# General Procedure for Inverse of an Integral Transform

Is there a general inversion formula or procedure for an integral of the form (where f is the function being transformed and g depends on the type of transform) $\int^{a}_{b} f(x) g(x,\xi) dx$ ? Inverses are defined in the conventional ways for functionals and integral transforms, respectively.

For instance, for the fourier transform. In the equation above, $a=∞,b=-∞$, $g(x,\xi)=e^{-2πix \xi }$. I know the inverse fourier transform is simple but I am concerned with a general procedure or process.

I am most interested in the cases where $a,b=±∞$ although a simple inverse for $\int^{a}_{b} f(x) dx$ is also something I am curious about (as far as this part of the question is concerned, if the inverse of an indefinite integral is the derivative, what is the inverse of a definite integral-- I am sorry if this is too elementary).

Feel free to use complex analysis or any other branch of math if it helps to answer the question. Also, you can repost on another site if it will help.

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What do you mean by an "inversion formula or procedure"? The problem you pose is a natural infinite dimensional generalization of inverting a matrix. If there were an "inversion formula," that would be great, but... There is no such thing as an inverse for $\int_a^b f(x) dx$ (with fixed a and b), since a lot of functions have the same integral. –  Michael Renardy Apr 9 '12 at 22:29
This is a very interesting question, but to broad to have a real answer. A suggestion is that you formulate is as big-list question, then you have to make it community-wiki. –  Marc Palm Apr 11 '12 at 7:57

I think what Michael is saying in his comment is that the question is way too general: you are trying to deal with arbitrary linear operators on an infinite dimensional vector space, so invertibility (let alone a simple formula) is a subtle question that can't be answered in a simple fashion. You wonder about the inverse operation of definite integration over a fixed interval $I$; integration yields a real number (assuming the integral even exists!), so for $a\in\mathbb{R}$, "$\int_I^{-1}a =$ any function $f$ with $\int_If=a$", so to speak (sorry for the horrible notation). This is satisfied by loads of functions, so the inverse doesn't exist.

So if you have a specific operator in mind, you could try to see if it's invertible and if you can find a nice expression for the inverse. The general question doesn't make much sense.

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Probably a "general procedure" would be as follows: Find appropriate Hilbert spaces $X$ and $Y$ such that the operator $Lf(\xi) \int_a^b f(x) g(x,\xi) dx$ maps boundedly from $X$ to $Y$. Often the space $X = L^2([a,b])$ and $Y= L^2([c,d])$ works (e.g. if $g\in L^2([a,b]\times [c,d])$). Then there always exists the Moore-Penrose pseudoinverse of $L$. However, if the range of $L$ is not a closed subspace of $y$, it is an unbounded operator defined on $\text{range}L\oplus \text{range}L^\bot$ which is a dense subspace of $Y$.

This works for $Lf = \int_a^b f(x)dx$ as a mapping from $L^2([a,b])$ to $\mathbb{R}$ and it is a nice exercise to work out the pseudo-inverse.

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Have you got references for this nice exercise? If I'm doing this right it ought to be the linear transformation that takes $1$ to the constant function $f:x\mapsto\frac{1}{b-a}$. –  Emilio Pisanty Apr 11 '12 at 11:55
Sorry, $\frac{1}{\sqrt{b-a}}$. –  Emilio Pisanty Apr 11 '12 at 11:57

If we ignore the second part, I still think the first part has value. What is the procedure for inverting a general integral transform?

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As previous commenters have mentioned, the question is asking for too much. But perhaps the following rephrasing of the question, which is moving the difficult part to something slightly more concrete will be of use. Given integral operators A and B with kernels a and b, i.e. (Af)(x) = int_R a(x,y)f(y)dy and (Bf)(x) = int_R b(x,y)g(y)dy, their composition is the integral operator AB with kernel c(x,y) = int_R a(x,z)b(z,y)dz. So long as c(x,y) = delta(x-y), then B is a (right) inverse of A. –  Aaron Hoffman Apr 11 '12 at 1:58