- #1

Pere Callahan

- 586

- 1

Hi,

I am having a question again (I should not work on so many things at the same time).

I came across an operator A acting on C^2 - functions f:R^2->R:

[tex](Af)(x,y)=(\Delta f)(x,y)-f(x,y)+f(0,y)[/tex]

My goal is to find the adjoint operator (acting on measures) with respect to the inner product

[tex]\left\langle f,\mu\right\rangle=\int_{\mathbb{R}^2}{f(x,y)d\mu(x,y)}[/tex]

For the case that [tex]\mu[/tex] has a density with respect to lesbesgue measure, i.e

[tex]d\mu(x,y)=g(x,y)d^2\lambda[/tex] I figured out that the adjoint, acting on [tex]\mu[/tex] and hence on the function g can be written as

[tex](A^*g)(x,y)=(\Delta g)(x,y)-g(x,y)+\delta_0(x)\int_\mathbb{R}{dt\, g(t,y)}[/tex]

However, the [tex]\delta_0(x)[/tex] makes me think that this is not quite the right approach

My interpreation for this is that [tex]A^*[/tex] acting on a "nice" measure (having a density) gives something which does not have a density.

I further concluded that the "eigenmeasure" for [tex]A^*[/tex] which I am interested in, cannot have a density so it seems necessary to find out how [tex]A^*[/tex] acts on for example a measure including a point mass at zero. I don't know how this can be done. More specifically, if we only consider the laplace part for now: What is [tex]Z(x,y)d^2\lambda[/tex] in

[tex]\int_{\mathbb{R}^2}{(\Delta f)(x,y)\delta_0(x)d^2\lambda(x,y)}=\int_{\mathbb{R}^2}{f(x,y)Z(x,y)d^2\lambda(x,y)}[/tex]

I would take this [tex]Z(x,y)d^2\lambda[/tex] to be [tex](\Delta^*\delta_0(x)d^2\lambda(x,y))(x,y)[/tex] if it happened to exist in some way or another...

Thanks for your help

-Pere

I am having a question again (I should not work on so many things at the same time).

I came across an operator A acting on C^2 - functions f:R^2->R:

[tex](Af)(x,y)=(\Delta f)(x,y)-f(x,y)+f(0,y)[/tex]

My goal is to find the adjoint operator (acting on measures) with respect to the inner product

[tex]\left\langle f,\mu\right\rangle=\int_{\mathbb{R}^2}{f(x,y)d\mu(x,y)}[/tex]

For the case that [tex]\mu[/tex] has a density with respect to lesbesgue measure, i.e

[tex]d\mu(x,y)=g(x,y)d^2\lambda[/tex] I figured out that the adjoint, acting on [tex]\mu[/tex] and hence on the function g can be written as

[tex](A^*g)(x,y)=(\Delta g)(x,y)-g(x,y)+\delta_0(x)\int_\mathbb{R}{dt\, g(t,y)}[/tex]

However, the [tex]\delta_0(x)[/tex] makes me think that this is not quite the right approach

My interpreation for this is that [tex]A^*[/tex] acting on a "nice" measure (having a density) gives something which does not have a density.

I further concluded that the "eigenmeasure" for [tex]A^*[/tex] which I am interested in, cannot have a density so it seems necessary to find out how [tex]A^*[/tex] acts on for example a measure including a point mass at zero. I don't know how this can be done. More specifically, if we only consider the laplace part for now: What is [tex]Z(x,y)d^2\lambda[/tex] in

[tex]\int_{\mathbb{R}^2}{(\Delta f)(x,y)\delta_0(x)d^2\lambda(x,y)}=\int_{\mathbb{R}^2}{f(x,y)Z(x,y)d^2\lambda(x,y)}[/tex]

I would take this [tex]Z(x,y)d^2\lambda[/tex] to be [tex](\Delta^*\delta_0(x)d^2\lambda(x,y))(x,y)[/tex] if it happened to exist in some way or another...

Thanks for your help

-Pere

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