I am reading Sobolev space by using Fourier transform approach. Here I have some questions that treated to be "obvious" by textbook but I can not understand it.
We define operator $\Lambda^s:=(I-\triangle )^{s}$ where $\triangle$ denotes the Laplace operator. Then the author write that $(\Lambda^su)^\wedge(\xi)=(1+|\xi|^2)^s\hat{u}(\xi)$ where $\wedge$ denotes the Fourier transform. I can compute myself that if $s$ is an integer, let's use $N$, then $(\Lambda^Nu)^\wedge(\xi)=(1+|\xi|^2)^N\hat{u}(\xi)$ by just repeating the fact that $\triangle u^\wedge (\xi)=-|\xi|^2 \hat(u)(\xi)$. But I don't understand the case $s$ is non-interger.
I need to compute $(\Lambda^su\Lambda^{-s}v)^\wedge(\xi)$. Here for simplification let's assume $u$ $v$ are $C_c^\infty(R^N)$. First of all, my professor says that the Fourier transform of $uv$ is $\hat{u}\ast\hat{v}$, i.e., the convolution. I search this online but only in Wiki I found this result without prove. Can somebody print out where I can find a prove of this? I tried it myself but i can't figure it out. Secondly, my textbook gives that $$ (\Lambda^su\Lambda^{-s}v)^\wedge(\xi) = \int_{R^n}(1+|\xi|^2)^s\hat{u}(\eta-\xi)(1+|\eta|^2)^{-s}\hat{v}(\eta)d\eta$$ but I am keeping get $$ (\Lambda^su\Lambda^{-s}v)^\wedge(\xi) = \int_{R^n}(1+|\eta-\xi|^2)^s\hat{u}(\eta-\xi)(1+|\eta|^2)^{-s}\hat{v}(\eta)d\eta$$ which just comes from the definition of convolution...Can you please explain to me why they don't have $(1+|\eta-\xi|^2)^s$ but just $(1+|\xi|^2)^s$? It is a huge deference...
Thx!
Assuming that your Fourier transform is defined by $\mathcal{F}(\phi) = \int_{\Bbb R^n} \phi(x) e^{-i\bf{x}\cdot \bf{\xi}}\, d\bf{\xi}$, it is by definition that for non-integer $s$, $(1 - \triangle)^s$ is the operator with Fourier multiplier $(1 + |\xi|^2)^s$ (if the Fourier integral had kernel $e^{-2\pi i\bf{x}\cdot\bf{\xi}}$ instead of $e^{-i\bf{x}\cdot\bf{\xi}}$, then the Fourier multiplier of $(1 - \triangle)^s$ would be defined as $(1 + 4\pi^2|\xi|^2)^s$). This is motivated by the calculation you made for integer $s$. For $1 < p < \infty$ and $s\in \Bbb R$, we can define the Sobolev space $H^{s,p}(\Bbb R^n)$ as the set of all $f\in L^p(\Bbb R^n)$ such that $(1 - \triangle)^{s/2}f\in L^p(\Bbb R^n)$, and set $\|f\|_{H^{s,p}(\Bbb R^n)} = \|(1 - \triangle)^{s/2}f\|_{L^p(\Bbb R^n)}$.
It's actually $(u * v)^{\wedge} = \hat{u}\hat{v}$, not $(uv)^{\wedge} = \hat{u} * \hat{v}$. In any case, the textbook answer is incorrect and your answer is correct up to a factor; there should also be a factor of $(2\pi)^{-n}$ in front of the integral. I'll show that
\begin{equation} (uv)^{\wedge}(\xi) = (2\pi)^{-n}\int_{\Bbb R^n} \hat{u}(\eta - \xi)\hat{v}(\eta)\, d\eta \end{equation}
Starting with the formula $\hat{u}\hat{v} = (u * v)^{\wedge}$, replace $u$ by $\mathcal{F}^{-1}(u)$ and $v$ by $\mathcal{F}^{-1}(v)$ to get
\begin{equation} uv = (\mathcal{F}^{-1}(u) * \mathcal{F}^{-1}(v))^{\wedge} \end{equation}
By the Fourier inversion formula, $\mathcal{F}^{-1} = (2\pi)^{-n} \mathcal{F}R$, where $R(g)(\xi) := g(-\xi)$. In particular $\mathcal{F}\mathcal{F} = (2\pi)^nR$. Therefore
\begin{align} (uv)^{\wedge}(\xi) &= (2\pi)^{-n}R(\mathcal{F}R(u) * \mathcal{F}R(v))(\xi) = (2\pi)^{-n}R\int_{\Bbb R^n} \hat{u}( - \eta) \hat{v}(-\xi -\eta)\, d\eta\\ &= (2\pi)^{-n} \int_{\Bbb R^n} \hat{u}( - \eta) \hat{v}(\xi -\eta)\, d\eta\\ &= (2\pi)^{-n}\int_{\Bbb R^n} \hat{u}(\eta - \xi) \hat{v}(\eta)\, d\eta. \end{align}
Using this formula, we obtain
\begin{equation} (\Lambda^su \Lambda^{-s}v)^{\wedge}(\xi) = (2\pi)^{-n}\int_{\Bbb R^n} (1 + |\eta - \xi|^2)^s \hat{u}(\eta - \xi) (1 + |\eta|^2)^{-s}\hat{v}(\eta)\, d\eta. \end{equation}