Lecture 59

Welcome back. In the last lecture, we posed ourselves a problem that how to reconstruct a function on R 3 from the data given by the slices. That means I have what we are going to have is that we have the average over all the planes of the function and we want to reconstruct the function. So, mathematically what we would like to say is that if P is a plane in R 3, we define Radon transform R f, where our f is a function in R 3 is a Schwarz class function we can take defined by R f. This is going to be a function on set of planes P, this we like to define P of f, that is the average of the function over the planes. Now, the first thing what we need to discuss that what do we mean by over a plane P. So, now any plane is going to determine by a vector on the unit sphere that is gives the direction and distance from the origin. So, we can define let gamma is in S 2, S 2 is the unit sphere in R 3. So, this is in R 3 such that x square plus y square plus z square this is equal to 1. And then you take t belongs to R, then with the plane which depends on t and this gamma, this is equal to all set of all x belongs to R 3 such that x dot gamma this is at a distance t. So, the unit vector is orthogonal to it and distance t from the origin. So, easily one can see that P of t gamma this is equal to P of minus t of minus of gamma, this by definition. Now, how does one represent write down the function element in this point in this plane. Now, choose E 1 and E 2 such that gamma E 1 and E 2 they are orthonormal bases for R 3. That means, they are lying on the unit sphere and each of them they are mutually orthogonal, their dot product is 0. So, now if I take x in P t gamma, then this is going to be x can be written as t gamma plus u 1 E 1 plus u 2 E 2. Now, for f in Schwarz space of R 3, we define integral P over a plane P t gamma f this to be equal to integral over R 2 f of t gamma plus u 1 E 1 plus u 2 E 2, then this is u 1 and u 2 are scalar variable d u 1 and d u 2. That is the basically the integral Riemann integration defined on R 2 with each fix t and gamma. So, this is what we mean by the integration over the plane of a function on R 2. So far so good. So, now we have defined this, but what why should we believe that this is independent of the choice of the basis what we are taking. So, suppose someone is taking suppose this is my gamma and someone is taking E 1 and E 2 here, somebody else can also take some E 1 and E 2 prime, then why this integral over this P t gamma is going to be the same for any or choice of the orthonormal basis. So, which we need to show that it is independent of the choice of basis. So, integral is independent the choice of basis. Let gamma E 1 prime E 2 prime be an orthonormal basis for R 3, then we can find a rotation R which is going to take my E 1 to E 1 and E 2 to E 2 prime. So, consider rotation R such that R of E 1 is equal to E 1 prime and R of E 2 is equal to E 2 prime and therefore, if u is equal to u 1 E 1 plus u 2 E 2, then we denote R u this is equal to u prime. So, now if we write down then that is a trivial F of t gamma plus u 1 E 1 prime plus u 2 prime E 2 prime d u 1 prime d u 2 prime this we can write the notation F of t gamma this is R of u and then this is d u 1 prime d u 2 prime make a change of variable. So, what you are going to get this is R 2 F of t gamma plus u and then this is d u which is d u 1 into d u 2. So, this is independent of the choice of the basis. One of the interesting observation for this what one can see is the following. Now, I want to claim integral over R 3 of F of x d x this is equal to integral minus infinity to infinity integral over P t gamma F. Now, x in R 3 is through the standard basis x is equal to x 1 1 0 0 plus x 2 0 1 0 and x 3 of 0 0 1. So, what I what get the rotation R such that R of 1 0 0 is equal to gamma R of 0 1 0 is equal to E 1 R of 0 0 1 this is equal to E 2. Then this is also going to be an orthonormal basis gamma E 1 and E 2 and therefore, what we do is that if we write integral over R 3 F of x d x this is equal to integral over R 3 F of R x and what I will get again the Jacobian is determinant is mod of the determinant is 1. So, therefore, what we get this by the change of variable and this if we rewrite this is going to be F of x 1 gamma plus x 2 E 1 plus x 3 E 2 and then this is going to be d x 1 d x 2 d x 3. So, now, this minus infinity to infinity over R 2 F of x 1 gamma plus x 2 E 1 plus x 2 x 3 E 2 d x 2 d x 3 and then d x 1 this is nothing, but minus infinity to infinity this is what we have defined x 1 of gamma over F and then d x 1 this is with the same to be consistent with our notation this is P t gamma F d t that is what we. So, this is going to be a very useful observation for us. So, now, another thing what we would like to observe is that this Radon transform R F this is a map on set of planes. Now, plane are determined by t and gamma. So, in other words I can think for a given function F in the Schwarz space R F is a map from R cross S 2 to C. So, we can consider R F a function on R cross unit sphere that is S 2 because it is determine every plane is we can write it in terms of the P t gamma for a fix vector gamma and t. So, R F can be thought of this. So, now, this is at this point one can define the Schwarz space of R cross S 2 this is the function which is a function on t and gamma such that supremum over t belongs to R and gamma belongs to S 2 mod of t to the power k d l by d t l F of t gamma this is finite. So, this is what is the Schwarz space we can define this. So, now, what is our goal? We want to reconstruct. So, goal want to reconstruct F from Radon transform that is what we want to get and. So, an easier question would be if for two functions F and G if the Radon transform what we are getting the image is same is it necessarily true that F is equal to G that is the first thing. So, now, that is what in mathematically we say uniqueness if R F is equal to R G then F is equal to G is it true second thing is that the reconstruction problem that is the reconstruction which we say that can F be expressed in terms of R F can we have a formula by knowing R F we can determine our F that is what is the two main questions. So, we would like to address now for the uniqueness what we know is that from our Fourier analysis knowledge that if F hat is equal to G hat for all points in R 3 then by inversion we know that F is going to be equal to G. So, that suggest us to look at convert this problem into the Fourier analysis problem and try to solve it. So, now, remember that R F is a function on R cross S 2 now for each gamma belongs to S 2 R F dot gamma is a function on R. So, now, we can talk of the Fourier transform of this function for a each fixed gamma provided if F is in Schwarz space R F has to be integrable remain integrable having some moderate decay then we can talk of the Fourier transform. So, that is the first thing we will see. So, let us write a proposition if F is in Schwarz space of R 3 then for each fixed gamma in S 2 R F dot gamma which is a function on R belongs to the Schwarz space of R. Therefore, I can talk about the R F hat at xi gamma for fixed this is equal to F hat of xi gamma remember that this F hat is 3 dimensional Fourier transform and this is this Fourier transform is 1 dimensional Fourier transform and moreover this actually one can write that R F to avoid confusion xi comma gamma this is equal to integral over R, R F of t comma gamma e to the power minus 2 pi i xi dot xi t here this is the both xi and t are in R then this is d t and F hat of xi gamma this is equal to integral over R 3 F of x e to the power minus 2 pi i xi gamma dot x dx. So, that is the 3 dimensional Fourier transform what we are talking about. So, now first to prove that this R F for each fixed gamma we would like to show that R F Radon transform is a Schwarz class function on R. So, now what do we know? We know that F given to us that F is in the Schwarz space. Therefore, 1 plus mod t to the power n 1 plus mod u to the power n mod of F of t gamma plus u this supremum is lesser equal to a n and also its derivative. So, therefore, when we are writing that in order to show that R F is a Schwarz class function we need to show that R F t comma gamma this is lesser equal to some supremum over t is less than finite that is what we want to show this is what we want. Now, what is my let us write 1 plus mod t to the power n R F of t comma gamma this is lesser equal to or rather one more step 1 plus mod t to the power n and this is R 2 F of t gamma plus u du. Now, if I take the mod then I have mod over here which is lesser equal to 1 plus mod t to the power n and then this is integral R 2. This is definitely lesser equal to some constant a n by 1 plus mod u to the power n 1 plus mod t to the power n and du because F is in the Schwarz space of R n that is what we have seen. So, now, this is lesser equal to a n du by 1 plus mod u to the power n and for large n. So, this is finite which is independent of u. So, therefore, and this exactly the same kind of argument we can give for the derivative with respect to t. So, we can get that actually R F hence R F dot gamma this belongs to the Schwarz space of R. So, now, what we would like to then find out R F hat at xi comma gamma this is equal to integral minus infinity to infinity. Now, this is R F is integral over P over the plane P t gamma F then e to the power minus 2 pi i xi t dt. So, now, as you can see that the t, t I can write this t I can write this as t gamma dot or this is gamma dot t gamma plus u. Remember that u is u 1 e 1 e 2 there u 2 e 2 there basically the gamma e 1 e 2 are the orthonormal basis therefore, this gamma dot gamma is 1 this is the unit vector. So, this is just gives me t because gamma dot u is 0. So, now, this I can represent it minus infinity to infinity this is P t gamma F and then this I can replace minus 2 pi i xi t in place of t we can write t gamma plus u I can put this as inside now this d this is and then this is dt this all we had the observation that this is going to be R 3 F of x e to the power minus 2 pi i xi gamma dot x dx and this is nothing, but the Fourier transform of xi into gamma xi is a scalar. So, once we have that as you can easily one can see a corollary if R F is equal to R G then F is equal to G the proof is M because this is R F of t R F hat at xi comma gamma this is equal to R G hat at xi comma gamma this implies that F hat Fourier transform in 3 dimensional xi gamma is equal to G hat of xi gamma. Now, this is true for each gamma varies over the sphere S 2 unit sphere and this is true for all the all xi in R for all xi in R and now we can vary this I fix a gamma then I will get that F hat of xi gamma equal to G hat of xi gamma. So, similarly I will vary another gamma every gamma on the sphere then I am going to get this. So, this implies because if I take any x in R 3 then this is going to be given to me a direction and the distance from the origin that means some xi and gamma x equal to xi dot gamma. So, this implies F hat of x is equal to G hat of x for all x in R 3. So, by uniqueness of the Fourier transform this will imply that F is equal to G. So, this gives us the solution of the uniqueness problem what we have posed that if two data they agree whatever the data we are getting if they if they agree then the original function both the functions has to be equal. So, in the next lecture we will try to find a formula to reconstruct our F from the given data of the Radon transform. Thank you.