# Integral equation $f_i(x) = \sum_j (u_{ij} \ast f_j)(x)$, where $u_{ij}$ are tempered distributions, using FT

Consider the following "generalization" to the integral equations, where I am interested in solving for functions $$f_1, f_2, ..., f_n$$, and $$u_{ij}$$ are some known tempered distributions. $$f_i(x) = \sum \limits_{j=1}^n (u_{ij} \ast f_j)(x)$$ To make this well defined, assume that $$u_{ij}$$ are tempered distributions, $$f_i$$ are in Schwartz space $$S(\mathbb{R}^n)$$, and the equation is to be satisfied for all $$x \in \mathbb{R}^n$$. I am interested in how it can (or cannot) be solved using Fourier transforms. This is the reason why I assume that $$u_{ij}$$ are tempered.

Let me provide the approach I would take to solve for functions $$f_i$$ and then I finish with a question on where I get stuck.

As $$f_i \in S(\mathbb{R}^n)$$, its Fourier transform is well defined by an integral. If $$u_{ij}$$ were Schwartz functions as well, then the Fourier transform of both sides of the integral equation would give the following, where I use $$\widehat{f \ast g} = \hat{f} \cdot \hat{g}$$.

$$\hat{f_i}(k) = \sum \limits_{j=1}^n \widehat{u_{ij} \ast f_j}(k) = \sum \limits_{j=1}^n \widehat{u_{ij}}(k) \cdot \hat{f_j}(k)$$

Then, I would just solve this system of linear equations for $$\hat{f_i}(k)$$ and then compute the inverse transform as $$\hat{f_i} \in S(\mathbb{R}^n)$$.

The problem when $$u_{ij}$$ are tempered distributions is that $$\widehat{u_{ij}}(k)$$ makes no sense in the usual function sense. So, to solve the original equation, I consider weak form which is produced when taking inner product with some test function $$\phi \in S(\mathbb{R}^n)$$. Here, to simplify, I first use that inner product is linear and the distributional definition of convolution with a Schwartz function, where $$\tilde{f_j}(x) = f_j(-x)$$.

$$\langle f_i, \phi \rangle = \langle \sum \limits_{j=1}^n u_{ij} \ast f_j, \phi \rangle = \sum \limits_{j=1}^n \langle u_{ij} \ast f_j, \phi \rangle = \sum \limits_{j=1}^n \langle u_{ij}, \tilde{f_j} \ast \phi \rangle$$

Now, take (distributional) Fourier transform which means assume that the test function is already a Fourier transform, $$\hat{\phi} \in S(\mathbb{R}^n)$$. As $$f_i \in S(\mathbb{R}^n)$$, we can use the usual identity that $$\langle f_i, \hat{\phi} \rangle = \langle \hat{f_i}, \phi \rangle$$. However, to transform the right side use that by properties of inverse transform, $$\tilde{f_j} = \hat{\hat{f_j}}$$ and we can use that $$\hat{f} \ast \hat{g} = \widehat{f \cdot g}$$. Finally, use definition of distributional Fourier transform. So, we get the following.

$$\langle u_{ij}, \tilde{f_j} \ast \hat{\phi} \rangle = \langle u_{ij}, \hat{\hat{f_j}} \ast \hat{\phi} \rangle = \langle u_{ij}, \widehat{\hat{f_j} \cdot \phi} \rangle = \langle \widehat{u_{ij}}, \hat{f_j} \cdot \phi \rangle$$

So, calculations can be combined as follows, where in the last step we also use definition of multiplication of a tempered distribution with Schwartz function.

$$\langle \hat{f_i}, \phi \rangle = \sum \limits_{j=1}^n \langle \widehat{u_{ij}}, \hat{f_j} \cdot \phi \rangle = \sum \limits_{j=1}^n \langle \widehat{u_{ij}} \cdot \hat{f_j}, \phi \rangle$$

Question 1: Are all of the computations so far correct/reasonable?

Now, to get information about the function $$\hat{f_i}(k)$$, my intuition is that we might consider family of test functions $$\{ \phi_n \}_{n=1}^{\infty}$$ such that they converge in distributions to delta distribution at $$k$$. In that case, I believe that I can make sense of it and prove that $$\langle \hat{f_i}, \phi_n \rangle \to \hat{f_i}(k)$$ as $$n \to \infty$$ by taking some scaled approximation of identity and just computing the limit directly by hand by u-substitution in the integrals.

Question 2: How to make sense of limit of the right hand side of the integral equation as test functions tend to delta distribution? How can one argue that such limit exists? If it does exist, how to compute it (in a sense that is there a simplified result, for example, if $$u_{ij}$$ were Schwartz functions, we would have $$\widehat{u_{ij}}(k) \cdot \hat{f_j}(k)$$ on the right hand side)?

You don't really need this Dirac approximation I think. You have proven that :

$$\langle \hat{f_i}, \phi \rangle = \langle \sum_{j = 1}^n \hat{u_{ij}} \hat{f_j}, \phi \rangle$$

for every Schwartz function $$\phi$$. This means that the equality :

$$\hat{f_i} = \sum_{j = 1}^n \hat{u_{ij}} \hat{f_j}$$

holds in $$\mathcal{S}'$$ (the tempered distributions).

Your computations are correct : indeed, you can also prove them by approximating $$u_{ij}$$ in $$\mathcal{S}'$$ by Schwartz functions. For Schwartz functions, the equality is easy to prove as you noticed, and since the Fourier transform is continuous you have the desired result.

On the other hand, there is no reason for $$\hat{u_{ij}}(k)$$ to have a meaning when $$k$$ is fixed and $$u_{ij}$$ is simply a tempered distribution. (For instance, $$\hat{u_{ij}}$$ could be a Dirac mass.)

You can note that your system reads : $$(\hat{U} - I) \hat{F} = 0$$ (with $$I$$ the "identity") and that $$f_i \equiv 0$$ for every $$i$$ is always a solution. There might be no other, for instance when $$\hat{u_{ij}} = 2 \delta_{ij}$$ (a constant function, equal to $$1$$ when $$i = j$$, $$0$$ if $$i \neq j$$) ; everything might be solution (when $$\hat{u_{ij}} = \delta_{ij}$$) ; or something in between (for instance, when $$\hat{u_{ij}} = \delta_{ij}$$ when $$|x| \leq 1$$, $$\hat{u_{ij}} = 2 \delta_{ij}$$ when $$|x| > 1$$, then the solutions are all the functions $$f_i$$ such that the support of $$\hat{f_i}$$ is in $$B_1$$).

Given all these possibilities, it seems difficult to give the general form of the solutions unless you have more information about $$(u_{ij})$$.

If you know that $$\hat{u_{ij}}$$ is a function for every $$i, j$$, however (for instance if $$u_{ij}$$ is also a Schwartz function, or even $$L^2$$) then you can consider the equality in Fourier at every point (or almost every) and compute the solutions this way. But you should also verify at the end that the solutions you compute this way are also Schwartz.