Lecture 14: Basic Hilbert Space Theory

Key Takeaways

foreign okay so uh last time we introduced pre-hilbert spaces as Specter spaces uh that come equipped with a hermitian inner product our mission inner product is linear is uh so it's a pairing between two elements that gives you a complex number that's linear in the first entry um [Music] and if you switch the entries that's equal to the pairing of the original two entries uh taking the complex conjugate of that and it's positive definite meaning if I take the inner product of a vector with itself it's not negative and zero if and only if um the element is zero and we proved the koshy Schwartz inequality so let me actually just recall that we defined for if H is a pre-hubert space we defined what we called uh or using this Norm notation although I haven't proved it to Norm yet to be the inner product of V with itself um raised to the one-half power and this is a non-negative real number uh this is a non-negative number so taking the one-half power is Meaningful um and we proved at the end of last time the koshy Schwartz inequality that uh for all you and B and H if I take the inner product of u and v take its absolute value this is less than or equal to normal View times the norm of V okay so now let's uh use this to actually prove or to prove that this thing that I've been uh denoting what this Norm notation is in fact a norm on a pre-hilbert space so theorem with h is a the Hilbert space then this uh thing here defined in this way is a norm on H okay so remember we have to prove to or we have to prove three things for for something to be a norm we have to prove it's positive definite and then we also have to prove homogeneity and the triangle inequality so um note that this quantity here equals zero if and only if the inner product of V with itself is zero which by the positive definite quantity of a hermitian inner product implies that V equals zero so that proves that this function here on H is in fact positive definite um now if Lambda is a complex number V is in h if I take so we kind of saw this at the uh and the proof of the koshy Schwartz inequality but if I take Lambda times V an inner product Lambda times V this is equal to Lambda times Lambda bar a scalar pulls out of the first entry unphased the second if a scalar is in the second entry it comes out with a complex conjugate and then of course for a complex number times this complex conjugate I get uh the norm of that or the length of that the absolute value of that complex number squared so taking the square root of both sides of this which is equal to this I get that the ABS the this quantity here is equal to the absolute value of Lambda times this Norm of V which proves homogeneity of dysfunction on H okay so that all All That Remains to prove that this is an actual Norm so I don't have to keep being uh or at least trying to be um careful with what words I'm using um will prove that this is a norm uh okay so now we need to prove the triangle inequality yeah you let UV be an H then I compute U plus v squared this is equal to U plus v u plus v which is equal to um just how we computed it for uh you know when we had a t here and a t here when we did the proof of the koshy Schwartz inequality and we'll use this identity quite often that the norm of U plus v squared is Norm U squared plus Norm B squared plus two times the real part of u and v the real part of the inner product of u and v okay now this is less than or equal to RM squared plus v squared plus the absolute value of the real part now the absolute value of the real part of a complex number is less than or equal to the absolute value of that complex number so this is less than or equal to the absolute value of the inner product of u and v and by koshy Schwartz that let that is less than or equal to so still what I have before plus 2 uh times the inner product of u and v and um I guess let me go to the next four oh the proof is essentially done what I had before is equal to normal View plus normal V squared I.E so I had started off with the norm of U plus v squared and I had proved that's less than or equal to this thing here squared so taking the square root of both sides I get uh the norm of U plus v is less than or equal to the normal view plus the normal V okay so this thing that I defined before is in fact a norm and I'm not just using a that notation to denote an imposter foreign using the koshy Schwartz inequality we can also prove that taking the inner product is [Music] um you know it's a function on H cross h that's continuous Okay so let me label this as continuity of the inner product so let me state it as the following if we're in a pre-hilbert space and un converges to U and VN converges to B in a pre-hilbert space with Norm as defined uh as before right so now we have a norm on a pre-hilward space and so we can Define convergence uh since it's a norm space um free Hilbert space H would Norm this then you and inner product VN converges to you and our product fee okay um all right so why is this so last time and the previous proof I mean you know we really didn't use the full strength of the coachy Schwartz inequality we could have just gotten by with the real part uh having just proven that the real part of the inner product um is an absolute value less than or equal to the inner product of or the product of the Norms of u and v here we'll actually use the full koshy Schwartz inequality so the proof is quite simple If U N converges to U and VN converges to B I.E let me just spell this out for you u n minus U converges to zero and Norm of VN minus B converges to zero as n converges to as n goes to Infinity then um we'll use the squeeze theorem to show that this quantity converges to U in a product with UV so I have to show that this quantity here let me not put that we're not in 18 100. typically in 18100 when I first started teaching the squeeze theorem I always include the lower bound but it should be clear that this is always less than always bigger than or equal to zero since it's absolute value or of a complex number um okay so this is equal to u n minus u v n plus um let's see U and a product v n minus V and the triangle inequality now just for the you know modulus or absolute value of complex numbers that's less than or equal to u n minus U being Plus absolute value of U v n minus V okay and this is less than or equal to by the koshy Schwartz inequality um Norm of u n minus U times the norm of VN plus Norm of U is Norm of VN minus V and now um the VNS are so the VNS are converging to V and therefore the Norms of the VNS are converging to the norm of B right so um recall that uh active VN converges to V then form of VN converges to Norma B okay all right um and again so let me add since one can prove as you do in real analysis the following reverse triangle inequality since if I have two vectors A and B in in H so this just works in any button space not necessarily A pre-operate spaced the absolute value of the difference in Norms is less than or equal to the norm of the difference okay so this is a convergent uh uh sequence of real numbers so it must be bounded so I can say that's less than or equal to times soup of n or let's use a different letter k times Plus form of v n minus V times Norm of U now this is a this is something converging to zero times a fixed number this is something converging to zero times a fixed number and therefore this goes to zero as n goes to infinity and therefore uh the thing which we started with so this quantity here an absolute value is less than or equal to something converging to zero and it's also not negative so therefore I get that so this is by The Squeeze theorem I.E okay okay so that's uh so on uh pre-hilbert space we can define a norm using the inner product and this inner product is continuous with respect to this Norm okay because once we have a norm we have convergence so we can talk about uh once we have a norm we have a notion of distance so we can talk about convergence of sequences and you know uh things like that so now um a pre-hilbert space is what is a Hilbert space this is simply a pre-hilbert space which is complete with respect to this Norm so a a Hilbert space h is a pre-hilbert space which is complete with respect to the norm uh with respect to this Norm which again I remember was defined as I stick a vector the norm of a vector here is equal to the inner product of the vector with itself raised to the one-half power all right so this is the new terminology and as we'll see um or at least we'll explicitly spell out in the in a form of a theorem but we will see probably by the end of this lecture that really there's for um a reasonable Hilbert space you're either one of two things so let me first give the examples the basic examples of Hilbert space so for example c n which is a set of n-tuples how do I denote this six numbers where the inner product of two uh vectors Z and W is equal to some J equals 1 to n g z J W J okay um so this is an example of a Hilbert space finite dimensional one meaning it's a bionics it's a vector space of finite Dimension um the other example [Music] the space middle L2 this was the space of sequences um such that each of these is a complex number and this uh series is convergent is finite um some of the absolute value of the AK squares this is a Hilbert's face as well with what's the inner product of two um elements of little L2 this is the sum from k equals one to Infinity of 8K BK okay all right um and note so you know this is uh pretty clear but note that uh the norm I get uh by um you know from this inner product this is simply which was a little L2 Norm right okay so um you know these are two basic examples of you know a Hilbert space uh we will in fact show that um every separable Hilbert space can be in a inner product and and therefore a link preserving way be mapped so that's usually called a an isometric isomorphism can be mapped isometrically to either see in or little L2 okay so these two kind of if one wants to go about uh you know um categorizing all of the possible Hilbert spaces as far as separable Hilbert spaces which are kind of the only reasonable ones of course come you know one can come up with wild examples of Hilbert's faces which are not then these are the only two I shouldn't say two because you know this one's indexed by its Dimension but these are the only two types that you come up with either a finite dimensional uh CN or it's isometric to little L2 okay but um so but I will still you know write down another example for um since we went to all that work if e is a measurable subset of r 2 of e the big L2 of e which remember this was a space of measurable functions F from E to C such that F squared e is finite this is also um this is a Hilbert space what's the inner product with the the inner product of two elements in L2 defined to be kind of just the analog of little L2 where we replace the sum with an integral times G okay the label integral of f times um the complex conjugate of G [Applause] okay now um you know I wrote down Capital L2 I wrote down little L2 um but what about the other little LP and and big LP spaces or any of those Hilbert spaces um you know so of course if I Define the inner product in this way um as I did before then that only induces you know the little L2 and the big L2 Norm right but is there perhaps some other kind of magic inner product out there that I can put on little LP or big LP so that I would get out uh the little LP or big LP Norm when I Define the norm according to you know how I've been doing it um so the question you know on hand is uh the other little LP or big LP spaces also Hilbert spaces right so again it's clear that if I were to define the inner product in in the way I did in the two examples then that only is going to give the the L2 Norm I'm asking now is there some magical inner product I can Define on these spaces that spits out the little LP or big LP Norm right um so the answer is no and there's in fact you know a way to determine whether or not a space is a Hilbert space because if you think about it the way you know we've come about um you know introducing uh what a pre-opered space is and the Hilbert space is we first had a norm in our hand and then we defined an inner product well if that's the way you're building your space and you know automatically that it's a Hilbert space because you had an inner product first and then you defined the norm second let's suppose you're only given the data you're given is some Norm on the space when can you determine if that Norm comes from an inner product right that's the question that uh you know is underlying this question I wrote on the board if I have a norm space right so all my initial data is the norm when can I tell that that Norm comes from an inner product and um it so you'll prove this basically by direct calculation but this is the following parallelogram law which is the following if H is a pre-hilbert space then all u v and H if I take the norm of U plus v squared and I add the norm of U minus V squared this is equal to twice the norm of U squared plus the norm of V squared right so this is a condition um that is stated purely in terms of the norm right moreover uh if H is a Norm space satisfying star then H is a pre-hilbert space okay in other words uh although it seems like you know we defined Hilbert spaces or pre-hilbert spaces initially in terms of an inner product in fact you can say a norm space is a pre-hubert space if and only if it satisfies this parallelogram law okay so if you have on your hands just the norm then as long as that Norm satisfies this identity that Norm can be derived from an inner product so using this theorem you can check that uh the answer is only for P equals 2. okay so in other words if you you can come up with u and v that so that this inequality is not satisfied when p is not equal to two okay okay so this um so now we have the notion of a Hilbert space where this Norm uh given in terms of an inner product the space is complete now we have an inner product so we can start talking about um vectors being orthogonal to other vectors or or the normal sets um which you know when I first started lecturing about this stuff I started I was already using the terminology this thing being orthogonal to this thing or something like that um so of course if you don't remember what those words mean um from linear algebra I'll brief I'll quickly remind you so suppose we're in a hill pre-hilbert space we say that two elements u and v are orthogonal if their inner product is zero right and we if uh instead of saying orthogonal I want to write this in words I'll write you Purp V okay so throughout H is going to be uh pre-hubert's face if I don't actually write it if H is a pre-hilbert space a subset um which I'll denote by Eden Lambda Lambda and Capital Lambda so just some subset indexed by some indexing set Capital Lambda we say this set is orthonormal if we're all Lambda each one of these vectors and the subset has unit length and I need to uh different indexed elements implies that they are orthogonal okay so um maybe this notation scares you because what's this indexing set typically we just use the natural number so let me just make that remark although I'll make a few remarks in general about what the normal sets that are not necessarily indexed by the natural numbers will mainly be interested and you know a finite set or accountably infinite set okay so mainly so although uh an orthonormal uh subset of H is you know could be a very crazy type of uh subset mainly we're going to be interested only in finite or countable um accountably infinite uh orthonormal sets so what's uh what are some examples okay so um simplest examples ah if so this is uh an example of a set of orthonormal or an orthonormal subset of C2 or let's say one this is also an example of an orthonormal subset of C3 um let's say uh using notation from before if I denote by E sub n this is uh the sequence consisting of zeros up until I hit the end spot and then 0 afterwards and entry uh which is an element of little L2 then this is an orthonormal subset of L2 okay one other example is let's look at the functions one over square root of 2 pi times e to the I and X and let's think of these as elements of L2 and minus pi to Pi then um this is an orthonormal subset of my right o-n instead of orthonorm instead of writing out uh orthonormal but this is an orthonormal subset of L2 right since uh if I take the inner product of two of these guys I get let's say I just look at uh two different ones um M does not equal n so then I take the inner product of i m x with inner product of i n x complex conjugate so that's the inner product on big L2 and this is equal to 1 over 2 pi times minus pi to Pi of now if m equals n then this I just get the length of e to the i m in n x squared the length of e to the i m x is 1 and therefore I would just get one here DX over 2 pi which gives me 1. but now they're eight they're not equal so I get e to the I M minus n x DX and now the integral of this creature so here e to the I let's say I'm going to come up with some e to the I y when Y is a real number this is by definition cosine of y plus I sine y now you can check that fundamental theorem of calculus still holds for uh e to the i m minus n over X and this equals e to the i m minus n X over I times M minus n and uh e to the i m minus n times x is 2 pi periodic so when I evaluated at minus Pi I get the same value if I evaluate it at Pi so this equals 0. so that's a or it's the normal subset of big L2 so you know this collection of vectors and and I'm calling them vectors but this collection of elements in big L2 you know is still countable even though I'm indexing it by the integers rather than the natural numbers foreign okay so we have the following [Applause] now you know most of what I'm going to say with regards to um accountable subsets of um so accountable orthonormal sets still carries through to possibly uncountable orthonormal sets where now you know a infinite sum over an uncountable number of elements has to be defined in a precise way but um I will really just stick mostly to the accountable case and if you're interested you can always look that stuff up um so I should say you know what I haven't said is that whether or not the lp spaces let's say over an interval a b or over R are separable or not or even over a or even over a yeah so let's stick to either close and bounded interval over or r um now why are these spaces separable and meaning they have accountably dense uh subset so including big L2 the reason is the following so what you proved um in the assignment is that the continuous functions are dense and little and big LP for p between P equals uh 1 to Infinity strictly less than infinity okay they're not dense in L Infinity so as long as you stay away from that they're dense okay now um so continuous functions are dense in LP now what's one way to approximate a continuous function you can going back to your introductory analysis class hopefully you covered What's called the virus drops approximation theorem which says that for every continuous function you can approximate it uniformly on the interval by a polynomial okay so that shows that polynomials are dense and all the lp spaces of course not L Infinity now how do you go from the set of polynomials which is uncountable to a count of which is dense in LP but is uncountable to a countable dense subset which is density no p make everything rational okay the set of polynomials with rational coefficients is in fact countable okay and so and you can approximate every polynomial with real coefficients by on a closed and bounded uh interval by a polynomial with rational coefficients is not too difficult to to believe and therefore the polynomials with rational coefficients okay are dense and the lp spaces as long as I'm not in L Infinity um and therefore all the lp spaces are separable um now the little LP spaces are also separable because except for L Infinity also which is not separable um because first off a dense subset of the let's say little O2 let's make things definite is the Subspace consisting of all sequences which terminate after some entry in other words it's zero after that okay you know convince yourself that this Subspace of all finitely terminating sequences this is dense and little LP for every uh for p between one and infinity not equal to Infinity okay so finitely terminating sequences are dense and little LP um unfortunately that's again an uncountable uh I mean it's a it's a Subspace so it's going to be uncountable so how do I get now accountable thing again I replace everything by rational numbers so if I uh I can approximate every sequence which terminates after a certain point consisting of real numbers by a sequence full of just rationals terminating after a certain point by just choosing the rationals very close to those real numbers right we're using here and I said didn't say this explicitly a minute ago that you know we still have we have the density of the rationals and the reels for every reel I can find a rational very close to it okay so um you know this is kind of thinking that goes in goes on so but now the set of all finitely terminating sequences with rational coefficients this is a countable set okay and that countable set is dense and little LP and therefore little LP is separable as long as p is between one and Infinity uh excluding Infinity okay so um you know I said at the beginning that we're going to be mainly interested in in separable spaces without actually saying why little LP or little L2 and big L2 are separable um but uh so I just gave you the argument by word of mouth now instead of actually writing it down Okay so um we have the following vessels inequality for orthonormal accountable orthonormal subsets so if the n is well let me just put n here if this is a countable meaning it's either finite or it's countably infinite but it's countable uh orthonormal subset a pre-hilbert's face h and for all U and H if I look at the sum of squares U and a product e sub n this is less than or equal to the norm of U squared okay um so our discussion here of orthonormal subsets is taking place within a pre-hilbert space um don't need Hilbert spaces to talk about these Concepts uh yeah but when we're in a Hilbert space and we have a certain orthonormal subset that will be important that that we're in the Hilbert space okay so uh the proof is uh let's do the finite case first so uh suppose I have finite collection of orthonormal or finite orthonormal subset of H or a fine or I'll often say finite collection of orthonormal vectors in H no end subset of H again o n standing in for orthonormal then let me just record a few identities which are pretty easy to verify some from n equals 1 to n of U inner product e n e n um if I take the norm of this thing squared let's compute this out this is the inner product of with the understanding that n is going from one to capital n let's use a different index here and this is equal to n m u e n uh now I have this sum here of I should leaving out e in of u b m complex conjugate times the inner product of e n with e m that's the two vectors that are taking the complex complex conjugate of this is a number here it just comes out this number here gets hit with a complex conjugate when it comes out now the inner product of En with em is 0 when n does not equal m and is equal to one because it's equal to the norm of E N squared when n equals m so all I pick up from this double sum which is just a finite double sum is when n equals m and N going from one to n uh and therefore this is equal to u e n squared okay [Applause] foreign that's one formula I want to have another one is that if I take the inner product of U with n equals 1 to n of the sum and you know maybe you recognize what this uh sum here actually is I'll say so in a minute so this is equal to sum from n equals one to n of so U and a product e sub n times this number this number comes out and gives me a complex conjugate okay and therefore zero which is bigger than or equal to the norm of U minus n equals 1 to n of u e n e n now if you remember back from linear algebra from calculus that if I have um orthonormal vectors and I have a vector U this quantity is nothing but the projection of U onto the span of those orthonormal vectors okay so what I'm looking at here is if you like the norm of the part of you that's orthogonal to these finitely many vectors okay okay so this thing's bigger than or equal to zero we use that formula of how to compute the norm of something plus something and this is equal to Norm of U squared plus Norm equals one to n u e n e n squared minus 2 times the real part of U inner product sum from n equals 1 to n of U e n a n okay and now we know what all of these things are this is equal to uh I'm not even going to go over this is equal to this no this quantity here as is this inner product it's also equal to this thing here so the real part is equal to this thing here because this is a real number and I get since this comes with a 2 I cancel one of those so I get normal View squared minus sum from n equals 1 to n of U e n Square okay which is exactly what I wanted to prove for the finite case but the infinite case then follows from the finite case [Applause] um and by letting capital N Go to Infinity so infinite case suppose e n equals 1 to Infinity is orthonormal subset h then we know that for all n capital n we have that sum Over N equals 1 to n of u e n this is less than or equal to the norm of U squared so I can just send capital n to Infinity to get that the sum equals 1 to Infinity of Norm of U E N squared is less than or equal to U squared okay okay um so orthonormal subsets we can Define as you know a collection of vectors that have unit length and are mutually orthogonal to each other um now just orthonormal subsets list any old orthonormal subset is not really the most useful thing if we're trying to study the entire Hilbert's or pre-hubble pre-hilbert space h because we may miss something if we leave out certain orthonormal vectors um so but a more useful um type of orthonormal effect is a maximal orthonormal set or which is defined as follows an orthonormal subset e Lambda Lambda and capital Lambda of a pre-hilbert space h maximal so again if you know having a possibly uncountable collection of what the normal subsets indexed by some indexing set makes you uncomfortable replace this with you know in where n is going from one to capital N or n is going from one to infinity and so accountable collection if you like but I'm stating this so that um you know that uh more something more General is true so this is maximal if what um The Following holds If U is an H and um U is orthogonal to everything in this orthonormal subset this implies that U is zero so um an example of course is you can check that this collection here is is a maximal orthonormal subset of C2 a non-example right you can kind of maybe see this coming the one we had a minute ago this is not maximal since there's a vector that has inner product zero with both of these but is non-zero right since this should be C3 since um let me just write it kind of loosely this way since this Vector is 0 1 0 is orthogonal to both of these but it's non-zero right maximal means if you're orthogonal to everything in your collection then it has to be zero but this is non-zero and it's orthogonal too um uh everything to these two vectors there another example is you know again with the notation from before this is where this is the sequence that is zero except for the end spot where it is one this is maximal orthonormal subset of little L2 okay now what we're going to see very shortly is that um if we have a countably infinite maximal subset of a Hilbert space then that set serves kind of the same purpose as a basis as a orthonormal basis does in linear algebra right I mean if you look at here already you know this orthonormal set is maximal and it also forms a basis for C2 right this was not maximal and you see it doesn't form a basis for C3 right so maximal is gonna kind of give us a a condition which is is kind of equally useful as being a basis but it won't be a Hammel based these subsets won't be a Hammel basis in the sense that every Vector can be written as a finite linear combination of the elements of a maximal orthonormal subset but what is true is that we can write it as possibly an infinite sum involving uh uh the or the maximal or the normal subset you know which is in most cases just as uh good if you want to use that all right okay so um first off when does every uh pre-hilbert space have a maximal uh uh subset so let me State this as a theorem in fact it'll state two theorems the first is uh every should say non-trivial because we could have the Hilbert space the speed the zero Vector Hilbert's pre-hilbert space has maximal or the normal subset whether it's separable or not it has a maximal orthonormal subset and the way you prove this is using so I'm not going to give her proof of this I'm going to give you a proof of something a little less strong but about as useful as as we'll need um one proves this by uh the by using Zorn's Lemma by you know taking a set your set that you're going to put a partial order on to be the collections of orthonormal bases are not orthonormal bases a Max of orthonormal subsets and then ordered by inclusion and then one can do a Zorn's Lemma argument and apply Zorn's Lemma to to obtain a maximal um orthonormal subset but that's kind of hands off and maybe that scares you a little bit because uh Zorn's Lemma is a equivalent to the Axiom of choice so [Music] if you don't like using the Axiom of choice maybe you have a problem with using it to construct a maximal orthonormal subset of a pre-hilbert space but uh we can actually do this by hand if the Hilbert space is separable so this is the theorem will actually prove every uh non-trivial um separable meaning it had meaning the pre-hubert space has a countably dense subset every non-trivial separable pre-hilbert space accountable maximal orthonormal subset okay which as I said this is kind of the main uh types of orthonormal subsets we'll be interested in just because you know defining infinite sums is easier to do over a countable uh index than it is an uncountable index but uh anyways so we're going to prove and actually construct this essentially by hand um using uh the process that's the name of this little section the Graham Schmidt process so if you from remember from linear algebra if you have a collection of um vectors you can always find and orthonormal collection of vectors that span uh the Vector space that is spanned by the original set of vectors okay and that's what we'll do so since H is separable let be a countable dense subset of H and it's uh this is a non-trivial pre-hubert space so we can always make sure that the first one is a non-zero Vector okay so uh you know accountable means dense remember means that for any element in h there exists an element from this sequence uh or from this collection that's within Epsilon of uh that Vector from h [Applause] so now I'm going to make the following claim this is essentially a the gram Schmidt process which will improve which will approve by induction we're all in natural number there exists another integer a natural number M of n and and orthonormal subset 81 up to em such that the following is true um first is uh the span of E1 up to e m of n equals the span of B1 up to VN and so you see you know n is changing so maybe for each time I change N I get a different orthonormal subset or a wildly different orthonormal subset from uh the integer before it but um so the property of these subsets are that I'm just simply adding a a vector or not so and E1 to e m of n so this collection is equal to um the previous collection Union either the empty set if VN is in the span of the one up to B n minus 1 and some new Vector e of M sub n and I'll tell you what e of M sub n is it otherwise okay so what I'm saying here is that I have this countably infinite list of B's and for each N I can come up with a finite orthonormal subset that spans the same span as V1 up to V in and at each stage all I do is add a vector or not depending on if the next V is in the span of the previous or not okay so I hope that's clear um Okay so to prove this by induction so proof of claim this is by induction so let's do the base case n equals one uh let's take we take E1 to B B1 over length of V all right so now we're started now we've got our first Vector in this list that we're building up inductive step so um let's call this what I want to prove star so suppose the star holds for uh so this whole claim here I shouldn't say Just A and B but the whole claim holds for n equals K and now I want to prove it holds for n equals K plus one so I wanna what kind of vector do I need to add to the previous uh collection of vectors to now span V1 up to V K plus 1. okay so uh if B K plus 1 is in span of B1 up to V K then E1 up to e m of K I should say the span of these vectors equals span of V1 up to V k equals because V K plus 1 is in this span this span is also equal to the K plus 1. and therefore uh this case is handled and we're in this spot where we don't add anything to the previous uh collection okay this proves uh what we wanted to for n equals K plus one um by not adding anything so this was in the case that VK plus one is in the span of V1 up to VK so now let's do the more interesting case of V1 not equal or not in the span of V1 up to VK so now suppose be K plus one is not in um span of V1 to VK I Define w k Plus 1. to be V A plus 1 minus its projection onto the previous uh list of orthonormal vectors so some from J equals one M of K B K plus 1. e j then first note that this Vector cannot be so w k plus 1 this this Vector cannot be zero right otherwise this would imply that V K plus 1 is equal to this quantity here and therefore since uh and therefore V K plus 1 is in the span of the ejs for J between one of between 1 and M sub K which is equal to the span of the V J is for one up to K so but we're assuming this is not in the span of should say this is equal to the span of E1 up to a m sub K right that's what we're assuming that's the inductive hypothesis right so then this uh Vector does not equal zero and Define e M sub K plus 1 or M to be the vector WK over its length for plus one now this is a unit vector um and and what and um and if I take so now I claim this is orthogonal to E1 up to e of M sub k if I take e M K plus 1 and a product e sub N Sub J so let's take this is equal to Simply by definition 1 over length of M K plus 1 times the inner product of V K plus one minus sum from J equals one to M sub K so remember this this is supposed to be the projection onto the E1 up to e m sub K and V K plus 1 minus that is supposed to be orthogonal to those guys so we'll see that this ends up being zero and let me change this J to an L since I am or no and I get one over times the sum uh or times V A plus one Isabel minus now Isabel inner product with the sum right now this is the sum from J equals 1 to M sub K of this number here times the inner product of e sub J inner product e sub l right um now the inner product of e sub J and E sub L is zero unless J equals L and one other and one if J equals l and so I only pick up uh the J equals L part of this sum when this inner product hits e sub l uh in particular I only pick up the coefficient in front since these have unit length so I get V sub Plus One V sub l equals zero then um this guy now does in the job so I would write a little bit more but I'm running short on time okay um so we've proved the claim and now let's use this to conclude that this collection of all of these Els um are maximal so so we let s to be uh V sub n which equals and s is an orthonormal subset of h So eventually so you know I may not be adding any more vectors after a certain point and therefore just have a finite collection that's also possible but I may also have a accountably infinite collection so this is a orthonormal set by B and so uh we now show X is maximum okay all right here is where so we haven't used anything about the nature of this subset of H all we've used is that you know it's accountable so we could do this step by step where we construct these orthonormal vectors that span each finite uh equal to the span of each finite collection of vectors now to show its maximal this is where we use the fact that this collection of these were dense there are dense in h so um suppose U is an H and um for all l U of e sub l equals zero so this is either a finite set so L is between 1 and N or it's a countable set and L is going from 1 to Infinity so this equals it either this is equal to E 1 up to E M sub n for some n or B1 is countably infinite okay so since v j this is a dense subset of H we can find a sequence of elements from this collection so V of J sub k equals 1 to Infinity such that v j of K converges to U as K goes to Infinity okay now by property a the fact that this span is equal to that span V of J sub K is in span of E1 up to E M of J sub k and therefore my vessels inequality and the fact that U is orthogonal to all of these uh uh orthonormal vectors I get that V of J k squared now V of J sub of J of K this is in this span so you can show that in fact um this Norm is in fact equal to the sum of the coefficients that I get from taking the inner product of okay if I have a vector in the span of a collection of orthonormal vectors a finite collection of orthonormal vectors then the norm squared is equal to the sum of the coefficients just like an RN and Since U is orthogonal to each of these this I can write as and by vessels inequality this is less than or equal to um the sum of these things squared is less than or equal to squared and this goes to zero because K goes to Infinity since the V of J sub K is converge to U but we started off with the norm of V of J sub K which implies 0 and therefore U which is the limit of these must be zero proving that this is a maximum subset okay um all right so next time we'll prove that these maximal orthonormal countable count of these maximal countable orthonormal subsets in fact form a pretty good analog of uh bases that you find in finite dimensional linear algebra we'll stop there