 So, we now continue with the second chapter in this course. The second chapter is mean convergence and some applications to PDEs. So, what is mean convergence? Mean convergence means convergence in L2 norm. Remember in the very first chapter we talked about the different modes of convergence of a Fourier series. We talked about point wise convergence, we talked about mean convergence, we talked about Cesare convergence. Cesare convergence will be taken up in the next chapter. The first chapter concerns point wise convergence, now we talk about mean convergence. So, first we shall discuss what is convergence in mean? We shall introduce the appropriate function spaces and cover the requisite preliminaries. We shall discuss two applications of the Parseval formula. The most important formula that we shall derive in this chapter is the Parseval formula. As an application of the Parseval formula, we shall give Hurwitz's proof of a classical result in geometry called isoperimetric theorem. We shall prove the maximum modulus theorem in complex analysis via Parseval formula. Then we will take up Abel's summability and the Poisson kernel and we shall give some applications to solving the Laplace's equation on a disk and the heat equation. So, this is the theme of the second chapter. So, let us begin. So, the second mode of convergence we talked about is convergence in mean. We need to recall some rudiments of Lebesgue theory, theory of Lebesgue integration. If you are not familiar with Lebesgue integrals, you can pick up any book on real analysis. My favorite book is Royden real analysis. There is another beautiful book, Goffman and Pedrick, a first course in functional analysis. A beautiful book. Somewhere in the future slide, the exact reference is given to you. So, let I be an interval on the real line and P be a real number between 1 and infinity where I include 1 and I exclude infinity and Lp of i, Lp of i is a set of all measurable functions on i such that the integral mod fx to the power p dx is less than infinity. There is a pth power of the modulus of f is integrable over i. Even the norm, the Lp norm is defined to be the integral over i, the pth power of mod ft, the whole thing to the power 1 by p. So, you must have seen these things in your real analysis courses. This is the Lp norm. This is the, this thing that you see is the Lp norm of f. With respect to this Lp norm, Lp of i is a Banach space. Of course, we have to show that if f satisfies this integrability condition and if g also satisfies this integrability condition, then f plus g will also satisfy this condition. In other words, first we have to prove that Lp of i is a vector space, complex vector space and then we have to show that this is a norm. The triangle inequality is the one that is going to be non-trivial. For that, you have to show that with respect to this norm, it is a complete norm linear space. Every Cauchy sequence converges, that is what we are required to show. This is done in standard courses on measure theory. This is called the Ries-Fischer theorem, that Lp of i is a Banach space, that is a Ries-Fischer theorem. You might also wonder why is it that I have excluded p over here. We also have the notion of L infinity of i, but the definition is slightly different. We would not use L infinity of i in this course and so, when we need it at that stage of the game, we will define it. The only Lp spaces that we are going to use in this particular chapter is when p equal to L2 of i is what we are going to be concerned with in this particular chapter. And secondly, the interval in question is going to be minus pi pi. So, you may right away restrict yourself to L2 of minus pi pi, if you like. So, let us continue with the discussion, but with p equal to. So, for p equal to what is the norm, the L2 norm is norm g equals integral a to b mod gt squared dt the whole thing to the power half. Of course, the integral may be plus infinity, in which case the function is not in L2. For example, if I take g of x equal to 1 upon root x, 1 upon root x is integrable on 0 to 1. Integral mod gx dx is finite for gx equal to 1 upon root x, but square it mod gx squared is 1 upon x, can you integrate 1 upon x from 0 to 1, it becomes plus infinity. So, this 2.1 becomes plus infinity for this particular choice of gx. So, gx is an example of a function in L1 of 0 1, but it is not in L2 of 0 1. For example, L1 of 0 1 and L2 of 0 1 are different spaces, they are not equal, but L2 of ab is going to be contained in L1 of ab, a and b are finite real numbers. In particular, every L2 function from minus pi to pi is also an L1 function. So, if I take a function in L2 of minus pi pi, I can compute its Fourier coefficients and I can write down its Fourier series. Now the question is that will this Fourier series converge to f in L2? Surprisingly, the answer is yes. If a function f is in L2, then obviously it is also in L1, the Fourier coefficients are defined, the Fourier series are defined and the Fourier series converges to f in L2. The theory is nice and elegant. By the way, this inclusion is an exercise for you using the Cauchy-Schwarz inequality. Use the Cauchy-Schwarz inequality and prove this inclusion, L2 of ab is contained in L1 of ab, where a and b are finite real numbers. So, we want to study the convergence or the partial sums of the Fourier expansion with respect to L2 norm. So, let us now look at some preliminaries on L2. So, the space of functions g from ab to r for which the integral 2.1 is finite, integral mod gt squared dt that should be finite. That is denoted by L2 of ab, it is a vector space. It is evident that if f is in L2 and I multiply f by a constant, then cf will also be in L2. What is not immediately clear is if f is in L2 and g is in L2, then f plus g is also in L2. We have to show that fx plus gx mod squared has finite integral. So, let us just expand it mod fx squared plus mod gx squared plus 2 times mod fx mod gx. I am going to get less than or equal to because I apply the triangle inequality. And then apply the elementary inequality to ab less than or equal to a squared plus b squared. We get mod fx squared plus mod gx squared plus mod fx squared plus mod gx squared less than or equal to 2 times mod fx squared plus mod gx squared. Integral of mod fx squared is finite, integral of mod gx squared is finite. Therefore, integral of mod fx plus gx the whole squared is also finite. So, f plus g is also square integrable if f and g are square integrable. So, certainly L2 of ab is a vector space. We have to show triangle inequality. Triangle inequality is usually proved using the Cauchy-Schwarz inequality. I will leave it to you or consult any of your favorite books on real analysis. Now, let us discuss completeness of L2. The completeness of L2, the space L2 of ab endowed with the L2 norm is complete. L2 of ab is Cauchy-Complete with respect to the norm 2.1. What is the norm 2.1? Integral mod gt squared dt the whole thing to the power half. What are the last clause mean Cauchy-Complete mean? It means if gn is a sequence of functions such that norm gn minus gm goes to 0 as mn go to infinity, then there is a g in L2 of ab such that the sequence gn converges to g with respect to the L2 norm. We shall not prove this theorem 16 here. It is standard in real analysis courses. For example, you can see the book of Goffman and Pedrick functional analysis. What is the problem with continuous functions? Why do we have to work with Lebesgue integrable functions? Why cannot we work simply with continuous functions? Suppose if you take a continuous function on a closed bounded interval, it is surely integrable. Its square is integrable. So, what is the problem? Why not just work with c of ab? So, c of ab is a subset of L2 of ab and I can again endow c of ab with the same norm 2.1 namely integral mod gt squared dt square root. But the problem is that c of ab is not going to be complete. One can find a sequence gn of continuous functions such that norm gn minus gm goes to 0 as mn goes to infinity, but there is no continuous function g such that gn converges to g. There is a L2 function g such that gn minus g goes to 0, but that L2 function may not be continuous. So, Cauchy sequences may not converge if I restrict myself to continuous functions only. Cauchy completeness is very important. It is essential for many important existence results in analysis. So, we must insist on Cauchy completeness. It is exactly to ensure Cauchy completeness that we need to work with Lebesgue integrable functions not simply continuous functions. L2 of ab has one more important feature. It is a inner product space. What are the inner product of fng? The inner product of fng is integral from a to b f t gt dt equation 2.2 the displayed equation in the slide. Here we are assuming that fng are real valued. What happens if fng are complex valued in this formula 2.2? We must make a small modification. What is the modification? We must put a bar on this function g where the bar signifies complex conjugation. Now, when fng are equal we get norm g squared is the inner product of g with itself. So, it is an inner product space. As it always happens with an inner product space there are certain obvious and pleasant consequences. The parallelogram law norm f plus g squared plus norm f minus g squared is twice norm f squared plus twice norm g squared. The familiar parallelogram law and two elements fng in L2 are said to be orthogonal if the inner product of fng is 0. That is if the integral 2.2 vanishes of course, if g is complex you must put a bar. Third the triangle inequality norm f plus g less than or equal to norm f plus norm g. Fourth is an exercise. Take L2 of minus pi pi take L2 of minus pi pi 1 sin x cos x sin 2 x cos 2 x dot, dot, dot they are all in L2 of minus pi pi. They are mutually orthogonal to each other. So, it is an orthogonal system. They are pairwise orthogonal any two of them are orthogonal. That is an elementary exercise which is said to compute the integral. Now, let us come to another important example. Look at equation 2.3 take x squared minus 1 to the power n, n equal to 0, 1, 2, 3. But n is 0 it is simply the constant function 1. If n is 1, 2, 3, etc this is a polynomial of degree 2n. And this polynomial of degree 2n I am differentiating how many times n times. I am going to get a polynomial of what degree? Polynomial of degree exactly n. You divide this polynomial by 1 upon 2 to the power n n factorial. And this is a normalization. And with this normalization this polynomial is called the nth Legendre polynomial. Legendre polynomial prove that these Legendre polynomials are pairwise orthogonal in L2 of minus 1, 1. This Legendre polynomial is a system of orthogonal polynomials. This 1 upon 2 to the power n n factorial has been chosen in such a way that pn of 1 becomes 1. If n is odd pnx is an odd function. If n is even pnx is an even function easy to check. Now, the next problem is a beautiful exercise in Rohl's theorem. The polynomial pnx has n distinct real roots in minus 1, 1. This minus 1, 1 is a fundamental interval. In this fundamental interval this nth Legendre polynomial has n distinct roots. These roots of the Legendre polynomial are extremely important in numerical analysis. In numerical quadrature called Gaussian quadrature. There are many places where this Gaussian quadrature is discussed. My favorite is S Chandrasekhar, Radiative Transfer, Dover Publication New York 1960. I am a great fan of Chandrasekhar's writings. They are charming. His writings are always delightful. So, I would recommend S Chandrasekhar. You can read any book on numerical analysis and you will be able to find some discussion on Gaussian quadrature and how the zeros of the Legendre polynomial play a role there. The Legendre polynomials have many other properties. For example, in linear algebra you are familiar with the Gram-Schmidt's process. Take an inner product space. Take a vector space v with an inner product. Take a bunch of vectors v 1, v 2, v 3 etcetera and subject it to the Gram-Schmidt's process. If you take a linearly independent set of vectors, I am going to get an orthonormal system out of that. Now, let us do the following. Take L 2 of minus 1, 1. Take L 2 of minus 1, 1. It is a vector space. 1 x, x squared, x cubed, the monomials, they are linearly independent and they are all in L 2. Subject it to the Gram-Schmidt's process. What happens? What comes out of the Gram-Schmidt's process? The Legendre polynomials. So, the Legendre polynomials can be studied from a variety of different perspectives. There are many other types of orthonormal orthogonal systems of polynomials such as Shebyshev's polynomials, the Hermite polynomials, the Laguerre polynomials. You will see more information on this in this link that I provided to you in my home page. Look at the notes for this course MA 207 and you will find a lot more information on Legendre polynomials as well as other systems of orthogonal polynomials. Then comes least square approximations. A trigonometric polynomial. What is a trigonometric polynomial? A trigonometric polynomial of degree at most n is simply a linear combination alpha 0 plus summation j from 1 to capital N alpha j cos jx plus beta j sin jx. This is called a trigonometric polynomial of degree at most n. Of course, if alpha n or beta n, one of them at least is non-zero, then I will say that it is a trigonometric polynomial of degree exactly n. I say at most n because it may happen that alpha n may be 0 and beta n may be 0 and this sum simply goes from 1 to n minus 1. Suppose if f is in L2 of minus pi pi, then as I said I can calculate the Fourier series for f and I can do the partial sums of the Fourier series and these partial sums are examples of trigonometric polynomials. So, the nth partial sum Sn fx is a trigonometric polynomial of degree at most n. Now what we do is the following. We take two trigonometric polynomials of degree at most n. One of them is Sn fx and the other one is a general qnx and I find out these norms. That means, if I try to approximate f of x by a trigonometric polynomial of degree at most n, one such approximate, one such possible approximation is the nth partial sum Sn fx and you compare the error over here with the error caused by choosing any other trigonometric polynomial of degree at most n. The remarkable thing is that of all possible trigonometric polynomials of degree at most n, if I select Sn fx, then the error will be the least. This is why it is called the least square approximation. So, the theorem says that Sn fx is the best approximation among all trigonometric polynomials of degree at most n, least square approximation. The proof is very easy as a matter of fact. So, that theorem has been stated very clearly here. f belongs to L2 of minus pi pi and qnx is a trigonometric polynomial of degree at most n. Then norm fx minus Sn fx is less than or equal to norm fx minus qnx. Equality holds if and only if qnx equal to Sn fx. Let us look at the trivial equation fx minus qnx, add n subtracts, add n subtract Sn. So, fx minus Sn fx plus Sn fx minus qnx. This difference Sn fx minus qnx, I am simply calling it rnx. Obviously, rnx is a trigonometric polynomial of degree at most n. We shall prove that fx minus Sn fx, this first piece is orthogonal to 1 sin x cos x, sin x cos nx. If it is orthogonal to each of these, then it will be orthogonal to linear combinations of these. In other words, this will be orthogonal to every trigonometric polynomial of degree at most n. In particular, it will be orthogonal to rnx. Let us prove that. Well, cos jx is orthogonal to 1 sin kx cos kx when k not equal to j. That is a first observation. That is a trivial observation. Cos jx is orthogonal to 1. It is orthogonal to sin kx as well as cos kx k not equal to j. So, when I take fx minus Sn fx and multiply it with cos jx, what is it going to be? It is going to be fx times cos jx integral, but that is pi times aj from the definition of Fourier coefficient minus integral minus pi to pi Sn fx cos jx. But the nth partial sum contains all sorts of terms, but they will all be orthogonal to cos jx except the jth term. What is the jth term? Aj cos jx. That is the only thing that will survive in this integral. And so, what you get is aj cos square jx. The aj comes out of the integral, the cos square jx you have to integrate, multiply and divide by 2. 2 cos square jx is 1 plus cos 2 jx. The integral of cos 2 jx from minus pi to pi is 0. We simply get 0. So, fx minus Sn fx is orthogonal to cos jx for every j from 1 to n. And so, it is going to be orthogonal to linear combination of 1 cos x cos 2 x that are the cos nx sin x sin 2 x sin nx. So, it is orthogonal to Rnx. And so, by the Pythagorean identity, we will get norm fx minus qn x squared equal to norm fx minus Sn x squared plus norm Rn x squared. What are the Pythagorean law? If 2 vectors in a vector space, if u and v are 2 vectors in a vector space and if they are orthogonal, then norm u squared plus norm v squared is norm u plus v the whole squared. Because fx minus Sn x is orthogonal to Rnx, norm of fx minus Sn x squared plus norm Rn x squared is the same as the norm or the sum. The sum is exactly fx minus qn x squared. All right. So, we see at once that norm fx minus qn x larger or equal to norm fx minus Sn x. Why? Because I am knocking off this term non-negative term. So, I am going to get an inequality. So, that is the first important observation. And when will equality hold? This inequality will be equality if and only if Rnx is 0. Rnx is 0 if and only if qn x equal to Sn fx. And so, the least square approximation has been established. So, now what we want to do is that we want to deduce some corollaries from this least square approximations and the first corollary will be the Bessel's inequality. And from Bessel's inequality, we are going to get another relation called the Parseval formula. The Parseval formula will tell you that if A0, An, Bn are the Fourier coefficients of an L2 function, then equality will hold. Of course, you will ask why not directly prove the Parseval formula? Why are we proving the inequality first and then proving equality later? This will be a stepping stone for proving the Parseval formula. That is why we are proving this inequality first. The Bessel's inequality comes first and we are going to use the Bessel's inequality to prove the Parseval formula. The least square approximation that we have proved will play a crucial role in establishing the Parseval formula. We will continue this in the next capsule. I think it is a good place to stop here. Thank you very much.