Lecture 28 – part b – Chapter 5, printed slides 102-108, Proof of Open Mapping Theorem

okay so let’s carry on so we’re given the asymmetric about zero and its convex and we’re assuming that the balinese centered on X radius R is a subset of a and we prove that the ball with the same radius ends on the origin is a subset of a well all you have to do is let let Y in a weave norm Y smaller than R then X plus y is of course in this ball the ball in the center on x-ray DSR so today but then a symmetrical about the origin that should be clear we know that whole ball is you know so anything whose distance 2 x’s lesson little R is going to be an A and this is one of those points so that is that ok ok so then a symmetric about the origin so minus X is you know and let me see which I want I think I want minus X plus y minus X plus y which is equal to minus X minus y is also a day again X minus y is in be e of x are again okay finally we use a convexity to average X plus y with minus X plus y why is a half of X plus y plus minus X plus y that’s a convex combination of two points in a what I mean by convex combination that be t 1 plus 1 minus 2 times another but tears are half here so why is it as well I guess the miners expert didn’t matter so much what we wanted was that X plus y was in that minus X plus y was in and then average those two any questions on on that proof okay so we can we found a little ball you can reflect that little ball in the origin then you can take the midpoints of the points between them and that gives you the same livable around the origin and let me remind you before we prove the earth mapping theorem on top of this i just want to remind you of the closed sets version of the bear category theorem the most useful to us version the bear category theorem so the bear category theorem in terms of

clothes sets bc t bear category theorem in terms of clothes sets if you let X be a non-empty complete metric space and suppose you’ve got some clothes subsets my brief natural number and that the union of the ball is X then at least one of these clothes sets must have no no tu interior so then there exists an end in the natural numbers with interior of en non-empty in the next few theorems this is the version of the bear category and we’re going to want to use and it’s also the one you need it for that exercise on the question sheets so it is probably this is the most commonly used version of the bear category anyone need more time on that slide okay here we go the open mapping theorem let x and y be balanced faces and let t be a continuous linear operator and continuously a map from X to Y and suppose it surged yet too so that the preimage of every we’ve got the pre medium every open set is open then for some reason rather the image of every open set turns out to be open a very surprising result will prove that and then we’ll do the benefits or morphism theorem afterwards so come back to this five point 10 in a bit so we’re on the proof or the open mapping theorem let me make sure I get the notation again as in the is on the handouts so proof of the open mapping theorem so we’re given x and y a banach spaces t is in B of X Y so it’s a continuous linear map and T of X is equal to Y let’s have a look at the following subsets of why the following closed sets en contained in y en is equal to the closure in Y of T of the closed ball in let’s have the open ball it doesn’t really matter letter the close ball in X send on the origin radios n which is also equal to by the way n times the closure in Y of the image of the unit ball though it doesn’t really matter for our purposes you can you find that you can scale up and down by n it’s a homeomorphism of the space you can check that last claim then

well they’re closed by definition because they’re the closures of sunsets that’s a good start they’re also the image under linear map of a symmetric convex set so each en is itself it’s closed it’s closed my and its convex because it’s the image under linear map of a convex set and symmetric about the origin because its image under any map of asset which is symmetric about the origin so that’s a good start I claim the union of these things as a whole of why the union from N equals one to infinity of e n is equal to Y because why the equal to t of x is equal to t of the union for men equals one to infinity of the ball in X close ball in X Center zero radius n but it is true that the image the you the image of a union is the union of the images doesn’t work for intersections but it’s true for unions well that’s a subset of the unity ends even before you close them so even before you close them up it’s already the whole of Y so certainly of time you close them up it’s still white because we were only working in subjective were at the moment so what’s our first conclusion using the bear category theorem that’s right at least one of these has not left interior at this stage you have a choice you could notice that the e ends are all scales out copies of e 1 so in fact you can conclude the e1 of non empty interior or you can just work with en it doesn’t really matter all observed since the year is n times e1 e 1 itself has not empty interior this is the usual argument about scaling up or down by multiples of n is a homeomorphism of Y so it doesn’t change anything now what next it’s got non empty interior but I actually would like to get that interior to be in a special place so what could we say now so there are nothing the interior since e 1 is symmetric and convex at is symmetric about zero and its convex zero is in the interior of e1 so there exists some real number are weird the ball my and I’ll go for the I’ll go

for the closed ball I can do this and if its interior if you’re in the interior then to some closed book on the origin also ball in Y symptoms area radius R contained anyone so what what have I got so far we have that closure let’s rewrite it it’s in it’s in the closure in Y of T of the clothes ball in X now I claim so anything in Y who’s norm is less than R let’s scale up by one of our that’s what I do thus the closed unit boarding why you can scale by whatever are and that’s contained in the closure of T of the all in x radius naught I have to scale up by one of our notice all the three p+ so I haven’t done any harm but now we think back to the open mapping lever the open mapping lemma asked you to approximate things in the destination set closed whose nor more than two big the other mapping lemma said as long as it for things in the codomain whose normal that most one in other words points here you’re supposed to be able to find things in the domain whose norm isn’t too big we’re going to use one over R which are close to their element notice we did have to close up here we didn’t claim that you actually mapped onto everything but but if you’re in the closure that means you can get as close as you like so you can take any positive number alpha so taking EG any alpha is strictly between autumn one so take any Alpha in naught 1 and M equals 1 over R then for all Y in the closed unit volume why it exists an X in X with norm X no more than M and such that norm of T X minus y well wising the closure of the T of these things so we can definitely get within alpha I think only needed less or equal to alpha in the other mapping gonna if you want to be more specific you can take EG out for equal to half that will do fine so then the conditions of the open mapping never satisfied and t that open mapping and we finished okay it’s one of the best results of the area you have a mapping theorem and by the way what else does the open mapping lemma say under these conditions what can you remember the other two things you have a mapping them has said so it said that these why would be complete however that was already part of the conditions of the theorem because we needed it to do bear category theorem so if we hadn’t known why I was completed

and used that already we would have now deduce why I was complete from the open battery level but it was already complete and the other figures of course part of the aerobatic level shows the fig is surjective which is part of it being it out of that but again we already do it was surjective because that’s part of the conditions here so we’re only using one one bit of the open mapping letter here and because we already had surjective an already had y complete but we we needed to use the bit that said that the mapping with an open mapping any questions about that proof so where do we use the fact that X was complete and where do we use the fact that why was complete because you really do need both for this result there are some where you can get away with weakening it so anyone told me one or the other where do we use that why was complete when we use a beer category theorem we use a bear category theorem my so we need Y to be complete to get that to work where to be user X was complete it’s hidden we didn’t mention that X was complete during the proof so it must be hidden in something we applied if you have a mapping lemma we needed X to be complete for the open mapping leather so that’s where we use the completeness of X so perhaps I could add a little comment there note here that X is complete exit complete so we can apply the open mouth another right so moving 50 yard or in fat moving swiftly back here’s a benefice a morphism theorem every continuous linear isomorphism between Banach spaces automatically hazard and inverse which is continuous well here you are x and y Banach spaces you can linearize both of X to Y then suppose it t is continuous you need that then the inverse is automatically continuous as well and so it continues in both directions and the proof it’s notice that here that T to bite us what is continuous if it only have teased that open mapping so its immediate from the open mapping theorem so once you’ve got the open mapping theorem the bad a Kaiser more than fear and comes for free and this is one of those automatic continuity results where you you don’t think you’ve assumed enough to know that something should be continuous and yet it turns out that you get it for free and I like those results there’s a whole area called automatic continuity theory it’s not hugely active at the moment there was some very very big problems and was a lot of excitement around it over the last you know 40 or 50 years and there were various big breakthroughs and and my research supervisor guard Dale’s played an important role in in solving some of these problems so not quite so active these days the automatic continuity theory but even so there’s there’s a lot of spin off areas developed from it

which are still very active okay so we’ve got open mapping theorem barrier height of autism theorem what else we got left in chapter 5 first of all some exercises so this is one of these directions it’s a bit tricky and but anyway you should still be able to do it the if you if either of them is not assume complete the proof breaks down but that doesn’t mean that you found a give in to the example yet so here’s a little a little testy exercise for you is to give examples in both cases so first of all suppose X is completed wise incomplete give an example of a continuous linear surjection where the inverse is discontinuous and then do it the other way around slightly easier if you allow both of them to be incomplete and then there’s the example that’s on the question tude and that’s already on I got remember which question sheet that is question cheap five I think I think it’s chic five question one and we’ve already are we’ve already on to part two now we can do the banner space vector space isomorphism theorem so you’ve got banner spaces and a continuous linear map from X to Y suppose teeth surjective then isomorphic is a better space so that’s that’s a so that’s with a linear homeomorphism to the quotient moreover you can do a linear whole room of them of this form a tea till they’re capturing through the equation map as usual so you’ve got X and you’ve got the equation space X out of the colonel and you’ve got Y and you start with your surjective Lydia mat here then you factor through the quotient map t tilde now and then of this exercise about quotient maps of little 01 which it’s a combination of the combination of the exercises that I mentioned before on sheets 5 and sheet six so i’ll come back and say more about this treasure map it up over this is for this exercise see see earlier mentioned exercises on sheets five and six okay let’s have a another look at this and the the existence of a continuous linear map t tilde here that has this role was already discussed earlier the fact that you can factor through the quotient this way was was something that we already mentioned earlier and that you get a continuous linear map this way but it’s also we also discussed this earlier when we talked about vector spaces and so the fact that T tilde is a linearize of autism is just the vector space isomorphism theorem so the teacher is a linear isomorphism just from the vector space fence so if you just from knowing it’s a linear map that’s surjective you can quotient out the kernel and make this linear isomorphism so it’s a linear isomorphism but then from earlier results we had we said that you could make a continuous linear map this way so relax a continuous linear mat and it’s an isomorphism and then the inverse is automatically continuous by the banner pfizer morphism theorem so by the benefit of autism theorem because this is a linear isomorphism and it’s continuous the inverse is automatically continuous too because their barracks basis I’ll remind you that x over the

hurdle of t is a cause a closed subspace and when you question that a closed subspace then you’re still complete we knew that X with a banner space so the quotient by Colonel t is still valid space so this so that’s about our space yet save the kernel of tea why the better space this is a linear item autism it’s continuous from earlier remarks so the inverse is automatically continuous and you get the Banach space either morphism theorem any questions about about this version okay so that case we can move on to the closed rafter this one’s very going to be very useful to us when we when we try and prove automatic continuity results about Panik algebra first we need to look at the graph of a linear map now what is X plus in a circle why that’s the external direct Sun so X plus y is equal to X cross Y but the external direct Sun so it’s got these operations x1 y1 plus x2 y2 equals well the usual things let’s move that all across at x1 plus x2 y1 plus y2 etc so it’s the vector space direct sum now give them the X&Y normed spaces we’d like to put a Norman X plus y and here’s the word i’m going to put on it there’s lots of choices i’m going to put on what i call basically the one norm where you just add the norms of the coordinates so you take the norm of X in the first space and you add it to the norm of Y in the second space and that gives you a nice norm and as an exercise check that Norm this norm it uses the product apology what do I mean well X has got a topology because it’s a normal space why is got a topology because it’s a normed space and so X cross Y can be given the product apology but it’s also got a topology coming from the norm one you want to check their the same but that’s just checking that what you’ve got is coordinate wise convergence that if you’ve got a sequence all you have to do to check this is that xn yn tends to X Y if and only if x 7 x 2 x and y intends to why once you check that you’ve gotta coordinate wife converges so what’s a graph of a linear map when it’s what the graph of any function it’s a set of all pairs X in the domain and image of X in the codomain so it’s x TI x where x is in x that’s the subset of the product and notice this is exactly one of these for each point X in X so if you thought of this is an all new graph of a function it behaves like the graph of a function there’s exactly one point in this graph corresponding to each point of the domain but it could be that several different X’s give the same TX so that’s just like a graph contain you could have the same value it more in one place but there is only there is only one value above a particular X so each x

coordinate appears exactly once whereas some of the y coordinates may not appear at all some point of why may not turn up at all some points of why men you turn up for several different X’s and so on it’s just like a typical graph and will say that T is closed graph if the gravity’s is closed with Petra norm one which is the same as being equivalent Lee that’s closed in the product apology now here’s some notes about grass it’s easy to check what I’ve just told you how to do that normal one does give the usual product apology I couldn’t find a standard notation for the graph of a function so I’ve gone for graph of tea but i don’t think that standard i think some people use a guru some people use gamma some people use i couldn’t find any notation that lots of people agreed on for the graph of tea so I just said gravity and here’s your exercise this is a nice one this is useful to us when we’re checking some automatic continuity results on Banach algebras turns out you don’t normally check that the graph is closed you should normally check that if X n tends to X and T X n tends to Y over closed graph officially if X M tends to X in X and T X n tends to Y in why you need t x equals y that’s what you need to check to check with a clock closed graph because then you to have xn txn will be a typical sequence in the graph converging to something a pair XY in the product but you need that pair XY to be in the graph to say that X Y Z in the graph is to say that T x equals y so this is what you have to check in general to check you’ve got closed graph of course if cheese continuous you get that straight away so if T’s continuous then you know that if X n tends to X and T XM x to y mt x equals y by continuity of T so continuous maps automatically have closed graph just for that reason in this setting it is particularly easy in this metric space type setting but it turns out that you’re allowed to restrict to the special case when x is 0 you only have to check that if X n tends to 0 and txm tends to something in Y then t0 should be Y in other words why 30 so this is a special case of the one above okay i’m getting at a problem with with my writing now that’s online let’s try got it back again good so this is a special case of above and it turns out this is the decisive special case you only have to check a special case when x equals zero and note the teeth are equal zero and I’ve already told you how to do this exercise if T is continuous and she is closed graph just by checking what closed graph means using sequences so I leave that is a little exercise to check this this first ones or not it’s exactly the same as it’s very very similar to checking continuity at the origin for a linear map is enough if you get continuous with the origin and get continuity everywhere this is like sort

of checking closed graph at one special place gives you the closed graph everywhere else so closed graph theorem says it’s rather nice thing that if you want to check whether a linear map is whether a linear map is continuous is necessary and sufficient to chapter it has closed graph so we won’t have time for the whole prove here but let me just draw you the picture of what’s going on you’ve got a little am at X going to why you’ve got your linear map tea and you want to know is it continuous or not well we’ve also got the graph of T’s living somewhere contained units across y and we can map x 2 x TX and the inverse to that is a corner projection this is actually an isomorphism from exxon to the graph of T the graph of T is actually a vector space and here we can just use the projection map onto y so we can look at we’ve got to remember we’ve got X cross Y we can map to X and we can map to why using the coordinate projection onto X and the corner prediction on to y PX of x y equals x py try again p y of x y equals y and you can restrict these things to the graph so here i can just use py the map here which i call p x but when you restrict PX to the graph of t hear those have got those i’ve got slightly in each other’s way now let’s put that on the other side and what’s going to turn out is the graph of T if it’s closed either Banach space because we can see the gravity of the vector space if it’s closed it’s a Banach space because X plus y with the one norm is actually complete so then you’re a closed subset of a Banach space and so you’re complete then we can show that these maps here the one permits to the graph of tea or its inverse map which is a restriction of the corner production there can the PX is a continuous map PX is always continuous but it’s an isomorphism because that’s where you restrict it to the graph of tea and so inverse is automatically continuous which is going to be this map and there you compose it with py then you get T so it’s going to be almost immediate consequence of the banner Kaiser morphism theorem but we’ll do that more detail next time just to finish off the closed graph theorem and then we’ll there the uniform boundedness and I think we’ll be able to start the Bannock algebra stuff because uniform banners won’t take very long ok so we’ll stop there for today the it