R}s(1+\xi^2)^s\hat{f}\check{f}d\xi-\int_\mathbb{R}s(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi$$ $$=\int_{\mathbb{R}}s(\Lambda^sf)^2dx-\int_{\mathbb{R}}s(\Lambda^{s-1}f)^2dx$$

Therefore, $$\int_\mathbb{R}xf'(x)\Lambda^{2s}fdx=-\int_\mathbb{R}(\Lambda^sf)^2dx-\int_\mathbb{R}x\Lambda^sf\Lambda^s{f'}dx+ \frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx$$ $$=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int_\mathbb{R}(\Lambda^sf)^2dx -s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx \blacksquare$$

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.