General Procedure for Inverse of an Integral Transform - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:03:44Z http://mathoverflow.net/feeds/question/93595 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93595/general-procedure-for-inverse-of-an-integral-transform General Procedure for Inverse of an Integral Transform Karmon Euloid 2012-04-09T19:59:21Z 2012-04-11T07:56:22Z <p>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.</p> <p>For instance, for the fourier transform. In the equation above, $a=âˆž,b=-âˆž$, $g(x,\xi)=e^{-2Ï€ix \xi }$. I know the inverse fourier transform is simple but I am concerned with a general procedure or process.</p> <p>I am most interested in the cases where $a,b=±âˆž$ 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).</p> <p>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.</p> http://mathoverflow.net/questions/93595/general-procedure-for-inverse-of-an-integral-transform/93613#93613 Answer by Philip van Reeuwijk for General Procedure for Inverse of an Integral Transform Philip van Reeuwijk 2012-04-09T23:26:15Z 2012-04-09T23:26:15Z <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/93595/general-procedure-for-inverse-of-an-integral-transform/93718#93718 Answer by Karmon Euloid for General Procedure for Inverse of an Integral Transform Karmon Euloid 2012-04-10T23:50:31Z 2012-04-10T23:50:31Z <p>If we ignore the second part, I still think the first part has value. What is the procedure for inverting a general integral transform?</p> http://mathoverflow.net/questions/93595/general-procedure-for-inverse-of-an-integral-transform/93729#93729 Answer by Tom Copeland for General Procedure for Inverse of an Integral Transform Tom Copeland 2012-04-11T02:27:16Z 2012-04-11T02:27:16Z <p>Just to direct you towards the literature and help you clear up some of your thinking on the subject take a look at Wiki's <a href="http://en.wikipedia.org/wiki/Inverse_problem" rel="nofollow">Inverse Problem</a>. Then perhaps look at <strong>Methods of Applied Mathematics</strong> by Hildebrand and <strong>Principles and Techniques of Applied Mathematics</strong> by Friedman.</p> http://mathoverflow.net/questions/93595/general-procedure-for-inverse-of-an-integral-transform/93741#93741 Answer by Dirk for General Procedure for Inverse of an Integral Transform Dirk 2012-04-11T07:56:22Z 2012-04-11T07:56:22Z <p>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 <a href="http://en.wikipedia.org/wiki/Moore-Penrose_pseudoinverse" rel="nofollow">Moore-Penrose pseudoinverse</a> 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 $Y$.</p> <p>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.</p>