 Welcome to lecture 35 on measure and integration. In the previous lectures, we had been looking at the pth power integrable functions on a measure space xs mu and we have studied some general properties of these function spaces. Today, we will look at in slightly more in detail the special subspace, namely when p is equal to 2. So, we will look at today. The topic for today's discussion would be the space L2 xs mu, the space of complex valued square integrable functions. This is a special space in the sense that this can be viewed as a generalization of the Euclidean spaces. Recall, in Rn, the Euclidean space, we have the notion of dot product of vectors, which is related to the magnitude of the vectors and also help us to define the notion of angles on Rn. Let us see what we can do as far as L2 is concerned. We look at the space L2 of xs mu to be the space of all functions, which are defined on x, which are complex valued such that mod of f square d mu is finite. For such spaces for a function f in L2 of x, we have defined the norm of this, which is equal to integral of mod f square d mu raise to power 1 by 2. We said that this is very much like the magnitude in Rn. So, this is similar to magnitude in Rn. Let us just briefly recall what was the magnitude in Rn. For a vector say x in Rn, the Euclidean norm is defined as look at the components. If x is having components x1, xn, then look at mod of xi square sigma 1 to n and the square root of that. For a function f in L2, let us regard f as a vector with fx as the component. This is the xth component of f. How would we define the norm? Look at the xth component fx. Look at square. This is very much similar to what we have done for Euclidean. Look at the component square. Now, sum it up, but here the summation is over x and x is any x belonging to x, any indexing set. So, it is nothing but integral d mu and then the power 1 by 2. In that sense, this is a perfect generalization of the Euclidean norm to arbitrary spaces. Now, on L2, we have the notion of for a vector x belonging to Rn with components xn. We have the notion of what is called the dot product. So, dot product if x and y are two vectors with y components as yn, then x comma y, the dot product is defined as xi yi, i equal to 1 to n. If we are in a complex plane, so if x and y belong to cn, then the dot product xy is defined as sigma xi yi bar i equal to 1 to n. This notion of dot product is related to the magnitude in the following way that in either Rn or cn, either one, the norm of x square is equal to the dot product of x with itself. We know that the dot product in Rn or cn gives a notion of angle and orthogonality, which helps us to do geometry in Rn. So, the basic idea of today's lecture would be on the space L2 of x, we already have the notion of norm, the notion of distance. We will define the notion of inner product or the notion of dot product on L2 of x and show how it is related to the notion of distance. So, that helps us to define the notion of orthogonality or perpendicularity of two elements in L2. So, we can do geometry in L2 of x as mu. So, let us define what is the notion of the dot product in L2. So, for functions f and g in L2, keep in mind our spaces are complex valued. So, define the dot product of f with g or also called the inner product of f with g as integral of fx g bar x d mu x, where this g bar is the complex conjugate of the function g. So, g bar of x is gx bar. So, this is defined, so the inner product or the dot product between f and g for two functions f and g is defined as the integral of fx gx bar d mu x, which is perfectly similar to in complex c to the power n a i b i bar. So, the first thing we want to say that this is well defined and that follows from the holders inequality that we had proved. Recall holders inequality that said that if f is a function in Lp and g is a function in Lq, then f into g is a function which is integrable and the integral of fg is less than or equal to the L2 norm of. So, here is a typo. So, let me just say it once again what we are meaning here, that should be integral. So, for holder inequality we had that if f belongs to Lp and g belongs to Lq, then fg belongs to L1 and integral mod fg integral of mod of f into g d mu is less than or equal to the pth norm of f and the qth norm of g. So, for p equal to 2, so that will give us that integral of fg d mu and p is equal to 2, q is equal to 2. So, we have got that 1 over p plus 1 over q is equal to 1. So, the holder inequality will give us that this is less than or equal to L2 norm of f into L2 norm of g and this is nothing but our, so this is nothing but this is bigger than or equal to fg inner product is less than or equal to this. So, that gives us the Cauchy Schwarz inequality. So, this is also called the Cauchy Schwarz inequality namely the inner absolute value of the inner product between f and g is less than or equal to the norm of f into norm of g. So, that says that this inner product is well defined, it is a well defined quantity. So, for every f and g in L2, we have the notion of the inner product. So, f into f comma g which is the inner product is well defined quantity and this behaves perfectly similar to that of the inner product for the ordinary vectors in R n or C n. That means, this is a function which is defined on L2 cross L2 and has the following properties namely the inner product of f with f is always bigger than or equal to 0 and the equality holds if and only if f is equal to 0. So, let us look at this property that how is this true? So, inner product of f with itself is nothing but integral of mod f square d mu. So, this is always bigger than or equal to 0 and this will be equal to 0 if and only if our function mod f is equal to 0 almost everywhere. So, if and only if f is belonging to L2 as a treated as an element of L2, f is equal to 0. So, the first property is obvious namely the dot product of f with itself is always bigger than or equal to 0 and the second property says that the dot product or the inner product of f with g is same as the inner product of g with f. So, we are interchanging f and g and the complex conjugate of it. So, the inner product of f with g is same as the complex conjugate of the inner product of g with f and that is quite simple to verify from the definition. So, if we have got the inner product of f with g, so the inner product of f with g is equal to f g bar d mu and that is equal to, so inner product of f with g is equal to f g bar and that is equal to f bar g integral d mu bar. So, that is equal to and this is equal to g comma f bar. So, that verifies the property namely the inner product of f with g is equal to the inner product of g with f bar. Similarly, it is easy to verify using that the integral is linear, it is easy to verify that the inner product is linear in the first variable that means, alpha f plus beta g inner product with h is equal to alpha times the inner product of f with h plus beta times the inner product of g with h. Similarly, in the second variable this is conjugate linear because of the property of 2. So, f inner product with alpha g plus beta h is same as alpha bar of f g inner product f g plus beta bar of inner product of f h. So, in the second variable it is not linear it is a conjugate linear and finally, this property is obvious namely the L 2 norm of f is square root of the inner product of f with itself. So, all the properties that we have for the dot product in R n or C n are defined for the inner product in L 2. This is not very special for L 2. In fact, one can look at any vector space h over the field of real or complex numbers and if one has a function which is defined on h cross h taking values in the underlying field of real or complex having properties similar to that 1, 2, 3, 4, 5 of L 2. So, one can define what is called a inner product space. So, in general a inner product space is defined to be a vector space h on which there is a notion of inner product defined and what is a inner product it is a function defined on h cross h to R with those properties. So, once we have one has a notion of inner product that gives rise to notion of magnitude by the property that the magnitude of a vector u in h is nothing but is defined as the dot product of u with itself square root. And one verifies that Cauchy Schwarz inequality holds for this kind of inner product and that means this is a well defined norm. So, Cauchy Schwarz inequality will say that is a norm defined on it and once you have the notion of norm that gives rise to a metric on the underlying vector space and one can ask whether it is complete under that metric or not. So, if on a vector space inner product is defined. So, it becomes a inner product space and inner product space gives rise to a norm and if the underlying metric induced by the norm is complete one says h is a Hilbert space. So, that is the general definition of a Hilbert space. So, our L 2 is an example of a Hilbert space because L 2 x s mu is a vector space on which a notion of norm the L 2 norm is defined and that L 2 norm is related to the inner product just now we have seen and we have already seen as Rie's Fisher theorem which said that L 2 x s mu is a complete metric space in the L 2 metric. So, this is an example of a Hilbert space. Now, for once we have the notion of the inner product one can define the notion of two elements in L 2 to be orthogonal or perpendicular to each other. So, we say two elements f and g in L 2 are orthogonal to each other if the inner product between them is equal to 0 and that is what we have for vectors in R n also that the dot product is equal to 0. So, and we write this as f perpendicular to g. So, f perpendicular to g is defined as saying that the inner product of f with g is equal to 0. Now, we can also define the inner product f element f orthogonal to a subset s. So, writing it as f orthogonal to a subset s means that f is perpendicular to every element of s. So, f is perpendicular to s will mean that f is f comma h the inner product is equal to 0 for every element h of h in s. So, similarly we can define orthogonality of two sets also. With this one can prove what is called the Pythagoras identity namely if f and g are two functions in L 2 and f is orthogonal to g then the norm square of f plus g is equal to norm f square plus norm g square. So, let us just quickly verify the Pythagoras identity namely if f and g are two elements in L 2 and f is orthogonal to g then we want to compute the L 2 norm of square of this. By definition this is related to the inner product. So, this is inner product of f plus g with itself and now using the property of linearity what we will get is this is f comma f plus g comma f plus f comma g plus the inner product of g with itself. So, that gives you norm of f square plus norm of last term will give you norm of g square, but f is orthogonal to g that means f comma g is equal to 0 and g comma f is also equal to 0. So, these two terms the inner product of g with f and f with g both are equal to 0. So, we get what is called the Pythagoras identity namely the norm of f plus g square is equal to norm f square plus norm g square whenever f is orthogonal to g. Let us carry over this idea of orthogonality a bit further. So, let us take as any non-empty subset of L 2. We call as a subspace of L 2 those who have done a bit of linear algebra will recognize this definition L 2 is a vector space. So, we are looking at a vector subspace of L 2. So, a set S is a non-empty subset is called a subspace of L 2 if for any f and g in f and g in S. So, this should be in S not in L 2 and alpha beta in complex number alpha f plus beta g belongs to S then we say S is a subspace. So, this is not L 2 that is S. So, that means for any two elements alpha beta in S the linear combination alpha f plus beta g should be in S in that case it is called a subspace of S and a subspace of S is called a closed subspace if it is closed under the metric on L 2 that is L 2 metric. So, it should be a closed set that means what that means whenever we have got a sequence f n in S and f n converges to a function f in L 2 norm then the limit must also be inside S. So, in that is what is the definition of a closed source subspace it is a subspace and it is a closed set under the L 2 metric. So, that is the notion of a closed subspace for given a set S in L 2 will denote by S upper suffix perpendicular. This is also called the orthogonal complement of S to be all elements in the space L 2 which are perpendicular to all elements of S. So, given a set S we are looking at all elements in L 2 which are orthogonal to every element of S. So, that is called S perpendicular and this is called the orthogonal complement of S and the claim is that this is a closed subspace of L 2. So, let us verify this fact that S a subset of L 2 and S perpendicular is the set of all elements f in L 2 such that f perpendicular to H for every H in S. Then claim is that first of all S perpendicular is a subspace S perpendicular is a subspace. So, let us take some H and G belonging to S perpendicular and alpha and beta belonging to C. Then alpha H plus beta G comma let us take an element f for f in S perpendicular for F in S. This will be equal to alpha times H f plus beta times G f using the property of linearity in the first variable for the inner product. Now, because f belongs to S and H and G are in S perpendicular, so this quantity inner product is 0 and the second inner product is 0. So, this sum this inner product is equal to 0. So, that says if H and G belong to S perpendicular and alpha and beta are in C, then alpha H plus beta G is always orthogonal to every element of S. Hence, it belongs to S perpendicular. So, S perpendicular is a subspace of. The next, let us prove that this is a closed subspace. So, next we want to check that it is a closed subspace. So, let f n belong to S perpendicular and f n converge to f in L 2. We want to check, so claim that f belongs to S perpendicular. So, for that, let us take any element H belonging to S and compute, we want to compute f comma H and the claim is this f comma H. So, this is equal to f comma limit n going to. So, what is H? This is not true. Let H belong to S. Now, the claim is that since f n converges to f in L, converges to f in L 2. So, this is equal to limit n going to infinity of f n comma H. So, this is a very simple thing to verify, because if we look at the difference of the two. So, if we look at f H, so why is this true? This is true because we look at this inner product of f with H and inner product of f n with H and look at the absolute value of this. Then we can write this as, so this quantity is equal to absolute value of f minus f n inner product with H and with this by Cauchy-Schwarz inequality, this is less than or equal to L 2 norm of f minus f n and L 2 norm of H. This goes to 0. So, we get therefore, f with H inner product is equal to limit n going to infinity inner product of f n with H. Since each f n belongs to S perpendicular, H is in S. So, that implies that each term is equal to 0. So, this is equal to 0 for every H belonging to S. So, that implies that f belongs to S perpendicular. So, this proves that S perpendicular for any set S, if we look at its orthogonal complement, then that is a close of space of H. So, this is also called the orthogonal complement of set. We next state an important result which seems geometrically obvious, which can be proved for any Hilbert space. So, we will just look at it for our space L 2. We will not prove this result. We will just assume this result and the proof can be referred to in the book. So, the result says that if f is a L 2 function and S is a close subspace of L 2, then look at the number alpha, which is an infremum of all the distances, L 2 distances of f from G, where G is any element in S. Then, the theorem says that this infremum is attained at some point in S. So, that means there exists not only it is attained, then there is a unique function f 0 belonging to S such that this infremum alpha is equal to norm of f minus f 0. Further, if this f does not belong to S, then look at the difference of f minus f 0, that is always going to be perpendicular to S. So, that is the claim of the theorem. We will not prove it, we will just geometrically analyze this result a bit. So, here is a close subspace S. So, look at the close subspace S of L 2. This is a close subspace of L 2 and we have got a function f, which is outside it. So, what we are going to do is, we are going to look at any point inside S, A point G and look at the L 2 distance of this. So, look at the L 2 distances of various points. So, it says that there is a point which is there is a value called f 0 such that this distance L 2 distance of f from it is the minimum. So, this is 0. So, look at f minus f 0 and this says if I look at f minus f 0, that is going to be orthogonal to it. So, that is the theorem. So, there exists a unique point f 0 belonging to S such that alpha the infremum is equal to the distance of f minus f 0 and f minus f 0 is perpendicular to S. So, look at this vector f minus f 0, that is always orthogonal to this S. So, geometrically in a sense, given a point and given a subspace, it is kind of the projection. The projection gives you the minimum. So, this is the generalization of the projection theorem for finite dimensional spaces. So, this is also called the best approximation theorem for Hilbert spaces. So, let us just say look at once again. It says that if you are given a close subspace of L 2 and look at and you are given a function f in L 2, look at alpha the minimum of the distances between f and elements of G says there is a value. There is a function in S which where this value is attained and as a consequence of this, it also says that if S is the proper close subspace of L 2, then its perpendicular cannot be 0 because there is a if it is a proper, then there is an element f minus f 0, which is not 0, which is orthogonal to it. So, as a immediate consequence that if S is a proper subspace proper close subspace of L 2, then its orthogonal complement cannot be 0. It has to something else. So, that also means that if the orthogonal complement of something is 0, then S must be equal to L 2. Another way of stating the same thing is this. So, as I said, we will not be proving this theorem, but we will give some applications of this today. So, let us look at some properties of orthogonal complement before we go on to prove some general facts. So, let us take S 1 and S 2 be subsets of L 2, just subsets, then the following properties hold namely S 1 perpendicular is a close subspace that we have already shown and S 1 perpendicular and S 1 they intersect only at the most at 0. They are just sets S 1 perpendicular is a subspace, because S 1 may not be a subspace. So, it says if S intersection S perpendicular is always inside 0 and that is obvious, because if f belongs to S 1 intersection S 1 perpendicular, then that means the inner product of f with itself, because f belongs to S 1 and it also belongs to S perpendicular that must be equal to 0. So, that implies norm of f is equal to 0 and that implies f must be equal to 0. So, it says so hence S 1 intersection S 1 perpendicular is inside 0. So, as obvious consequence, if S 1 is also a subspace, then S 1 intersection S 1 perpendicular, they do not have anything common other than the vector. The second property says that if S 2 is a subset of S 1, then S 1 perpendicular is a subset of S 2 perpendicular and that is obvious, because if we take any element say H in S 1 perpendicular, then H the dot product or inner product of H with every element of S 1 is equal to 0 and in particular with S 2 is equal to 0. So, that also belongs to S 2. So, this property is obvious. The third property says that S 1 is a subset of S 1 perpendicular. So, orthogonal complement of the orthogonal complement always includes S 1. So, that property is again obvious, because if we take f belonging to S 1 and H belonging to S 1 perpendicular, then we know that. So, then it implies H comma f the dot product is equal to 0, because f belongs to S 1 and H belongs to S 1 perpendicular and that is equal to 0, but that means H is perpendicular to f and so that means, so this means that for every H in S perpendicular f comma H or H comma f is equal to 0, that means f is belonging to S 1 perpendicular perpendicular. So, S 1 is always a subset of S 1 perpendicular perpendicular. We want to show that, so in case S 1 perpendicular perpendicular equal to S 1, in case these two are equal, then the left hand side is an orthogonal complement of a subspace. So, this is a close subspace. So, this is a close subspace implies S 1 is a close subspace. So, implies S 1 is a close subspace and let us prove the converse part, namely the converse is also true. So, suppose S 1 is equal to S 1 perpendicular perpendicular, then the claim that S 1 is a subspace, S 1 is a close, then S 1 is a close subspace. So, that we have already shown, we want to prove the other way round, so this is not what we want to is. So, suppose S 1 is a close subspace, so converse is if S 1 is a close subspace, then we want to show S 1 is also equal to S 1 perpendicular perpendicular. So, to prove this, suppose not. So, let us take, let there exist f belonging to, we want to, this is a subset of this anyway. So, let us assume there is a S 1 perpendicular perpendicular f naught in S 1. So, let us assume that is so. So, in that case, let us apply our best approximation theorem. So, implies by the theorem, just now we stated which we did not prove that there exist an element f naught belonging to S 1 such that f minus f naught is perpendicular to S 1. So, that means f minus f naught belongs to S 1 perpendicular, but also we have got f minus f naught belongs to, so there is a element f is in not in S 1, so there is element in f naught in S 1 such that the difference is perpendicular to S 1. Now, let us observe that S 1, now let us observe that this element f naught belongs to S 1, f naught belongs to S 1, which is contained in S 1 perpendicular, perpendicular. So, S 1 which is belonging to S 1, so that implies, so we have got f belonging to S 1 perpendicular perpendicular and f naught also belonging to, that means f naught S naught, f minus f naught belongs to S 1 perpendicular perpendicular and f naught is also in the same thing and this being a subspace, the difference must also belong to, this belongs to S 1 perpendicular perpendicular. Now, the element f minus f naught belongs to S 1 perpendicular perpendicular and it also belongs to S 1 perpendicular. So, from here and here it follows, so it belongs to a subspace and orthogonal complement of it, that means f minus f naught must be equal to 0, implying f is equal to f naught. So, that means what, that means f belongs to f naught f and where is f naught, f naught is in S 1, so this f also belong, so we would start with an f in S 1 perpendicular perpendicular and we are getting that f is equal to S naught, where S naught in element is S 1, so that implies that f belongs to S 1. So, what we have shown is, whenever f belongs to S 1 perpendicular perpendicular, it also belongs to S 1, so these two are equal. So, that proves the fact that if S 1, that proves the fact that if S 1 is equal to S 1 perpendicular perpendicular, then S 1 is a closed subspace of it. And next, let us observe the fact that if S 1 and S 2 are two closed subspaces and S 1 is perpendicular to S 2, then S 1 plus S 2 is also a closed subspace. So, let us observe that S 1, S 2 are closed subspaces and S 1 is perpendicular to S 2. So, let us take two elements, so let us look at S 1 plus S 2. We want to show it is a subspace, so let us take an element say f plus g in S 1 plus S 2, where f belongs to f 1 g 1, where f 1 belongs to S 1, g 1 belongs to S 2 and let us take another element f 2 plus g 2 also belonging to S 1 plus S 2, where f 2 belongs to S 1 and g 2 belongs to S 2. Then for every alpha and beta, alpha times f 1 plus g 1 plus beta times f 2 plus g 2 is equal to alpha f 1 plus beta g 2 plus alpha g 1 plus beta g 2. And now, because S 1 is a subspace, so alpha f 1 plus beta alpha f 1 plus beta f 2, from here it was beta f 2 and alpha g 1 plus beta g 2. So, this element belongs to S 1 and this element belongs to S 2, so implies that this element belongs to S 1, so this element belongs to S 1 plus S 2. So, that proves that S 1 plus S 2 is a subspace. To prove it is a closed subspace, let us observe, so let f n plus g n belong to S 1 plus S 2, where f n belongs to S 1 and g n belongs to S 2 and f n plus g n converge to f in L 2. So, we want to show that S 1 plus S 2 is closed, that means we have to show that f belongs to S 1 plus S 2, so that is what we have to show. Now, let us observe that f n plus g n being convergent is Cauchy. So, f n plus g n is a Cauchy sequence is a Cauchy sequence. So, let us look at f n plus g n minus f m plus g m. So, this norm goes to 0 as n and m go to infinity, but now note that f n minus f m belongs to S 1 and g n minus g m belong to S 2. So, this implies by Pythagoras theorem that norm of f n plus g n f n minus f m square plus norm of the norm of f n minus f m square plus norm of g n minus g m square, this is equal to norm of f n plus g n minus f m minus g m square. So, that is by Pythagoras theorem and this goes to 0. So, that implies that norm of f n minus f m goes to 0 and norm of g n minus g m goes to 0. So, meaning what this says that f n itself is Cauchy and g n itself is Cauchy. So, that implies that f n is a Cauchy sequence is Cauchy implying that f n must converge to some h. Similarly, g n must converge to some g. All in L 2. Similarly, that implies that f n plus g n converges to h plus g. We know that this converges to f, so that implies that f is equal to h plus g. Now, note that f n is a sequence in S 1 and S 1 is closed. So, this h belongs to S 1 and g belongs to S 2. So, this belongs to S 1 plus S 2. So, this completes the proof that if S 1 and S 2 are closed subspaces and S 1 is perpendicular to S 2, then S 1 plus S 2 also is a closed subspace. Finally, we claim that if S 1 is a closed subspace, then we know that S 1 intersection, S 1 perpendicular is 0. So, in that case L 2 is equal to S 1 plus S 1 perpendicular and the reason for that is because this intersection is 0. So, there cannot be S 1 plus S 1 perpendicular is a closed subspace. So, there has to be an element. If it is not whole, then there must be an element outside which is not true. So, that means for every closed subspace S 1 of L 2, L 2 can be expressed as S 1 plus S 1 perpendicular. That means every element of L 2 can be represented as an element in S 1 plus an element in S 1 perpendicular and this decomposition will be unique because S 1 intersection S 2 is S 1 perpendicular, S 1 is a subspace. So, there is nothing common between them. So, this is also called sometimes the projection theorem. That means for every closed subspace S 1 of L 2, L 2 can be represented as S 1. One writes as a direct sum of S 1 perpendicular, namely these two are equal and the intersection of these two subspaces is equal to 0. Let us come to analyzing maps on the space L 2 which is a vector space. As on any vector space, one can analyze linear maps on the vector space taking values in the underlying field. Here, our vector space is L 2. Actually, it is a Hilbert space. So, look at a map which is a linear map T from L 2 to C. We say it is a boundary linear functional if it has the following properties. First of all, it should be linear as a. So, T is a linear map as a vector space L 2 to C. Secondly, we want that it is bounded in the sense that if there is a constant m such that norm of T f is less than or equal to m times the norm of f 2. So, this is called the boundedness of the linear map T. So, we say T is a bounded linear functional if T is linear on L 2 and absolute value of T f is less than or equal to a constant m times f, where m is a constant fixed and this happens for every f in L 2. It is quite easy to this condition boundedness is actually it implies that T is also continuous because if f n converges to f, then T f n absolute value is less than or equal to norm of m times the norm of f n minus f and that will go to 0. So, it is easy to verify that T which is if it is bounded, if a linear map is bounded if and only if it is continuous and because you are on a vector space continuity at 0 is enough. So, one can verify easily that every bounded linear map is continuous at 0 is equivalent to it. One way of defining a linear maps continuous linear maps is the following. Well, for fix any g in L 2 and look at the map T lower g defined on L 2 to be T g at f is equal to f, g for every f. So, that means that the value of T g at f is defined as f g inner product of f with g for every f and it is easy to see that this is a linear map because the inner product is linear in the first variable. So, that will give that it is a linear map and it is bounded because of the Cauchy Schwarz inequality. So, this is linear and by the Cauchy Schwarz inequality it is a bounded linear map. So, one way of constructing bounded linear functionals on L 2 is by taking the inner product of any element f with a fix element g and this is important theorem called these representation theorem which says that this is the only way of constructing bounded linear functionals on L 2. So, it says that T if T is any bounded linear functional then there is a unique g 0 belonging to L 2 such that T f is equal to f g naught. That means every linear functional on f or linear functional T on L 2 arises via inner product of f with a fix element g 0 and this g 0 is also unique. Let us just outline the proof of this. So, first of all let us observe that there are two cases look at suppose that there does not exist any g 0 with the required claim then what will happen then g 0 must belong to what is called the kernel of T of g. So, that means what and kernel of g is all elements say that which are mapped to 0 and this is a closed subspace of kernel of bounded linear functional is a closed subspace. So, if there is no g then this will be so that means that our required claim will hold with T equal to 0. So, that is essentially saying the kernel of T g is a closed subspace of g and if one possibility is kernel of T g is equal to the whole space then it is equal to 0 because if kernel of T g is equal to the whole space then T g will be identically 0. So, any g 0 will satisfy 0 equal to 0 will satisfy. So, let us assume that the kernel of T g 0 is a proper closed subspace of it then by the best approximation theorem g 1 there is a g 1 in kernel of T perpendicular that is the consequence of the best approximation theorem and thus g 1 will not be equal to 0. So, in that case one verifies that if you take g 0 to be equal to T g 1 divided by g 1 comma g 1 into g 1 if we this selection of g 0 is a required unique function such that T of f is equal to f g 0 for every f in L 2. So, essentially one applies the best approximation theorem to get an element g 1 in kernel of T perpendicular and why one is looking at kernel in g in kernel of T perpendicular is because if the required condition is to hold then that function g 0 has to belong to kernel of T perpendicular because if it is g comma f and that means for f in kernel that must be 0. So, that means the required g 0 has to be from here. So, let us pick up any element and then modify it and show that that is the required. So, this is what is called the Ries representation theorem. So, this is the Ries representation theorem. So, today what we have looked at is the space L 2 is a perfect generalization of the space of the R n or the space C n that means there is a notion of a inner product defined on it which gives which is related to the norm and which gives the notion of perpendicularity. So, we prove we stated one important theorem namely if s is any close subspace of L 2 and you take an element f in L 2 then there is a best approximator then there is an element g 0 in the space close subspace which best approximates with a minimum distance from f and as a consequence of this one consequence is the projection theorem namely every close if s is any close subspace of L 2 then L 2 is a direct sum of s plus s perpendicular and the second consequence is characterizing all bounded linear functionals on the Hilbert space L 2 namely linear the only way bounded linear functionals can be constructed on L 2 is by the inner product that means t of f if t is a bounded linear functional then t of f must be equal to the inner product of f with an element g 0 for some element g 0 with the inner product. So, that is characterization of bounded linear functionals. Thank you.