• Pere Callahan
In summary, PereIt appears that the eigenfunction for the adjoint operator with respect to the inner product is not unique for all possible measures.
Pere Callahan
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:

$$(Af)(x,y)=(\Delta f)(x,y)-f(x,y)+f(0,y)$$

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

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

For the case that $$\mu$$ has a density with respect to lesbesgue measure, i.e
$$d\mu(x,y)=g(x,y)d^2\lambda$$ I figured out that the adjoint, acting on $$\mu$$ and hence on the function g can be written as

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

However, the $$\delta_0(x)$$ makes me think that this is not quite the right approach

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

I further concluded that the "eigenmeasure" for $$A^*$$ which I am interested in, cannot have a density so it seems necessary to find out how $$A^*$$ 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 $$Z(x,y)d^2\lambda$$ in

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

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

-Pere

Last edited:
It maybe is a good idea to first consider the problem in only one dimension:

Here the operator is

$$(Af)(x)=(\partial_x^2f)(x)+f(0)-f(x)$$

Again I am interested in an eigenmeasure of the adjoint operator with respect to the same inner product as above (integration). I found

$$(A^*g)(x)=(\partial_x^2g)(x)+\delta_0(x)\int_\mathbb{R}{dt\, g(t)}-g(x)$$

where the function g represents the measure g(x)dx...How ever I have a strong feeling that this operator does not have an eigenfunction.. can somebody confirm this...?

Thanks

-Pere

Last edited:
So if nobody replies I do it myself

My last one dimensional DGL is obviously solved for $$x\neq 0$$ by

$$g(x)=A e^{-|x|}$$

(I only want finite measures (in fact probabilty measures) hence integrable functions.

How ever for x=0 this is not a solution, so something weird might happen there..any suggestions...maybe an Admin could move this into Differential Equations ... if it seems appropriate..

-Pere

edit:

I solved it again, this time via a Fourier Transform, and I changed my opinion concerning $$g(x)=A e^{-|x|}$$ not being a solution for x=0...the delta distribution seems to somehow compensate for the non-differentiability at x=0. So I'll try the same method for the original two-dimensional problem and ask again if it does not work out

Last edited:

## 1. What is an eigenfunction for an adjoint operator?

An eigenfunction for an adjoint operator is a function that, when acted upon by the adjoint operator, returns a scalar multiple of itself. In other words, the adjoint operator "eigenfunctions" are the functions that remain unchanged when operated on by the adjoint operator.

## 2. What is the significance of eigenfunctions for adjoint operators?

The eigenfunctions for adjoint operators play a crucial role in linear algebra and functional analysis. They allow us to decompose a complex function into simpler components, making it easier to study and analyze. In addition, the eigenfunctions provide a basis for the adjoint operator, allowing us to express any function as a linear combination of these basis functions.

## 3. How do you find eigenfunctions for an adjoint operator?

In order to find the eigenfunctions for an adjoint operator, we must first determine the adjoint operator itself. Then, we can use standard techniques such as solving for the characteristic equation or using the Gram-Schmidt process to find the eigenfunctions. In some cases, the eigenfunctions may also be obtained by inspection or by using specific properties of the adjoint operator.

## 4. Can an adjoint operator have multiple eigenfunctions?

Yes, an adjoint operator can have multiple eigenfunctions. In fact, most adjoint operators have an infinite number of eigenfunctions. This is because, in general, the adjoint operator operates on functions in an infinite-dimensional space, and there are an infinite number of possible functions that could be eigenfunctions for the operator.

## 5. What is the relationship between eigenfunctions and eigenvalues for an adjoint operator?

The eigenfunctions and eigenvalues for an adjoint operator are closely related. The eigenvalues represent the scalar multiples that the eigenfunctions are multiplied by when operated on by the adjoint operator. In other words, the eigenvalue is the "eigen" part of the eigenfunction. Without knowing the eigenvalues, we cannot fully understand the behavior and properties of the eigenfunctions for an adjoint operator.

• Calculus
Replies
2
Views
951
• Calculus
Replies
1
Views
951
• Calculus
Replies
5
Views
373
• Calculus
Replies
20
Views
2K
• Calculus
Replies
6
Views
895
• Calculus
Replies
4
Views
331
• Calculus
Replies
3
Views
1K
• Calculus
Replies
1
Views
918
• Calculus
Replies
1
Views
2K
• Calculus
Replies
3
Views
632