 Hello, so let us continue with the discussion of Fourier transforms, in the last capsule we proved the Riemann-Lebesgue Lemma for Fourier series again, but gave Riemann's own argument for that. The argument we gave in the first lecture was quite different, but the second argument is very convenient to give the analog of Riemann-Lebesgue Lemma for Fourier transform. So what is the Riemann-Lebesgue Lemma for Fourier transform again, theorem 45 on the slide, if f is in l 1 of r, then the Fourier transform f hat of chi tends to 0 as chi tends to plus minus infinity. First let us imitate the same argument we gave before, to show that if a function f is continuous on minus a a, then integral from minus a to a f of x e to the power minus i x chi dx goes to 0 as x goes to plus minus infinity. Call the integral i and you set x equal to y plus pi upon chi and proceed exactly as we did before. Now use Lousen's theorem as indicated in the last capsule to extend it from continuous functions from minus a a to l 1 functions on minus a a. Now let us assume that f is in l 1 of r. Remember that if a function is Lebesgue integrable, then given any epsilon greater than 0, you can take a large interval minus a a such that outside the large interval the contribution of the integral is terribly small. So, the integral mod f x dx outside the interval minus a a is less than epsilon by 2. So, now we need to only worry about the closed interval minus a a. So, let us see how to deal with the closed interval minus a a. So, let us write down integral over r f of x e to the power minus i x chi dx absolute value. In other words, I am trying to estimate the Fourier transform f hat of chi. So, mod f hat of chi is what you see on the left hand side. Break this integral into two integrals from minus a to a that is the second piece and the first piece is integral from r minus minus a a. The first piece take the absolute value inside the integral use triangle inequality e to the power minus i x chi is a unit complex number and you get that that contribution is less than epsilon by 2. Now, take the second integral mod integral minus a to a f of x e to the power minus i x chi d chi and we know that that piece goes to 0 as chi tends to plus or minus infinity. So, there is a chi not bigger than 0 such that for mod chi bigger than chi not that absolute value integral minus a to a f of x e to the power minus i x chi dx is less than epsilon by 2 and accordingly we conclude that mod f hat of chi less than epsilon when mod chi bigger than chi not that was pretty easy because we did the detailed analysis for the Riemann-Lebesgue Lemma for Fourier series and a large part of the argument simply carried forward over here and that completes the proof of the Riemann-Lebesgue Lemma. Now, we want to estimate the Fourier transform in L2 norm. We could take a function f in the Schwarz class and we want the estimate for the Fourier transform in L2 norm. Why do we want the estimate in L2 norm because of the pleasant feature that L2 of r is a Hilbert space and so hopefully Hilbert space techniques can be brought into play. However, the argument is not so straightforward. We need a certain result known as the Plancherel's theorem or the Parseval's formula for Fourier transform. We have the Parseval formula for Fourier series and we have a Parseval formula for Fourier transforms. Suppose f and g are in the Schwarz class S then equation 4.11 in the display that is the Parseval's formula or Plancherel's theorem integral minus infinity to infinity f of t gt bar dt equal to 1 upon 2 pi integral over the real line f hat of chi g hat of chi bar d chi where the bar signifies the complex conjugation. Again we need to employ the e to the power minus epsilon chi square trick. So, let us try first why this has to be done. Let us try the naive approach. So, take the right hand side of 4.11 and let us put in the definition of Fourier transform f hat of chi is integral over the real line f of x e to the power minus i chi x g hat of chi is integral over the real line gy e to the power minus i chi y dy, but since I am taking the complex conjugation it will become e to the power plus i chi y dy. So, these 2 integrals over the real line combine to give me an integral over the plane namely integral over r squared f of x gy bar e to the power minus i chi x minus y dx dy. Now I want to switch the order of integration. So, the integral over r 2 will be done later and the integration with respect to the chi variable will be done first and I get the integral over the real line e to the power minus i chi into x minus y d chi. Again we ran into the same oscillatory integral for which we need to introduce the x minus epsilon chi square trick. We have now gained sufficient experience how to do this. Well we need the Fourier transform of the Gaussian along the way. So, the right hand side can be written as limit as epsilon goes to 0 1 upon 2 pi integral minus infinity to infinity f hat of chi g hat of chi bar e to the power minus epsilon chi squared. The limit was inside the integral I can pull it outside the integral because f hat and g hat are in the schwar space they decay very rapidly. Now that the limit has come outside the integral I will put in the definition of f hat and g hat as I described while I was doing the previous slide I get f of x gy bar dx dy it will be a integral over r 2 and what I will get is x p of minus i chi into x minus y and now I have the e to the power minus epsilon chi squared d chi. And so the inner integral now can be computed it is a Fourier transform of the Gaussian what you see here is a Fourier transform of the Gaussian that is another Gaussian and the epsilon will come in the denominator and there will be a root pi by root epsilon 1 root pi cancelled out and 1 upon 2 root pi epsilon times. Now the usual change of variables x minus y equal to root 4 epsilon z this simplifies to e to the power minus z squared dy will be 2 root epsilon dz and the 2 root epsilon cancels out a root pi remains the denominator and I will get e to the power minus s squared d s take the limit inside the integral and you will get simply g x bar and you get simply integral e to the power minus z squared dz which is the root pi the root pi and the root pi cancels out you get f x g x bar dx and to take the limit inside and outside the integral you need to appeal to the dominated convergence theorem from the theory of Lebesgue integrals that is a standard argument. So that is how you prove the Plancherel's theorem or the Parseval formula in the Parseval formula we must take f equal to g if I take f equal to g then this says that integral minus infinity to infinity mod f t squared dt equals 1 upon 2 pi integral minus infinity to infinity mod f hat chi the whole squared d chi. So 4.11 simplifies when g equal to f to the following formula 4.12. So 4.12 follows immediately from 4.11 taking f equal to g. So theorem 47 is the important link that we wanted to extend the Fourier transform as an operator on L2 of r how does it how does this estimate 4.12 help us to achieve this. So let me first record the thing as theorem 48. So the Fourier transform extends as a bounded linear operator on L2 of r further it is a linear isomorphism on to L2 of r in the earlier case from L1 it was into L infinity but here in the context of L2 the Fourier transform maps L2 on to L2 so it is surjective in this case. Now we shall use the fact that the Schwarz space s is dense in L2 of r the Schwarz space is dense in L1 it is dense in L2 in fact it is dense in Lp for any p between 1 and infinity 1 included in infinity excluded. We shall not prove this theorem that the Schwarz space is dense in L2 it is a version of the Luzin's theorem if you like. So now that we established 4.12 let us denote by script f the mapping from s to s given by the Fourier transform. Now the Fourier transform maps s on to s thanks to the inversion theorem. So this f if you restrict it to the Schwarz space I get a mapping from s on to itself. Now I want to extend this map to a operator on L2 on to L2 for that we observe that being linear norm of the Fourier transform of f minus Fourier transform of g is less than or equal to root 2 pi norm of f minus g and f and g are in the Schwarz class. When I say norm I mean L2 norm in this particular context and thanks to the inversion theorem we also have the inverse Fourier transform of f minus inverse Fourier transform of g less than or equal to 1 upon root 2 pi the L2 norm of f minus g and which shows that both the Fourier transform and its inverse as a map from s to s are uniformly continuous with respect to the L2 metric. So what we have here is we got a metric space capital X which is L2 metric which is L2 of the real line and we got a subspace y which is dense in x what is the subspace the Schwarz class s. So now we are in the following situation we have a complete metric space x and we are going to dense subspace y and we have a mapping from y to y which is uniformly continuous then t extends continuously as a map from x to x. I repeat that in our context x is L2 of the real line and y is the Schwarz class and this transformation t is the Fourier transform operator and we know that the Fourier transform maps s onto s and it is continuous with respect to the L2 topology is uniformly continuous with respect to the L2 norm. So we want to prove lemma 4.9 once we prove lemma 4.9 then we would have established theorem 48 and this equality 4.12 will continue to hold by the density argument by the density of s in L2. So let us prove the lemma. So we need to define t as a map from x to x. So take an element little x in capital X because y is dense in x I can select a sequence y n in capital Y converging to x. Now look at the sequence ty n. Now ty n is going to be a Cauchy sequence because t is uniformly continuous because ty n minus ty m or the distance between ty n and ty m is less than or equal to epsilon for all n m greater than or equal to n naught because of uniform continuity of. So because ty n is Cauchy and the metric space x is complete remember L2 of r is a Hilbert space it is complete. So ty n converges to some point let us call that point tx and we want to say that that tx is the image point that we are looking for but there is one small problem that we need to address and the problem is the following. It may happen in fact it will happen that there are several sequences converging to x. Say I take two sequences y n converging to x and z n converging to x and y n and z n are sequences in y. Correspondingly ty n is a Cauchy sequence t z n is also a Cauchy sequence question is will ty n and t z n converge to the same limit the answer is yes then this assignment tx to x is a well-defined assignment and we get a well-defined mapping. If ty n and t z n converge to different limits then we are going to be in serious problem now we shall show that that does not happen indeed ty n and t z n will both converge to the same limit to do this we use an interlacing argument which is very frequently used in analysis. So, y n converges to x and z n converges to x we know that. So, let us interlace the sequences y n and z n as follows first put y 1 and then z 1 and then y 2 and then z 2 and y 3 and z 3 we interlaced the two sequences y n and z n the interlaced sequence again converges to x correct because y n and z n individually converge to x. So, ty 1 t z 1 ty 2 t z 2 etcetera is a Cauchy sequence and so it converges as as above. So, both the subsequences ty 1 t y 2 t y 3 and t z 1 t z 2 t z 3 they must converge to the same limit. So, the limit t x is independent of the choice of the sequence y n. So, this assignment of t x to this domain point x is a well defined assignment and we get a well defined map from x to x and this extension is also denoted by the same letter t now we have to show that this extension is also continuous that part I am leaving it as an exercise for you. So, now with this lemma out of the way we can see that the Fourier transform is now extended as a continuous linear map from L 2 onto itself and the estimate 4.13 holds because it holds for the Schwartz class and so it continues to hold for L 2 of r. Now, the next item that we need to discuss is the notion of convolution of two functions and we want to prove the convolution theorem for Fourier transform. The convolution theorem roughly says that if I take two functions f and g the Fourier transform of f star g is a product of the Fourier transforms f hat into g hat. So, let us describe what is the convolution of two functions. Let us begin by taking two absolutely integrable functions f and g on the real line. The convolution f star g is defined as follows f star g of x is integral over the real line f of y g of x minus y dy and the result is a function of x. Check that f star g equal to g star f and check that the convolution is also absolutely integrable. So, the next theorem says that if f and g are both in the Schwartz class and the convolution f star g is also in the Schwartz class and further f star g hat is f hat into g hat equation 4.14 in the slide. We shall not prove the first part of the theorem. We shall not prove that f star g is again in the Schwartz class. The proof is not difficult. It just gets involved and the proof follows routine arguments. By way of analogy, you may have seen the convolution theorem for Laplace transforms in undergraduate courses in ODE's. There is only one small piece of warning that I would like to give you that in undergraduate courses on ordinary differential equations, the definition of convolution that you see in the context of Laplace transforms is slightly different because when you compute Laplace transforms, you are ignoring the function on the left hand side and the Laplace's integral is an integral from 0 to infinity. So, what happens is that the convolution theorem for Laplace's transform should be written as the Laplace transform of the heavy side function u times f star u times g will be Laplace of f into Laplace of g. That is how the Laplace transform convolution theorem ought to be stated. Okay, let us get to the proof of the convolution theorem. So, let us start with the Fourier transform of f star g. Fourier transform of f star g is what put in the integral. Integral over r e to the power minus i x chi f star g x dx. Next step is obvious, no prizes for guessing. It is going to be the definition of the convolution integral over r f of y g of x minus y dy and I am going to switch the order of integration. I am going to get integral f of y dy integral g of x minus y e to the power minus x chi. Put x minus y equal to z in the inner integral and we get f star g hat chi equal to integral f of y dy over r integral over r e to the power minus i y plus z chi g z dz. So, here e to the power minus i y chi comes out of the integral and that is simply the product of the two Fourier transforms. And so, the proof of the convolution theorem is complete. Let us use the convolution theorem to discuss some solutions of partial differential equations. The heat equation. We already discussed the heat equation in the context of Fourier series in the first part of the course. We return back to the heat equation in the context of Fourier transforms. We solve the initial value problem for the heat equation ut minus u xx equal to 0 and the initial distribution of heat is u of x comma 0 equal to f of x. Now, what we will do is that we will take the Fourier transform and we shall obtain an ODE for the Fourier transform. We will solve the ODE and we will use the inversion theorem to get the integral representation for the solution of the heat equation. The advantage of this over the previous method is that in the previous method we used Fourier series and there we needed the fact that f of x is a periodic function. So, in the previous discussion the heat equation was solved for periodic initial conditions and the solution sought was also periodic. Here we do not have periodicity the initial condition need not be periodic. We shall discuss the wave equation also utt minus u xx equal to 0. We shall discuss the wave equation in one space dimension. We shall discuss the conservation of energy for the wave equation. We shall discuss the principle of equipartitioning of energy in the context of wave equation in one space dimension using Parseval formula. So, I think with this let me close this capsule here and we will take up in the next capsule the solutions of initial value problems for these classical PDEs. Thank you very much.