Mixing
Fourier Restriction: extension operator identity
$begingroup$
Consider the extension operator:
$$
Eg(x)=int_S g(xi)e^{2pi i xcdot xi}dsigma(xi).
$$
For simplicity we consider the 2D-case, where $S$ is the paraboloid $ximapsto xi^2$, $xiin [-1,1]$. (We will frequently use an abuse of notation: $g(xi_1)=g(xi_1,xi_1^2)=g(xi)$, where $xi_1in [-1,1]$ and $xiin S$.)
We know $EE^*f=f*widehat {dsigma}$. But what is $E^*E$? I claim that $E^*E=I$, and here is my formal computation.
begin{align*}
Eg(x)
&=int_{-1}^1 g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_1^2)}J(xi_1)dxi_1\
&=int_{-1}^1 int_{mathbb R} delta_0(xi_1^2-xi_2)g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_2)}J(xi_1)dxi_2dxi_1.
end{align*}
Hence, formally,
$$
widehat {Eg}(xi_1,xi_2)=g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)
$$
Thus $widehat {Eg}$ is a distribution in $mathbb R^2$ supported on the surface $S$, in the sense that if $hin C_c^infty(mathbb R^2)$, then
begin{align*}
widehat {Eg}(h)
&=int_{mathbb R^2} g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)h(xi_1,xi_2)dxi_2dxi_1\
&=int_{mathbb R} g(xi_1) J(xi_1)h(xi_1,xi_1^2)dxi_1\
&=int_{S} g(xi)h(xi)dsigma(xi).
end{align*}
Then $E^*Eg$ is the restriction of $widehat {Eg}$ to $S$, so $E^*Eg$ is exactly the function $g$ defined on $S$. In other words, $E^*E=I$.
The place I am not sure of is the last line, because I do not know in general, how I could restrict a distribution to a submanifold.
distribution-theory harmonic-analysis fourier-restriction
asked Jan 29 at 6:12
Dong LiDong Li
795413
$endgroup$
add a comment |
$begingroup$
Consider the extension operator:
$$
Eg(x)=int_S g(xi)e^{2pi i xcdot xi}dsigma(xi).
$$
For simplicity we consider the 2D-case, where $S$ is the paraboloid $ximapsto xi^2$, $xiin [-1,1]$. (We will frequently use an abuse of notation: $g(xi_1)=g(xi_1,xi_1^2)=g(xi)$, where $xi_1in [-1,1]$ and $xiin S$.)
We know $EE^*f=f*widehat {dsigma}$. But what is $E^*E$? I claim that $E^*E=I$, and here is my formal computation.
begin{align*}
Eg(x)
&=int_{-1}^1 g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_1^2)}J(xi_1)dxi_1\
&=int_{-1}^1 int_{mathbb R} delta_0(xi_1^2-xi_2)g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_2)}J(xi_1)dxi_2dxi_1.
end{align*}
Hence, formally,
$$
widehat {Eg}(xi_1,xi_2)=g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)
$$
Thus $widehat {Eg}$ is a distribution in $mathbb R^2$ supported on the surface $S$, in the sense that if $hin C_c^infty(mathbb R^2)$, then
begin{align*}
widehat {Eg}(h)
&=int_{mathbb R^2} g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)h(xi_1,xi_2)dxi_2dxi_1\
&=int_{mathbb R} g(xi_1) J(xi_1)h(xi_1,xi_1^2)dxi_1\
&=int_{S} g(xi)h(xi)dsigma(xi).
end{align*}
Then $E^*Eg$ is the restriction of $widehat {Eg}$ to $S$, so $E^*Eg$ is exactly the function $g$ defined on $S$. In other words, $E^*E=I$.
The place I am not sure of is the last line, because I do not know in general, how I could restrict a distribution to a submanifold.
distribution-theory harmonic-analysis fourier-restriction
asked Jan 29 at 6:12
Dong LiDong Li
795413
$endgroup$
add a comment |
$begingroup$
Consider the extension operator:
$$
Eg(x)=int_S g(xi)e^{2pi i xcdot xi}dsigma(xi).
$$
For simplicity we consider the 2D-case, where $S$ is the paraboloid $ximapsto xi^2$, $xiin [-1,1]$. (We will frequently use an abuse of notation: $g(xi_1)=g(xi_1,xi_1^2)=g(xi)$, where $xi_1in [-1,1]$ and $xiin S$.)
We know $EE^*f=f*widehat {dsigma}$. But what is $E^*E$? I claim that $E^*E=I$, and here is my formal computation.
begin{align*}
Eg(x)
&=int_{-1}^1 g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_1^2)}J(xi_1)dxi_1\
&=int_{-1}^1 int_{mathbb R} delta_0(xi_1^2-xi_2)g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_2)}J(xi_1)dxi_2dxi_1.
end{align*}
Hence, formally,
$$
widehat {Eg}(xi_1,xi_2)=g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)
$$
Thus $widehat {Eg}$ is a distribution in $mathbb R^2$ supported on the surface $S$, in the sense that if $hin C_c^infty(mathbb R^2)$, then
begin{align*}
widehat {Eg}(h)
&=int_{mathbb R^2} g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)h(xi_1,xi_2)dxi_2dxi_1\
&=int_{mathbb R} g(xi_1) J(xi_1)h(xi_1,xi_1^2)dxi_1\
&=int_{S} g(xi)h(xi)dsigma(xi).
end{align*}
Then $E^*Eg$ is the restriction of $widehat {Eg}$ to $S$, so $E^*Eg$ is exactly the function $g$ defined on $S$. In other words, $E^*E=I$.
The place I am not sure of is the last line, because I do not know in general, how I could restrict a distribution to a submanifold.
distribution-theory harmonic-analysis fourier-restriction
asked Jan 29 at 6:12
Dong LiDong Li
795413
$endgroup$
Consider the extension operator:
$$
Eg(x)=int_S g(xi)e^{2pi i xcdot xi}dsigma(xi).
$$
For simplicity we consider the 2D-case, where $S$ is the paraboloid $ximapsto xi^2$, $xiin [-1,1]$. (We will frequently use an abuse of notation: $g(xi_1)=g(xi_1,xi_1^2)=g(xi)$, where $xi_1in [-1,1]$ and $xiin S$.)
We know $EE^*f=f*widehat {dsigma}$. But what is $E^*E$? I claim that $E^*E=I$, and here is my formal computation.
begin{align*}
Eg(x)
&=int_{-1}^1 g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_1^2)}J(xi_1)dxi_1\
&=int_{-1}^1 int_{mathbb R} delta_0(xi_1^2-xi_2)g(xi_1)e^{2pi i (x_1,x_2)cdot (xi_1,xi_2)}J(xi_1)dxi_2dxi_1.
end{align*}
Hence, formally,
$$
widehat {Eg}(xi_1,xi_2)=g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)
$$
Thus $widehat {Eg}$ is a distribution in $mathbb R^2$ supported on the surface $S$, in the sense that if $hin C_c^infty(mathbb R^2)$, then
begin{align*}
widehat {Eg}(h)
&=int_{mathbb R^2} g(xi_1)J(xi_1)delta_0(xi^2_1-xi_2)h(xi_1,xi_2)dxi_2dxi_1\
&=int_{mathbb R} g(xi_1) J(xi_1)h(xi_1,xi_1^2)dxi_1\
&=int_{S} g(xi)h(xi)dsigma(xi).
end{align*}
Then $E^*Eg$ is the restriction of $widehat {Eg}$ to $S$, so $E^*Eg$ is exactly the function $g$ defined on $S$. In other words, $E^*E=I$.
The place I am not sure of is the last line, because I do not know in general, how I could restrict a distribution to a submanifold.
distribution-theory harmonic-analysis fourier-restriction
distribution-theory harmonic-analysis fourier-restriction
asked Jan 29 at 6:12
Dong LiDong Li
795413
asked Jan 29 at 6:12
Dong LiDong Li
795413
asked Jan 29 at 6:12
Dong LiDong Li
795413
asked Jan 29 at 6:12
Dong LiDong Li
795413
asked Jan 29 at 6:12
Dong LiDong Li
795413
795413
add a comment |
add a comment |
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