 Hello, so we continue reviewing those parts of functional analysis that are required in the study of Fourier series and generalized Fourier series. So, let us recall if x and y are Banach spaces, we are interested in studying linear transformations t from x to y, but that is not enough because these spaces are infinite dimensional and for an infinite dimensional space, linearity alone is not sufficient, we want those linear transformations that are continuous. So, when is the linear transformation t from x to y continuous, the theorem gives you 4 equivalent conditions. Condition 1, t is continuous, condition 2, t is continuous at the origin, condition 3 there exists a constant m bigger than 0 such that norm t x is less than or equal to m times norm x for all x in x and condition 4, t is uniformly continuous. Obviously, 4 implies 1 is completely trivial and 1 implies 2 is also completely trivial. What about 3 implies 4? Simply replace x by x minus y, simply replace x by x minus y and use linearity, what does that give you norm t x minus t y less than or equal to m times norm x minus y that will immediately confirm that t is uniformly continuous. So, 3 implies 4 is also completely trivial. Let us look at 2 implies 3, little bit of work, but it is fairly easy to prove that 2 implies 3 simply take epsilon equal to 1. Then there is a delta bigger than 0 such that norm x less than delta implies norm t x less than 1. So, now take a x not equal to 0, take an x not equal to 0 because this inequality 3 is completely trivial when x equal to 0 because both sides are 0. So, we might as well confine our attention to x not equal to 0. If x is not equal to 0, then x upon norm x is a unit vector and so delta x upon 2 norm x, I am calling it y, y is a vector with norm equal to delta by 2, which is certainly less than delta. So, because it is less than delta norm of t y is less than epsilon, epsilon has been taken to be 1. What is y again? y is delta by 2 norm x, delta by 2 norm x is a scalar, it is a positive scalar, t is linear. So, positive scalars will come out and you will get t x with the norm outside. So, you get norm t x less than or equal to 2 norm x upon delta and so 2 by delta is the m that we are looking for. So, the equivalence of these things is completely established. Because of 3, you say that instead of continuous linear maps, you call them bounded linear maps. So, bounded linear maps the same as continuous linear maps in functional analysis. We are only interested in studying normed linear spaces and Banach spaces, we are not going to be studying topological vector spaces here. The Bayer category theorem, this is one of the most important results in general topology on which the entire structure, the edifice of functional analysis rests on this very foundational theorem, the Bayer category theorem. As you know in functional analysis there are four pillars, the Hahn-Bannach theorem, the Banach-Steinhausen theorem, the open mapping theorem and the Krugus graph theorem. The open mapping theorem, close graph theorem and Banach-Steinhausen theorem, all of them rely on the Bayer category theorem. The Hahn-Bannach theorem is a theorem about convexity. So, let us look at the Bayer category theorem. What do the Bayer category theorem say? If you take a complete metric space x and you write this complete metric space x as a countable union of closed sets, then at least one of the sets must have non-empty interior. If a closed set has empty interior, you say it is no wear dense. So, a complete metric space cannot be written as a countable union of closed no wear dense sets. That is the way you state the theorem without using any symbols. Note that a closed set E has empty interior precisely when the complement is a dense open set. E is closed. So, x minus E is obviously open. E has empty interior means the complement is dense. With this simple observation, the proof is quite easy. So, let us prove this by contradiction. Suppose the theorem is false, then take x minus E n to be g n. So, g n is a open set and it is dense because E n has empty interior. Each g n is a dense open set and we are getting that intersection of g n from 1 to infinity is empty. We shall arrive at a contradiction by showing that the intersection is actually dense. So, that will be a contradiction. Empty set is not dense unless the space itself is empty. It is a trivial statement. So, let us take a arbitrary point in x. What does it mean to say that this intersection is dense? It means that any point is a limit point of this set. So, take a point p in x and I want to show that p is a limit point of this. So, since g 1 is dense, first I begin with g 1. g 1 is dense. The ball of arbitrary radius centered at p, I am calling it b 1 p. This ball must contain a point z 1 in g 1. Now, g 1 is an open dense set. g 1 is dense I have used. Now, I will use the fact that g 1 is open. So, I can find r bigger than 0, such that the closed ball s 1 of radius r 1 centered at z 1 is contained in g 1. And I am going to assume that this r is less than half. Now, since g 2 is dense, so with this r 1, I construct a ball with center at z 1 and this must intersect g 2 at some point say z 2. So, where is this z 2? z 2 is in b r 1 intersect g 1. This b r 1 is already contained in g 1. So, this z 2 is already contained in b r 1 and b r 1 is already contained in g 1. Now, we select an r 2 because both are open sets. We select an r 2 such that r 2 is less than one fourth. So, the closed ball s 2 and the b's are open balls and the closed balls I am using the symbol s 1 s 2 etc. The closed ball s 2 of radius r 2 centered at z 2 is contained in b r 1 intersection g. So, I get s 2 is contained in b r 1 and b r 1 is contained in s 1 and I keep going. Next g 3 is dense in x and so the ball b r 2 must intersect g 3 at some point z 3 and that z 3 lies in b r 2 intersection g 3. They are both open and g 3 being open, we can find the r 3 which is less than one eighth and the closed ball s 3 of radius r 3 centered at z 3 is contained in b r 2 intersect g 3. So, like that we keep going. We find a nested sequence of closed balls s 1 s 2 s 3 and these radii are going to 0 or the diameters are going to 0 by Cantor s intersection theorem. There is a point q which is contained in all the sets s n but for each j by very construction by very construction the s j is going to be contained in g j. Look at the construction s 3 is contained in b r 2 intersection g 3. So, s 3 is contained in g 3 earlier we got s 2 is contained in g 2 so on so forth. So, s j is contained in g j. So, this point q belongs to all the s j's. So, the point q belongs to all the g j's. So, the point q contained in the intersection of all the g j's. So, what have we proved? We have proved that we have taken a point p in the metric space. We took a ball around that point and now we found a q which is sitting in that ball and that point q is sitting in intersection of all the g j's. So, we have proved that the intersection of all the g j's is dense. Any point p is a limit point. We took a point p which is completely arbitrary. We took a ball around this p and we found a point q in this intersection contained in that ball which means that any arbitrary neighborhood of an arbitrary point will intersect this set. So, the set is dense and that is a contradiction. So, we have proved the Bayer category theorem. The Banach-Steinhaus's theorem our next result that we are going to establish is the Banach-Steinhaus's theorem which is a corollary of the Bayer category theorem. It is a simple corollary of the Bayer category theorem. It concerns a sequence of continuous linear transformations from x to y when x and y are Banach spaces. We shall need this Banach-Steinhaus's theorem only under a very special case where the target space is a complex numbers. We want to use the Banach-Steinhaus's theorem to prove the result that there exists plenty of continuous functions whose Fourier series diverge at specified point. So, for that application we need the Banach-Steinhaus's theorem for c-valued continuous linear transformations. So, a sequence of linear transformations Tn from x to c is said to be pointwise bounded. If for each x in my Banach space this mod Tn x where x is fixed and n is varying and the supremum must be less than or equal to mx. The sequence Tn x is bounded in absolute value and the bound will depend upon x. If I change the x this bound could change. What we are going to deduce is that this sequence of linear forms when the target space is c we call it a linear form. The sequence of continuous linear forms is uniformly bounded namely I can get rid of this x dependence and I can find a common m which will work for all the points in the unit ball that is if norm x is less than or equal to 1. So, for all the points x in the closed unit ball a common supremum will work. This is a common m that is why it is also known as the uniform boundedness principle. In short pointwise boundedness implies uniform boundedness on the unit ball. That is the simple way to state the Banach-Steinhaus's theorem. Let us get to the proof of the Banach-Steinhaus's theorem. If a sequence of continuous linear transformations Tn from x to c is pointwise bounded then it is uniformly bounded. Consider the family of closed sets. We plunge into the proof right away Ej equal to those points in x such that mod Tn x is less than or equal to j for all n's. These Ej's are closed because look at this set of all points x in x such that norm Tn x less than or equal to j that is a closed set and I am demanding this to be true for all n's. So, it is an intersection of closed sets so Ej is closed. So, the fact that Ej is closed is easy. Let us prove that the union of these closed sets is the whole of x. So, we have to show that union of Ej is x, union of Ej is trivially contained in x. So, the other inclusion take a point in x. We know that the sequence Tn x is pointwise bounded. So, mod Tn x is going to be less than or equal to mx for all n. Remember supremum mod Tn x is less than or equal to mx. So, I just have to take the j to be larger than mx. If I take the j to be larger than mx then I get that norm Tn x is less than or equal to j for all n and so that particular point x lies in Ej. So, any point of the Banach space lies in at least one of the Ej's. So, the Banach space x is a union of closed sets. So, with a Bayer category theorem at least one of the sets Ej say E capital J has a interior point. It has a non-empty interior. So, it has an interior point P which means there is r bigger than 0 such that the ball of radius r centered at P is contained in E capital J. Now, this particular point P I can apply the boundedness mod Tn P is less than or equal to mp. That hypothesis also it is our disposal. So, norm Tn P less than or equal to mp for all n, that also we have and the ball of radius r centered at P is contained in E capital J. Both these facts will now be used. Now, take a point y in the unit ball take norm y less than or equal to 1. If norm y is less than or equal to 1, multiply by r by 2, r by 2 y is in the ball of radius r by 2 centered at the origin. So, P plus r by 2 y lies in the ball of radius r centered at P. In fact, it lies in the ball of radius r by 2 which is contained in the ball of radius r and so that is contained in Ej which means that what does it mean to say it contains in Ej by definition it means Tn P plus r by 2 Tn y norm less than or equal to j. For all n use a triangle inequality we get norm Tn y less than or equal to 2 j by r plus 2 by r norm Tn P. But norm Tn P is going to be less than or equal to mp remember. So, all in all I get that norm Tn y less than or equal to 2 upon r times capital J plus mp. So, P is fixed and j is fixed and so all the norm Tn y's are bounded by a fixed number where y varies over the unit ball in my Banach space. The proof of the uniform boundedness theorem is thereby completed. Okay. So, now that we have finished the proof of the Banach-Steinhaus theorem we must use it to solve a problem in Fourier series. So, now what is that we go to apply we are going to prove the existence of a continuous function whose Fourier series diverges at the origin. So, you take a point in the interval minus pi pi and we can find a function which is continuous throughout minus pi pi, but the Fourier series will diverge at that chosen point. For simplicity we will choose the point to be the origin just for notational simplicity. Okay. So, remember from the very first chapter you take Sn fx take a 2 pi periodic function f and the nth partial sum of the Fourier series Sn fx. What is it given by? It is given by convolution with the Dirichlet kernel. Remember the Dirichlet kernel integral from minus pi to pi it is given by f of t dn x minus t dt where dn is the Dirichlet kernel. And now we are going to discuss whether or not the Fourier series converges or diverges at the origin. So, straight away I am going to put x equal to 0 and the Dirichlet kernel is an even function. So, I am going to get dn t. So, this particular object Sn f0 I am going to simply call it Tnf. So, what is Tnf? Tnf is integral from minus pi to pi f of t dn t dt. Evidently this is a linear map. Tn is a linear map. Now, suppose that the Fourier series of every periodic function converges at the origin. Suppose all continuous functions have convergent Fourier series we will arrive at a contradiction. Assume the contrary and we will arrive at a contradiction. If that happens that means that Sn f0 will converge to f of 0 as n tends to infinity. In other words Tnf will converge for every choice of f. In other words the sequence Tn of bounded linear transformations converges point wise that is it converges for each f. So, by uniform boundedness principle it will be uniformly bounded and we will arrive at a contradiction. So, Tnf is bounded for each f in the space of periodic functions by Banach-Stanhaus' theorem there is a constant m bigger than 0 such that mod Tnf less than or equal to m for all 2 pi periodic continuous functions of norm 1. In other words mod integral minus pi to pi f of t dnt dt less than or equal to m for all n and we shall now restrict ourselves to real valued functions because we would not need complex valued functions anyway. We will need some information about the way the integral of mod dnt behaves. So, that is display 7.4 integral minus pi to pi mod dnt dt is approximately c log n. This is an important piece of information that we need to establish, but how does it help how does the contradiction follow from here? Further I am also going to restrict myself to functions which are between minus 1 and 1. Let us see what is the idea. The idea behind this is to take f of x to be the signum function which we denote by sigma t. So, what is the signum function sigma t? The signum function sigma t is that function which takes the value plus 1 with the Dirichlet kernel dnt is positive and it takes the value minus 1 with the Dirichlet kernel dnt is negative. So, what will happen if I take sigma t instead of f of t? If I take sigma t instead of f of t then sigma t times dnt will exactly be mod dnt, is not that so? Because f of t basically is a sign of dnt. So, now for that signum function, we will get that integral minus pi to pi mod dnt dt is less than or equal to m because the Bernach-Steinhaus's theorem gave us that. But now the very next line I am saying that this is diverging when I take f to be a signum function, the left hand side becomes integral minus pi to pi mod dnt dt and so on the one hand I am saying that integral from minus pi to pi mod dnt dt is bounded and immediately in the next line I am saying that it goes to infinity. How is that possible? That will be a contradiction and that will prove the theorem. There is only one small problem and that small problem is I cannot take the f of t to be the signum function here because a signum function is not a continuous function. That is the problem. So, to say that let us take the f of t to be the signum function that is an illegal thing to do. The signum function is not continuous. So, we need to get around this difficulty. So, that is a little problem but we will still get around the difficulty by approximating the signum function by continuous function. Idea is that I cannot choose the f of t to be the signum function but because it is discontinuous I can approximate the discontinuous function by continuous function using Luzin's theorem for example and I will still get a similar bound like 7.5. So, I will still get the bound 7.5 through a devious argument and the rest of the proof will go through. Observe that the Dirichlet kernel oscillates in the interval minus pi pi. We must see where are the zeros of the Dirichlet kernel that are located and we must select the signum function accordingly and we must modify the signum function and that is a little bit of a delicate operation and I think that is best relegated to the next capsule. So, I think that is a very good place to stop this capsule. Thank you very much.