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I am interested in understanding the solutions $\phi$ of the following integral equation: $$0=\int_0^1 \int_0^1 \phi(x,y)(x-y)\thinspace dx\thinspace dy.$$ Equivalently, I am interested in understanding the kernel of the linear function $F:C^k(\mathbb{R}^2)\to\mathbb{R}$ where $F(\phi)=\int_0^1 \int_0^1 \phi(x,y)(x-y)\thinspace dx\thinspace dy$. (The particular function space for the domain doesn't matter especially much to me - I have been assuming that $k\geq1$, but if it is convenient to use a different domain, then that will work for my purposes as well.)

So far I know that any symmetric function is a solution, i.e. any $\phi$ satisfying $\phi(x,y)=\phi(y,x).$ However, I also know that not every solution is symmetric, as illustrated by the solution $\phi(x,y)=x+y^2$.

I also know that the kernel of $F$ is of codimension 1, and so its dimension is countably infinite. My questions are:

  1. Is it possible to give a concise description of the family of solutions to this equation?
  2. Is it possible to explicitly find a basis for the kernel of $F$? I mean this in the sense of finding a countable collection of functions whose span is dense in the kernel.
  3. More generally, in the (limited) literature I've read on integral equations, I have not seen sources which deal with integral equations of two variables. Could you point me to a source where I can learn more about equations of this type? My intuition is that this equation is of a simple enough form that it seems like somebody must have studied it before. Does anybody know of a source that discusses this equation?

Thank you in advance!

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    $\begingroup$ Since your equation constrains only one single number, $\phi $ can be almost completely arbitrary. For example, choose any $f(x,y)$; then $\phi (x,y)=f(x,y)-xN$ is a solution, where $N=12\int_{0}^{1} \int_{0}^{1} f(x,y) (x-y) dx\, dy$. $\endgroup$ Jan 18, 2021 at 5:59
  • $\begingroup$ @MichaelEngelhardt: Hmm, I think the fact that "$\varphi$ can be almost completely arbitrary" that you mention, is also captured in the question where the OP mentions that the solution space has co-dimension one. $\endgroup$ Jan 18, 2021 at 9:27
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    $\begingroup$ The claim in the question that the dimension of $\ker F$ is countably infinite is not correct: every infinite dimensional Banach space has uncountable dimension (that's a consequence of Baire's theorem). $\endgroup$ Jan 18, 2021 at 9:34
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    $\begingroup$ @JochenGlueck - yes, you're right, I was just making it a bit more explicit by giving an example of one basis vector one could take out to project onto the kernel of $F$. $\endgroup$ Jan 18, 2021 at 15:23
  • $\begingroup$ Thank you both for your replies! @MichaelEngelhardt, thank you for that construction, that is useful. $\endgroup$
    – mwalth
    Jan 18, 2021 at 19:19

1 Answer 1

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I think it's easiest to work in the Hilbert space setting for this problem, i.e., to consider $F$ is a functional on the space $L^2([0,1]^2)$, where $[0,1]^2$ is endowed with the Lebesgue measure.

Let $1 \in L^2([0,1])$ denote the constant function with value $1$, and let $h \in L^2([0,1])$ be given by $h(x) = x$ for all $x \in [0,1]$.

Choose your favourite orthonormal basis $(e_n)_{n \in \mathbb{N}_0}$ of $L^2([0,1])$ with the property that $e_k$ is orthogonal to both $1$ and $h$ for each $k \ge 2$ (for instance, $(e_n)_{n \in \mathbb{N}_0}$ can be chosen as a basis of orthogonal polynomials with respect to the Lebesgue measure on $[0,1]$).

Then $(e_k \otimes e_j)_{(k,j) \in \mathbb{N}_0^2}$ is an orthonormal basis of $L^2([0,1]^2)$, and it is easy to see that the following vectors are in $\ker F$:

  • $e_k \otimes e_j$ whenver $j \ge 2$ or $k \ge 2$.

  • $e_0 \otimes e_0$ and $e_1 \otimes e_1$.

  • $\frac{1}{\sqrt{2}}\big(e_0 \otimes e_1 + e_1 \otimes e_0\big)$.

Hence, these vectors form an orthonormal basis of $\ker F$.

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    $\begingroup$ I suppose by "$j,k\geq 2$" you mean "$j$ or $k$", not "$j$ and $k$". $\endgroup$ Jan 18, 2021 at 15:28
  • $\begingroup$ @MichaelEngelhardt: You're right, of course; corrected. Thank you for noting that! $\endgroup$ Jan 18, 2021 at 16:03
  • $\begingroup$ This is an excellent construction, and exactly what I was looking for. Thank you for your help! $\endgroup$
    – mwalth
    Jan 18, 2021 at 19:23

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