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GabS
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Is it possible to calculate the $L^2$ norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=C_{N,s}\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

For example,If $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. My main goal is to estimate $u$ where $u$ may not belong to $D^{s,2}(\mathbb R^N)$ but is smooth. This can be done using a cut off function multiplied with $u.$ Any reference is welcome.

Is it possible to calculate the norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=C_{N,s}\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

For example, $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. Any reference is welcome.

Is it possible to calculate the $L^2$ norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=C_{N,s}\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

If $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. My main goal is to estimate $u$ where $u$ may not belong to $D^{s,2}(\mathbb R^N)$ but is smooth. This can be done using a cut off function multiplied with $u.$ Any reference is welcome.

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GabS
GabS
  • 407
  • 3
  • 11

Is it possible to calculate the norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=C_{N,s}\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

For example, $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. Any reference is welcome.

Is it possible to calculate the norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

For example, $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. Any reference is welcome.

Is it possible to calculate the norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=C_{N,s}\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

For example, $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. Any reference is welcome.

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GabS
GabS
  • 407
  • 3
  • 11

$L^2$ norm of fractional Laplacian

Is it possible to calculate the norm associated to fractional Laplacian of $u$ and $s\in (0, 1).$

$$\|(-\Delta)^{s/2} u\|_2^2=\int_{\mathbb R^N}|(-\Delta)^{s/2} u|^2dx=\int_{\mathbb R^{2N}} \frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}dxdy$$

For example, $u(x)=(1+ |x|^2)^{-\frac{N-2s}{2}},$ can we calculate $\|(-\Delta)^{s/2} u\|_2$ just by using the above formula. I am aware $u$ is the some type of solution to the fractional Yamabe problem but I want a result independent proof. Any reference is welcome.