 Okay, so today we start with an important subject which is the theory of distributions. So we will go, this is an introductory lecture on the theory of distributions. The important, the theory of distribution is extremely important in mathematics, in functional analysis, in analysis in particular, because for instance we have seen that when we started PDE we have seen that sometimes it happens that solutions at some moment are not anymore regular. It may happen that they develop, say, a jump or a jump in the derivative and so on. This is a common phenomenon in non-linear PDEs. So once you have this situation then you don't know even how to write the PDE, because you would like to say ut equals, say, ux or something else, but then ut or ux are not defined at some point. So you don't really know how to really to interpret your PDE. However, with the theory of distribution there is, there is a way which very often can be applied and saying that, okay, your PDE is not maybe satisfied at each point, but it's satisfied in another sense. Which is, which will be the sense of distributions. And so once you have this, then one says that the PDE is satisfying the weak form somehow. And then so you have a sort of weak solution. Then you have the problem of uniqueness, existence and uniqueness of possible weak solutions. And then you have to prove that maybe your weak solution is bad, but not so bad. So maybe regular in some sense. Not C1, but almost C1 and so on. So you, this idea is the idea that your PDE develops, develops a singularity. And then you have to write down again the PDE in another form. And then you have another notion of solution, which is not point twice. Hence you have to study existence, uniqueness, and regularity of that new object. Okay? This is one of the applications. There are very many other applications. This is just one for PDEs. So let us come to the definition. So let omega be an open set. So we have the space C infinity C omega and owed with the following, following notion of convergence, convergence as follows. So take, so now we give a notion of convergence of this space. This is a strange, strange object, but the definition is the following. So if phi, phi, let me call L, maybe phi, excuse me, let me use capital N here so that I can use n small n here. So if phi n is a sequence, sorry, this is capital N dimension of the space. This is an index small n. I've changed notation. Okay? So if this is a sequence of elements of C and then you have an element of C infinity compact support, then we say that, we say that phi n converges to phi in C infinity C if the following two conditions are satisfied. One, there exists a compact set K such that the support of phi n is contained in K for any n to for any derivative operator d phi n d phi uniformly in K. So two conditions, well, you require that the support of all elements of the sequence is contained in a fixed K independent of n. Second condition on that K, on that compact K, you have uniform convergence of phi and all possible derivatives, first derivative converges to first derivative, second derivative mixed pure, et cetera, et cetera, third derivative and so on. Okay? All possible derivatives. So this is a symbol d that says, for instance, the partial derivative with respect to x i of phi n converges to the partial derivative with respect to x i of phi. Then the second partial derivative with respect to x i x j of phi n converges to the second partial derivative x i x j of phi. The third x i x j is K, converges to the third and so on for any possible derivative. Everything now is supported on K, also phi. I don't say anything about the topology on this space. I just give you the notion of convergence. I don't want to say anything about the topology which is quite complicated, inducing this convergence. I don't want to say anything on that. Just, this is enough for us. Modification. Okay, then notation, sorry. The omega is the space C infinity C endowed with such a notion of convergence. Endowed with... So from now on, I will try to use always script d omega instead of C infinity C. If it happens that I use C infinity C, I'm sorry they are the same. Okay? What it is important is the notion of convergence on this space, d, d, d, but it is script d. So something like this sometimes. Yes, also the function for the derivative operator of order zero is... I mean, if you take the derivative operator of order zero, then in particular you have that phi n converges to phi uniformly. So in this point two, it is also contained the convergence of phi n uniformly to phi. No derivatives. Now, modification. Let omega be an open set. Yes? Yes. Ah, no, you're... No. So he is mixing the notation d with this d. I mean, that's why I'm using script d. They are two different objects. This is an operator on phi n, this is a space. So let... Indeed I was trying... So this is a d, but it is script d. I don't know how... In latex it's d omega. So, okay. Now, other definition. So another definition is the following. So I don't rewrite everything. I just modify it. So let us consider the space of continuous function with compact support. Endowed with the following notion of convergence, if I have a sequence of continuous functions with compact support and I have a continuous function with compact support, I say that phi n converges to phi if there exists a compact set such that... So 0.1 doesn't change. 0.1 doesn't change. 0.2, I simply erase this uniform in K and let us use maybe the following symbol. No, c, c, c. Let us use a standard symbol. So this space is c, c, endowed with such a notion of convergence. So when I use this symbol, I mean this. What I see from here is that maybe sometimes d omega is called the space of test functions. Now, d omega is contained in c, c. In the sense that, of course, as I said, it's contained here, but also this embedding, say, is continuous in the sense that if I have a sequence converging here, then, in particular, it converges here. So pay attention to this strange arrow. This means that the inclusion is continuous. It is clear because if something is converging here, then not only functions, but also all derivatives are converging uniformly. In particular, functions are converging uniformly. Okay. So again, I don't say anything about the topology of this space. Okay. Definition. So let t be linear. We say that t is a distribution if t is continuous. By continuous now, since now I have only a notion of convergence, but I don't have a topology. So this means sequentially continuous, namely if vh converges to phi, namely if phi h is in the omega, phi h converges to phi in the omega, then phi n converges to phi n, then phi n converges to t. Okay. And so this is essentially the dual. Therefore, and you write, you write. So a distribution is nothing else, a linear continuous functional on d. Namely is an element of the dual of d. It would be better to give a topology and so on. We don't do it, but look at this. The notation reminds, of course, that the dual of d is distribution. Modification. Let t, definition, let t cc omega into r be linear. We say that t is a signed Radon, signed Radon measure with locally local variation, a distribution in omega, in omega, if t is continuous. So if t is continuous, what does it mean? As before, if you have a sequence phi n of elements of cc and a continuous function phi with compact support and you have convergence in cc, then you have this. So exactly the same conclusion here, this line, but where cc replacing d. Okay. Is this clear? Let me rewrite it. Phi n, an element of cc. Phi n converges to phi cc, phi. Okay. So these are the first definitions. Sorry? Radon? Radon, yes. Now, I have to erase everything because I don't have space. So I wait. Okay. So there is a useful remark. As usual in this linear context, the remark is that t is a distribution, a linear map. So remark, a linear map is in the prime if and only if t is continuous at zero, just only at zero. What does it mean that if phi n converges to zero in d, then t phi n converges to zero? Okay. So if you want to check that something is a distribution, then this criterion says it is enough to check. First you look at linearity. Of course, your map must, your function must be linear. Once you have linearity, it is enough to check continuity at zero, sequential continuity at zero. Okay. But the first theorem that we have is a criterion to control whether or not a linear map is a distribution which says the following. So t is in the prime if and only if for any k compact, there are two constants, capital M depending on k and small m depending on k such that such that t of phi is less than or equal to m and the max or m phi infinity, this for any test. So let me read it. So given any compact set, there is one constant mk which is this one depending on capital K, depending on the compact set that you have chosen. Then you have m integral number possibly zero, which is this. Now there is a symbol here. This is the order of derivative of the derivative operator. What does it mean? Well, d phi over dx i as order one, d phi over dx i dx j as order two, d phi over dx i j, et cetera, et cetera. Okay. So this is the order of the operator. So you have say, I don't know, four derivatives here you take the infinity norm, the sup norm of this derivative or the operator. Theorem one, theorem two is same kind of criterion but for cc. So t is in cc prime. So let t linear, t is in the dual of cc if and only if there exists a constant m, sorry, for any k compact contained in omega. So I have added this. A compact contained in omega here, sorry, and m is obviously positive. Okay. So for any k compact contained in omega, there exists m such that t phi less than or equal that m phi for any phi in the omega. And indeed there is something here which is not correct. A support of phi is contained in k and support of phi contained in k. Okay. So this is the criterion that you have. So this is, you see, this is sort of saying that for instance for this again the map t somehow is continuous in the sense of what we have seen up to now in previous lectures using the infinity norm of phi. So it's bounded somehow. But you have this constant on the support of the test. Okay. Let us prove theorem one and maybe a definition also. So if in theorem one m is independent of k then the distribution t then t is called then t is called of finite order and the order of t is the smallest of such smallest of such m is for any k compact. This is the, so I was saying this, so which is the the order is the smallest. So remark what is the order? So if t is, so take t, so what I want to say is that the following, so if you have a map from d in omega such that this is true, then the order of this t would be zero. So you have a map such that you have this, but here you don't have the max, you have that m is zero here. So you just have the infinity norm. Then that is a distribution of order zero. Okay. Distribution of order zero means something like this where m is zero and therefore here just have the infinity norm of phi. Okay. We will try, in the moment we will see examples. Maybe it is, let us try to prove theorem one, so prove theorem one. So assume that condition one holds, assume one, take a sequence phi n converging to zero and you want to show that phi t phi n want to show t phi n converges to zero. Okay. So what does it mean that phi n converges to zero in d? Then we have that the support of phi n are all contained in some h, so there exists that there exists h compact such that all the support of phi n is contained in h. So this is implied by the convergence of phi n to zero and by the definitional convergence in the omega. So that implies in particular that we have this and therefore once we have this compact set h, from one we can apply one with that compact set h. So we find the corresponding constant from one that exists mh and mh corresponding constant to the compact set h. Therefore such that t of phi is less than or equal, t of phi n is less than or equal than mh, the max of d less than mh d phi n infinity for any n. Okay. Do you agree? Is it okay for the moment? So let me repeat it. So I assume condition one and I take a sequence phi n converges to zero. I want to show that t is a distribution. Therefore using this remark I have to show that t of phi n converges to zero. So continuity at zero only. So by the convergence of phi n to zero in d it follows in particular the first condition is that all supports are contained in a fixed compact set h. So take this h insert h into condition one which we have by assumption. So given that h I can find mh and small mh such that this is true for any phi supported in k. In particular since all phi n are supported in h, this is true for any n. Okay. Is it true? Is it okay? So now we have used only for the first condition of convergence but we have also the second one. So what we have the second one was saying that all derivatives are converging to zero uniformly and therefore now this is fixed is independent of n. So I cannot increase the order of derivative and therefore this is going to zero. I repeated mh is independent of n therefore for any derivative operator this is going to zero and this implies that t of phi n is going to zero. But the max by the again using this since phi n converges to zero. Majes to zero in d and therefore t of phi n. Okay. Similar so this this proof just only we have shown for the moment this implication. Similar proof can be given for theorem two for measures instead. So I don't I don't do the proof for measures but just to the proof the second part of the proof for distributions. Okay. Now maybe a remark I have raised that before doing the second part so the other implication. So now I need to show this implication left right. But let me make some and since I have raised this before that before forgetting I would want to do a comment. So then it was there was that inequality that for any compact set the inequality was that so the supremum of t phi such that phi is in c c omega and phi infinity is less than or equal than one and the support of phi this is finite. So this number here for people knowing measure theory should recognize the so-called variation or maybe total total variation of the measure t. Oh okay you can you don't recognize here a measure because usually a measure is defined on sets. So this is another viewpoint of the same thing. We look at measures instead that as set functions as elements of a dual of a space. So measures can be you have two phases of the same thing the set phase where you measure sets mu of omega mu of a you have some additivity and so on or another viewpoint is the functional viewpoint looking at the measure as a linear continuous functional this is so that there is a sort of relay there is a big important theorem which relates the two viewpoints but remember that in principle you can look at measures either as set functions or linear continuous functional on c c in this case. There is a theorem but maybe you have you already know but doesn't matter that relates the two. I want only to say that this quantity is finite because you have that is it is less than m for measures because for measures you just have this without d. So this number and then you take the unit ball here then of course this number is less than m depending on the compact set. So fixed k you have mk bounding that number there. This is why the word local bounded variation the word local is means that your bound is finite on any compact set but in principle it's not finite everywhere just just finite on compact sets. So everything okay depends on this maybe condition one for which there is a compact set containing all the supports and so on. That strange notion of convergence that we've chosen. So this is maybe called the variation of your measure total variation of your measure on the compact set k. Why is this important? Well why is the word local important here? It's crucial because for instance the Lebesgue measure has local finite variation on compact set but it is not it's not bounded in the wall space Rn. You know everybody knows that if you have a compact set of Rn the Lebesgue measure is finite but the Lebesgue measure of the wall Rn is infinite. Now so let us do the so assume now that t is in the prime and assume by contradiction that this is not true. So I have to contradict this. Let us try to contradict without making mistakes. So there exists a compact set k such that for any constant m for any m in n then if phi is in the omega support of phi contained in k then t phi is larger than m max d less than m d phi infinity. So there exists okay so now we have so this this logically goes here so for any phi infinity there exists phi. We can find the test phi such that we have that thing. So there exists a compact set so now take m equal n an integer number and m equal n hence we can there exists hence there exists Wn in the omega support Wn contained in k such that t of Wn is bigger than n max d less than n d Wn infinity. So I've since I have the freedom of choosing capital M and small m then I choose any integer small n so they are equal in this case any integer okay therefore I can fix correspondingly for this n and this n I can fix correspondingly a test function with the support contained in k such that this is true. So here there is n here there is n and in place of phi I have this. So in particular so in particular t of Wn is different from zero so I now I normalize it so I define Vn equal to Wn divided by this number. So this is again an element of the omega also which is the advantage of doing this is that t of Wn is equal to 1 t of Wn is equal to 1 therefore surely t of Wn does not converge to zero hence t of Wn as a number does not converge to zero. I just I have to show what is the support notice also that the support of Vn is also contained in k for any n in n. So you see I have a sequence of test functions having all supports contained in a given compact set so if I show that so if I show that Vn converges to zero then I have a contradiction because t of Wn does not converge to zero. So to to reach a contradiction so to reach a contradiction it is sufficient to show it is enough to show that Vn converges to zero in the omega. Do you agree with this conclusion? Now now you have to use this this this inequality also so from this inequality it follows sorry this inequality it follows that it follows that what about so the d of Vn d of Vn the max the max d less than n d of Vn is less than one over n because this is simply binormalization so and remember that t of Vn is one okay well now if you fix fixing d derivative operator of any order of any order then d of Vn is less than one over n for any n bigger than d which says that Vn converges to zero in d hence is it okay so it works this concludes the proof so we have a criterion for recognizing whether or not something is a distribution so remember this criterion will be useful for you also for measures to check whether or not something is a measure with local finite variation total variation okay examples finally so I have so let me go in the order maybe the most important example but there are several but one basic example is the following so in the order so example one one definition more and then we go to the examples so definition let Tn be a sequence of distributions and T be a distribution we say that Tn converges to T in the prime if point wise it converges point point wise this is simply point wise convergence okay so let me go to the example so example one so let you be a function in l one lock omega so you know what does it mean l one lock right it means that the integral of the absolute that that u is absolutely integrable the absolute value of u is finite if integrated on any compact set okay this everybody knows and you also know what is convergence in l one lock maybe we don't know what is a topology in l one lock but we know what is convergence in l one lock right what is convergence in l one lock what does it mean that a sequence of elements u n converges to in l one lock to a function u in l one lock on any compact set you integrate so again we give a notion of convergence but we don't say what is the topology doesn't matter stay same principle as before we don't declare what is the topology but for us the convergence is enough okay now given u we can consider the following linear map tu canonically associated with u as follows tu is defined as follows on a test function phi this is the lebesgue integral so no mystery on this symbol this is the lebesgue integral okay it is not an abstract symbol meaning duality is exactly the lebesgue integral okay now this for any phi in the omega is this linear yes it is clearly linear is this well defined yes why is it well defined well defined well because this actually is nothing else the integral on the support of phi and the support of phi is some compact set contained in omega and u is integrable on all compact sets therefore this is well defined now in order to check that this is a distribution we have to check continuity what remains is continuity so take phi n converging to phi or to zero doesn't matter that in the omega we have to check that tu phi n minus tu phi converges to zero and this is less than or equal than phi n minus phi dx but what we know is the following that all supports of phi n are contained in some k so that phi n and the support of phi is also contained in k and k is independent of n and we have uniform convergence on this compact so this this object here converges uniformly to zero and therefore this by theorem on measure theory converges to zero so this example one says that to any function u in a one lock we can associate a distribution to you which which which has a representation is not an abstract linear continuous map but it is it is very concrete it is it has this representation is an integral is a linear integral on you linear with respect to you now we have therefore so this way we have a map this way we have a map theta say from l one lock into the prime sending you into tu why i'm saying this because in some sense i want to embed all l one lock functions inside distributions so in this way i want to generalize the notion of function at least l one lock function so what does it mean that i want to look this set inside this well there are various things that i have to check first that this map is injective so that i really embed injective injectively in an injective way l one lock inside something so if i take u different from v it is possible to check that u and v in a one lock omega then one checks that t of u is different from t of v from t of v sorry t of v because if by contradiction t of u is equal to t of v so what does it mean sorry what does it mean that two distributions are equal well it means that on any point they are equal so tu on any point phi is equal to tv on the same point phi for any phi but what is this what is this this is the integral of u phi the x and this because uh u and v are n one lock so this is the equal the integral of v phi the x and therefore we have that for any test phi the integral over omega u u minus v times phi is equal to zero there is a t lamb mind measure theory that ensures that then u equal to phi almost everywhere so u equal to v almost everywhere and therefore this contradicts this so this map here is injective let us see whether also it is let us see whether also it is continuous so it is a continuous injection so um what do we have to prove so i have to show that if u n converges to u in a one lock and i have to see whether or not so is it true that t of u n converges to t of u so what does it mean this remember convergence in the prime is point-wise convergence therefore this means that i have to check that for any test t u n phi must converge to t u phi for any test this is what i have to check but then again this is equal to the integral of u n phi the x and this is equal to the since these are and one lock functions the x okay but again now this is the same you see if you take the difference of this phi as compact support so the difference is the integration on the support of phi of u n minus u support of phi is compact u n is converging in a one lock so on that compact that integral is going to zero is this clear just very easy therefore we have a map sending embedding injectively this continuously inside this now one would like to find so in some sense we can now there is a new viewpoint of functions functions of this sort at least and one lock so very big space functions actually can be considered inside this so there is something larger than functions well there is something larger if i prove that there is something outside so an interesting question is now is there something here which is not here right so that this can be considered larger so injecting is not surjective so this linear map embedding this inside this is not surjective why so and you're you're starting now to understand that we are enlarging the class of objects and therefore also in view of applications to PDs we can look for solutions which maybe are not functions but distributions so this is a leading idea in PDs look for solutions which are not any more functions but other objects which solve the PD in another sense okay so let us go therefore to to example two so example two so let x be a point in omega and consider the following object defined as follows for any phi so this is called the Dirac delta at the point x is called the Dirac delta at the point x let us see whether or not this is a distribution then delta x is in the prime okay linearity is immediate phi plus psi equal phi plus psi at x equal phi of x plus psi of x we have to check continuity we have to check that if phi n say phi n converges to phi in the omega then that then phi n of x converges to phi x we have to check to check that if phi n converges to phi in the omega then well this is again very easy do you see it is immediate because if phi n converges to phi in the omega then there is a compact set containing all the supports where phi n converges to phi uniform you don't need derivatives here you just need the convergence in cc to have this as such indeed indeed this is a measure but anyway if phi n converges to phi then there is a compact set such that all supports are there and the convergence there is uniform okay now there are two possibilities either the compact set contain x or it does not contain x okay so if the compact set does not contain x then all phi and x are zero phi x is zero and therefore this is obviously satisfied if the compact set on the other hand contains x then we have uniform convergence there in particular point wise convergence in particular convergence at this point so I have not written anything maybe I leave you as an exercise to check homework where I am exercise so now the point is that that we want to show and maybe I leave you as an exercise again but this is much more interesting is that now delta x so we know it is in the prime but it is possible to show that it is not in a lock this is again home if it is true this is very interesting and is the explanation of the famous sentence the Dirac delta is not a function that every physicist says correctly Dirac delta is not a function what does it mean this so if you want to show this then well maybe we do this exercise but maybe it would be better to leave you as home and then if you are not able to do let me know so but another exercise is let us consider the sequence so omega equal r consider the sequence u k u n of x sinus of sinus sinus say of n x so these are this sequence is this point twice converging no this does not converge this point twice oh we have already seen that l 1 lock convergence implies convergence in the sense of distributions right because the function the map linear map theta was continuous now instead of having l 1 lock convergence we show that okay so l 1 lock implies convergence in the sense of distribution on the other hand this does not converge point twice but let us see whether or not it converges to something in the sense of distributions so these are functions in l 1 lock obviously so any any any of that function is a distribution by integration and therefore the point is is it true that this converging converges to something in the sense of distributions so the claim is that this converges so even if you don't have point-wise convergence then you have steel convergence in the sense of distributions do you agree with this do you agree with this or another homework is that or that is equal to zero now do you do you agree with this why do you really need you really need to invoke for the analysis for this let us try to do it elementally so what do we have to do this convergence in the prime means point-wise convergence okay so we have to check that so let us consider any test function phi and let us consider this so what is this well this is an l 1 lock function therefore we know that this is represented by rebegg integral okay you and phi dx therefore more concretely this is just sinus of nx phi x dx and now we would like to show that this as n goes to infinity converges to sort of mean value of this sinus function converges to zero yes you know that oh if you want this is correct correct yeah say this is an integral of an on an interval a b okay then make integration by parts what happens if i make integration by parts huh so if i make integration by parts i look this as a derivative so maybe there is a minus here so this is the derivative of the cosine so the cosine no there is okay derivative so there is a cosine 1 over n cosine of nx phi prime of x dx something like this okay there is there is no boundary term here the point is that there is no boundary term because phi is compactly support so if you want i can write everything in a big interval outside the where we're outside of which phi is zero so the the trick is exactly that that there are no boundary contributions so let me write again here this some k some some compact set k containing the support so k is any compact set let's say k is compact is the support essentially k contains the support of well now this is this is this is all because there are various ways to to look at this convergence but now you see this is in absolute value always less than or equal than one this is bounded in absolute value because phi is compactly is a test function so this is finite less than one this is finite by a constant bounded by constant then what remains is say if you want is the length of k the length of k is finite because k is compact and dividing by n this is so before this clearly goes to zero so maybe home i add you an exercise try to see what happens if you change u n x into k into n sinus of nx or also u n of x into n to some power big power alpha try to see whether or not still we have all these functions are in at one lock therefore we can ask exactly the same question is it true or not that that that functions considered as a distribution therefore the t the associated t converges to zero in the sense of distributions okay another homework let us see another example of distribution so let omega and let alpha be a multi index multi index and then consider the following object here phi is in the omega well then show that this is so for instance in one dimension for instance if n in if in one dimension if alpha is equal say one just one index only this would be the following quantity minus phi prime of x phi okay okay so this is not the Dirac delta because the Dirac delta was phi of x with the plus plus so this is another object but still it is a distribution so home home delta x alpha is a distribution of order home so this is home home so let u be a function in l one over n such that the integral of u is equal to one say u positive then try to show so define u k you define u k so this is capital n sorry so define u k x equal k to the n u k x so you scale with the index k k here is an index n small k is not any compact set so let let me call it small n u n and so n is an integer positive integer capital n is the dimension it is fixed so you scale it properly and so try to show that delta the Dirac delta at the origin is equal to the limit of u n in the prime of our n so what the second this i don't understand the question second line this n to the n u of n x so this is we have to check the following say for instance this is given in order to have the following in order to have the following normalization so u n x so if you integrate on r n u n x this is equal to the integral over n to the n maybe to the minus n actually and u to the n x so you do a change of variable n x equal to y and so this is equal to n to the n and then the x is equal n to the minus n dy so n to the minus y dy so this is equal to 1 yet in the prime why don't you say this i mean u n is in l 1 lock so you went you and yes i am i am i am identifying the things i mean once i have an l 1 lock function or yeah more precisely one should write t of u n but very often i identify u n with t of u n so i look as at u n as a distribution and therefore this is exactly well i mean that the proof of that is exactly as you said i mean if you if you don't understand the statement is statement says that this is t of u n to be extremely precise so you have to show that if you take a test function then you have the integral of phi of y and then you have n to the n and then you have u of n x the x the Fourier scale as before variables so you end up simply with phi of y over n and then you have u of y dy and then you show that this converges to phi of zero this is the i mean this is the meaning of the exercise so sorry you are right it would be better to write t of u n when i don't do this i mean i identify you with t of u whenever i am in one in l 1 lock is it okay