Mixing
A Question about Even Harmonic Functions and Fractional Laplacian
$begingroup$
I'm reading something with the following statement:
In the complex plane cut along the negative axis, the function $z rightarrow z^{1/2}$ is analytic, hence its real part is harmonic. Moreover, because it is even in $y$, its $y$-derivative on the positive axis vanishes; this means that the half-Laplacian of $mathcal{R}(z^{1/2}) = sqrt{x}$ is $0$ on $mathbb{R}_+$
Here the half-Laplacian is defined as:
$$-Delta^{1/2}u(x)=C,left(text{PV}int_{mathbb{R}^n}frac{u(x)-u(y)}{|x-y|^{n+1}}right)$$
where $C$ is some constant.
I'm having trouble with the statment in bold:
- How is $mathcal{R}(z^{1/2})$ even in $y$?
- Why does that imply the derivative is positive on the positive $x$ axis?
- Why is the half laplacian $0$ then?
complex-analysis harmonic-analysis
asked Jan 23 at 20:55
yoshiyoshi
1,242917
$endgroup$
add a comment |
$begingroup$
I'm reading something with the following statement:
In the complex plane cut along the negative axis, the function $z rightarrow z^{1/2}$ is analytic, hence its real part is harmonic. Moreover, because it is even in $y$, its $y$-derivative on the positive axis vanishes; this means that the half-Laplacian of $mathcal{R}(z^{1/2}) = sqrt{x}$ is $0$ on $mathbb{R}_+$
Here the half-Laplacian is defined as:
$$-Delta^{1/2}u(x)=C,left(text{PV}int_{mathbb{R}^n}frac{u(x)-u(y)}{|x-y|^{n+1}}right)$$
where $C$ is some constant.
I'm having trouble with the statment in bold:
- How is $mathcal{R}(z^{1/2})$ even in $y$?
- Why does that imply the derivative is positive on the positive $x$ axis?
- Why is the half laplacian $0$ then?
complex-analysis harmonic-analysis
asked Jan 23 at 20:55
yoshiyoshi
1,242917
$endgroup$
add a comment |
$begingroup$
I'm reading something with the following statement:
In the complex plane cut along the negative axis, the function $z rightarrow z^{1/2}$ is analytic, hence its real part is harmonic. Moreover, because it is even in $y$, its $y$-derivative on the positive axis vanishes; this means that the half-Laplacian of $mathcal{R}(z^{1/2}) = sqrt{x}$ is $0$ on $mathbb{R}_+$
Here the half-Laplacian is defined as:
$$-Delta^{1/2}u(x)=C,left(text{PV}int_{mathbb{R}^n}frac{u(x)-u(y)}{|x-y|^{n+1}}right)$$
where $C$ is some constant.
I'm having trouble with the statment in bold:
- How is $mathcal{R}(z^{1/2})$ even in $y$?
- Why does that imply the derivative is positive on the positive $x$ axis?
- Why is the half laplacian $0$ then?
complex-analysis harmonic-analysis
asked Jan 23 at 20:55
yoshiyoshi
1,242917
$endgroup$
I'm reading something with the following statement:
In the complex plane cut along the negative axis, the function $z rightarrow z^{1/2}$ is analytic, hence its real part is harmonic. Moreover, because it is even in $y$, its $y$-derivative on the positive axis vanishes; this means that the half-Laplacian of $mathcal{R}(z^{1/2}) = sqrt{x}$ is $0$ on $mathbb{R}_+$
Here the half-Laplacian is defined as:
$$-Delta^{1/2}u(x)=C,left(text{PV}int_{mathbb{R}^n}frac{u(x)-u(y)}{|x-y|^{n+1}}right)$$
where $C$ is some constant.
I'm having trouble with the statment in bold:
- How is $mathcal{R}(z^{1/2})$ even in $y$?
- Why does that imply the derivative is positive on the positive $x$ axis?
- Why is the half laplacian $0$ then?
complex-analysis harmonic-analysis
complex-analysis harmonic-analysis
asked Jan 23 at 20:55
yoshiyoshi
1,242917
asked Jan 23 at 20:55
yoshiyoshi
1,242917
asked Jan 23 at 20:55
yoshiyoshi
1,242917
asked Jan 23 at 20:55
yoshiyoshi
1,242917
asked Jan 23 at 20:55
yoshiyoshi
1,242917
1,242917
add a comment |
add a comment |
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