Lecture 60

Welcome back, so in the last lecture we have seen that the uniqueness of the Radon transform that if Rf is equal to Rg then f is equal to g. Now what we are searching for is that knowing the values of Rf which is given to us then can we reconstruct our f. So now this is the reconstruction problem and to in order to define the reconstruction problem, so first let us define what is the meaning of an integration over the sphere. So, we would like to get what is the integration over sphere of a function g, what is the meaning of that. Now what one can see is that if I take any element x in R3 then I can write in terms of the spherical coordinate in R3 means, so this is going to if I take any point x in R3 this is going to lie on a sphere which is centered at the origin and the radius R. So, this will imply that x belongs to S2 of R. Now if we know that if we write it in terms of the spherical coordinate, suppose this x is x1, x2, x3 then x1 this is equal to r sin theta cos phi, x2 is equal to r sin theta sin phi and x3 is equal to r cos theta where our r is the radius which is greater than 0 and theta varies over 0 to pi and phi varies over 0 to 2 pi. So, this what we know and also in terms of the polar coordinates integral of x dx what we can write it is in the terms of 0 to infinity, 0 to 2 pi, 0 to pi f of r sin theta cos phi, r sin theta sin phi and r cos theta and then this Jacobian is sin theta d theta d phi and then this is r square dr. So, now if I have a function g on S2 to C, we define the integral S2 of g of gamma, gamma is in S2 of d sigma gamma by our definition this is going to be 0 to 2 pi, 0 to pi g of gamma then this is sin theta d theta and then this is d phi. This is what is our integration over the sphere will look like this is the d sigma gamma. Okay, with this notation at place now let us try to address our reconstruction problem. For that let us define dual Radon transform this is r star of f at a point x in R3, this is going to be integral over S2 f of x dot gamma, gamma d sigma gamma. So, that is what is our dual transform. So, in a sense what you can see the usual justification for the dual is that if I look at r integral over S2 then if I take Radon transform f of t gamma and f of t gamma then this is d sigma gamma and then this d t then if you make this write down the definition this is p t gamma this is f of t gamma plus u d u f of t gamma d sigma gamma of d t. So, which means if we write this as r and then this is p t gamma and f of t gamma plus u and then integration over S2 over d gamma is nothing but r star f of r star f at x this is and this what we know is that is essentially is R3 f of. So, in this sense with this identification that is what it is called the dual. Now, we recall that the Laplacian is going to be del square by del x 1 square plus del square by del x 2 square plus del square by del x 3 square this is our Laplacian. So, with this notation the main theorem is the reconstruction formula what it is if I have delta that means the Laplacian then r star which is going to act on the function which is defined on r cross S2. So, that is I can talk of r f which is a function on r cross S2 and then this is equal to minus 8 pi square of f. So, what it says is that if I have r f then I can define the r star of r f r star of r f at x this is going to be equal to integral over S2 r f of x dot gamma comma gamma d sigma gamma. So, this I can say T gamma dot gamma. So, now, we can compute this thing and then we take the Laplacian then what we are going to get is our f. So, that is what is the reconstruction formula through radon transform. So, by looking at the images on 2 dimensional image now this formula is going to tell us that how to reconstruct our f. So, what is the proof we have seen that r f hat at S or xi comma gamma this is equal gamma this is what we have already seen. Therefore, and r f we know that is a Schwarz class function on r with all for a fixed gamma. Therefore, r f of T gamma this is going to be minus infinity to infinity r f hat of xi comma gamma e to the power 2 pi i xi T d xi this we can write as minus infinity to infinity f hat of xi gamma e to the power 2 pi i xi T d xi this is our r. Therefore, r star of r f radon transform this at the point x this is going to be integral over S 2 then r f T gamma this is going to be minus infinity to infinity f hat of xi gamma e to the power 2 pi i x dot xi gamma then this is d xi d sigma gamma is what by plugging in the definition of this. So, now we apply the Laplacian now all these are nice Schwarz class function and all one can push the Laplacian inside the integral. So, what here we get that this is minus infinity to infinity f hat of xi gamma then this is when we take then this is going to be e to the power 2 pi i. So, minus 4 pi square xi square e to the power 2 pi i x dot xi gamma then this is d xi d sigma gamma. We will pull out the minus 4 pi square outside what we have got is that S 2 minus infinity to infinity f hat of xi gamma then this is e to the power 2 pi i x dot xi gamma then this is xi square d xi and d sigma gamma. Now, if we have the xi is varying from 0 to infinity we can write this as the integral over R 3. Now, how to do that? So, now this we break the integral 0 to infinity f hat xi gamma e to the power 2 pi i x dot xi gamma then this is xi square d xi and d sigma gamma plus or this is again minus 4 pi square integral over S 2 minus infinity to 0 f hat of xi gamma e to the power 2 pi i x dot xi gamma xi square d xi then this is d gamma. Now, if we make a change a variable from on the second integral xi going to minus xi. So, this second integral is going to be 0 to infinity f hat of minus xi gamma e to the power minus 2 pi i x xi gamma then this is xi square d xi this is what the inside integral will look like. Now, we have another integral S 2 is sitting over d sigma gamma. Now, you make a change a variable on the sphere on the sphere then gamma goes to minus of gamma then this is going to be integral over S 2 0 to infinity f hat of xi gamma e to the power 2 pi i x dot xi gamma xi square d xi d sigma gamma. Now, with this observation this we can write that this is minus 4 pi square into 2 times S 2 0 to infinity f hat xi gamma e to the power 2 pi i x dot xi gamma xi square d xi d sigma then this is equal to minus 8 pi square and this is nothing but by the Fourier inversion formula this we can now polar coordinate to change into Cartesian coordinate this is f hat of a y e to the power minus 2 pi i x dot y then this is nothing but d y. So, this is nothing but 8 pi square f of this is plus f of x that is what the Fourier inversion. So, this formula gives us the method how to reconstruct f from the data given on the slices what we have got in dimension R 2 because the slices are. So, this is actually the mathematics behind the medical imaging in terms of the x-ray and tomography. So, essentially we get it on the plane and then we reconstruct the formula. It also Radon transform also has many other application in PDE and other thing. So, we can solve by using the Cauchy's problem using the Radon transform technique. So, which is a very powerful technique and which we are not going to discuss in this course. So, this in fact brings us to the end of this end of the course and now let us kind of summarize what we have done. So, we have started doing the Fourier series then we did on circle group T then we have Fourier series on finite abelian group then the Fourier series Fourier transform on R and Rn and we have seen various applications of that too. Now, what next I mean where does the subject what is the direction of this subject go well let us recall that what we have seen is that if f remain integrable and periodic then what we have seen that f hat of n as a sequence this is in little l 2 of z. Now, what we have seen that there is a sequence in little l 2 which is not a Fourier coefficient of a Riemann integrable function. So, naturally so the what we have seen that this is the Riemann integrable function on T this x 2 little l 2 and which is not on 2. So, one would like to ask that what is the larger space which can replace this Riemann integrable function such that the Fourier transform is on 2. In fact, that is nothing but the Lebesgue space l 2 of T we need to define the Lebesgue measure and through the Lebesgue measure one defines l 2 of T. So, that if we consider this is a all measurable functions on T and which is integral over mod of f of x dx is f of x square dx is finite. So, if we introduce the Lebesgue measure and consider this space then the Fourier transform is going to be on 2. Another nice connection what one would see that the Fourier series is going to be n from minus infinity to infinity f hat of n e to the power i 0 1 if I am taking n n x. Now, this is so one can ask that when does this Fourier series it becomes a power series. That means all the negative coefficient f of f hat of n is going to vanish. So, this is going to give us a bridge between the complex analysis and Fourier series. So, definitely all this can be extended to l p of T Lebesgue spaces or l p of R and l p of R n. So, then for example, for what we get that if f is in l 1 of R then we get that f hat belongs to C 0 of R. Because the Lebesgue space are also going to be functions which are integrable can going to be approximated through the simple functions. And one can show that the Schwarz space is dense in all those all those spaces. Therefore, the Fourier transform can easily be extended to all this spaces. And then one would like to know that what is going to be the boundedness behavior or f hat and so on and so forth. So, those questions can be asked. And also about the convergence of the Fourier series we know that S n is minus n to n f hat of n e to the power 2 pi n x. Then this S n f we know that how hard it is for the point wise convergence. But one can actually ask that whether this S n f whether this converges to f in l p. That means mod of S n f of x minus of f of x p d x whether this goes to 0 as n goes to infinity. So, we have seen for the Riemann integrable function for p equal to 2. And now for this general Lebesgue space one can ask this. In fact the answer is going to be true if p is greater than 1. And now one thing what one would like to see is that e to the power the building block is e to the power let us say i n. So, what it is happening here? So, one is seeing that in the Fourier series they are playing as the orthonormal family and which is going to span I mean this is orthonormal basis in the infinite dimensional means the sum is infinite here. So, it is and then when we go to R or R n then we consider this as the continuous homomorphism. Now for that matter if we take any arbitrary infinite group G which is so called locally compact abelian then this entire Fourier transform can be extended that is for the locally abelian. Now here as we can see that for the basis part e to the power 2 pi i xi x is not integrable. But however it is all these e to the power 2 pi i n x they are essentially is getting generated by a single function defined phi of n is equal to e to the power 2 pi i x and phi n x is equal to phi of n x then you are generating all this function. So, now from a single functions you are generating by getting the dilation you are generating the orthonormal basis. So, although e to the power 2 pi i xi x is not in R, but one can ask this question that if can we generate an orthonormal basis for L 2 of R from a single function just by the translation and dilation. So, that leads to the course called wavelet analysis which is a very powerful tool for some applied problems in engineering and other science. And next is that if we have a general group which is not abelian then what one should do to do the Fourier analysis first of all we need to define an integration in that group. So, which means there has to be a prescription how to talk about the integration and then next one would like to see this continuous homomorphism, but this continuous homomorphism circle will going to be replaced by a Hilbert space and mod t mod z is equal to 1 that is what the circle is essentially it should be some sort of a representation homomorphism which is to the unitary operator on Hilbert space. So, this basic whatever we have done all these questions will be in the higher setting, but for that one need to work out to mathematically to ask the appropriate questions. Thank you.