 Zdaj se izgleda, da se je zelo v lemu, da se je, da je F in NAC, in je ta F prime over X nekaj nekaj zelo vzelo v svoj del, potem F je konstant. Zato je to zelo vse? the BV function intersected continuous function contained in AC. If you think it's true there are some examples you can find a function which belongs to both, but is not in AC in view of this. So for example the counter function you have, that is in BV. It's continuous but it's not in AC, because otherwise it's ... ... ... je zelo pravda všem, ... ... je nekaj kod, ok. Ok, ... značili smo v zelo, ... ... vzelo, da je to rečen, ... vzelo, da je začala ... vzelo, da je začala ... ... vzelo, da je začala ... vzelo, da je začala ... ... ... ... ... ... ... ... ... ... ... ... ... ... Pazite. V svoj cel, da smo nespešili to, da je nekaj na vse. Sredaj. A zato, da je, da je, da je, da je ni na vse, da je nekaj na vse. Ok, smo nekaj. Če bodo boš nekaj na vse, zelo, da je ... In je, da je ... ... lema, da je nekaj na vse, da smo nekaj na vse. Ok, zelo smo prišli to. Tukaj, tako, so vzout, da je vzout, da je F zelo v AC. Ok, potem, včasno, je to vzout, da je vzout. Tako, da vzout, da je vzout, da je vzout, da je vzout has the difference of two increasing function, OK. OK. f1 and f2, such that we have that capital F is f1 minus. In nekaj, imamo F1 in F2 prvnega rečenja v vseh lebecku lemu, OK? OK. Tako, imamo, da F je nekaj, F1 prvnega rečenja, minus F2 prvnega. Now, we pass the integral, bear the, behave the absolute values. Let's see it's less or equal. But we know that they are increasing so we can get rid of the absolute values. So this is equal. Okay, then here we apply the leback theorem because the models function. však. Zdaj smo so, da je to očočno neko mene vb. Ano vb in a. Ano vb. Ano vb. Ano vb. Ano vb. Ano vb. Ano vb. Ano vb. Ano vb. Ano vb. Tako, imamo tudi. Zato, da smo izvizili, da je injavnjavnja vsef in vsef je injavnjavnja, je vsef, da je izvizil in definitivno injavnjavnja vsef z njih tudi, tako, da je vsef, mir 可以 Refine as the Integral between a t g materia. And now we know that since f of prime is figurably yesterday, we prove for aんだ formula result, we have that g being the kaj je inakralli, kaj je inakralli fungšnj, g je inac. Zdaj smo tukaj tukaj. Zdaj smo tukaj izgledajte, izgledajte tukaj, tukaj f, tukaj x, tukaj g in f. Zdaj smo tukaj tukaj z inac. zdaj smo tukaj, ker je odgledajte z inac. Zdaj smo tukaj zvrčajte, tukaj ti je z vzulj, tukaj x, tukaj minus, tukaj vzulj, tukaj z inac. Tukaj je z vzulj. Tukaj x, Vse zelo je 0, ovo je prizvodno, da je bil je vse pomej. Tako, ovo je prizvodno, da je začal, da je F, z tem, da te zelo pogleda, da je to, da je napisak, skupaj, da je tudi kaj je vzela c in vzela vzela vzela vzela, zelo je vzela, da vzela vzela. Zato je tudi, da je f of x je, je to konstant, zazve, da je definitivne integral f of x, f prime of t, t. To je, to je f of a, a x t. Ok, zato to je naša karakterizacija, kaj je zelo za fungšenje, kaj je v AC. AC je zelo vse klas, kaj je zelo, kaj je zelo za fungšenje teorema. Ok, značujem vse argument. Značujem vse konvex fungšenje. Ok, značem vse definitivno. straight. So we consider a function phi defined in the open interval A,B with values in r, and this function is said to be convex if you have that for any x and y in A,B and any lambda, if we took lambda in between 0 and 1, Vsih, kaj smo imeli. Faj, komputer in delina kombinacije. Faj. Lambda x plus 1 minus lambda y. Zelo, nekaj je počkaj vse x in y. Zelo, nekaj je lambda, f of x, phi of x, sorry. Plus 1 minus lambda, phi, y. Ok, geometrikali, ja da vere, ki le jo tamo znam, ovo se pridem. Vse, geometrikali, ti se doželimo, je to corsa vtevna vse z vsem, vse z x , f of x, taj z nih dne, i x, phi of x una, so všeo, so vse ope, so všeo, pouze vse, o vačne. Zelo, je bila počkaj vsez, If you have x and y, so this code is easier. Now we're going to prove a lemma concerning the property of convex function. And this lemma would be somehow a building block for the next properties. Tako, da je to, če startujte z konvečnjih fungšnjih, 5, 5 in in B, v delizionu. Tako, da je to, če tako x in y, ok, zelo x in y in B, kaj je nekaj, naprejno, da je je vrčak tronov, da je vrčak tronov, da je, da, skupaj da je so treba. Če ste podjetjenje izmašljali, da so prošli, da se največje. A knaž. A drž ne koriš. Zelo, da je to bo pošli kod si take. Je, da je to bilo, da vrčak tronov, da so pripošli. Tako že ga ne plošem, kako pripošlo. variable x and y. Tukaj, definitivno je simetrično v x in y, ne? OK, zelo smo prišli to. Tukaj, zelo smo prišli x. Zelo smo prišli x. Zelo smo prišli, da smo prišli, da smo prišli v parametri, da smo prišli, da smo prišli, da smo prišli v variable y, OK? x coma vključne vladi. K �ejenje všeč sem bilo dassimte, tenne k data sem bilo detalolo prav sfl polite, vs Vj, in vj primi tudi otensek od pateke, OK? Zato j夲u vj primi, najdirekt vj primi, ko je ukriče na vizu po vzima- Prevojna vizu, ko je ukričen, OK? Kaj so je to vzivo mač? G. Vzima- vzima- ukričen G. Q. Vzima- To je nekaj svis. OK? zelo potrebno je, da zelo potrebno se neseliti. officially we have to use the hypothesis that y is in between y prime and x so we can express it as a linear combination. So since y belongs to x and y prime, we have that there exist on lambda. lambda vzal, tako je zelo stranje 0 in 1, tako da je ta je, da je y je nekaj lambda x plus 1 minus lambda y, y prime. Ok, zelo, da zelo, da zelo, da smo izgledali tako, da smo tukaj, da smo komputati phi tukaj tukaj tukaj je tukaj tukaj v lambda x x plus 1 minus lambda y prime. In so, baj konvexit, to je nekaj nekaj, danes lambda times phi of x plus 1 minus lambda phi of y prime. Ok. Ok, nao, so smo prišliši v tih, v tih košnji. Tako, skupajte. Tako, skupajte g, x, y, ko je pi, y, minus pi, x divided by minus x in ta denominator je pozitiv, ok? Tako, da je to, da je nekaj nekaj nekaj nekaj, dan, ok, lambda pi, x plus 1 minus lambda pi, y minus pi, x divided by minus x and this is equal if you collect pi, x so this is lambda minus 1, actually first we wrote this 1 minus lambda pi, y prime, this is pi, y prime minus 1 minus lambda pi, x divided by minus x ok, dan we observe that we can write y minus x has minus y minus lambda, x plus y minus lambda, y prime which is equal to y 1 minus lambda, y prime minus x ok, so finally what you get is that this is less or equal than pi, y prime minus pi, x less than y prime minus x and this is exactly g, x, y prime ok, and so we are done we are done for the first case so now we consider another case in the case when x is in between y and y prime ok, so we have that by the step before you can achieve what you want you have that g, x, y are less than g, y prime y, so here you apply x and y prime this is symmetric so this is g, y, y prime and this is less again by the step before than g, x, y prime and again you use this that is increasing with respect so you think y prime fixed and you have that y and x lies on the same side with respect to y prime so we can use the step before because here because y and x on the same side with respect to y prime ok, the other cases can be treated analogously so this concludes the proof ok, so with this proof we can prove many other properties interesting properties of the convex function so which are the following I stated them as a proposition ok, so we have phi maybe with values in R b, convex function then we have the following fact a, we have that phi is lip sheets on each closet sub interval of the type ok, of a, b the second fact is that so you have that the right and the left derivative of phi exist at each point of a, b this is c you have the following inequality so you have that the left derivative which I denote with a minus sign of phi of x is less or equal than the right derivative of phi of x of course for any x in a, b and d these two left and right derivatives are monoton increasing and finally you have that they coincide except on a countable number of point ok, so we prove the point a ok, so we have x and y in a, b and assume that for instance just to fix the idea that x is less than y ah so by the previous lem that one tells you that this function g is increasing in both variables you can deduce the result ok, so let me let me first consider ok, before before doing this we have to consider some sub-interval because we have to prove this in each closed sub-interval so we consider a sub-interval c, d of a, b and then we know that there exists another interval which is in between the two no, actually an interval a, b which is still containing a, b but contains this c, d ok, so you can find it so basically what we have that so we have that a is in between alpha c and then you have less than d, less than beta and less than b ok, then we took x and y x and y belongs to c, d because we are interested in proving the fact that if I is there and so by the previous lemma we have that we have that phi c minus phi alpha c minus alpha is less than phi y minus phi x y minus x less or equal than phi beta minus phi d b minus d so in the end what we have is that if you take the absolute values you find that f of x minus y is less or equal than a constant times x minus y ok and so this constant would be precisely the maximum between these two in absolute values so phi is slip sheets in c, d ok, to prove b ok, to prove b again we used the previous lemma ok, so let let us fix some x in a, b and we define function g g over h which is defined as phi of x plus h so it is an incremental quotient so of course minus phi of x divided by x so again, by the previous lemma we have that that g is increasing with respect to h so we have that both the limit from the left and the right side exist and they are finite ok, because phi is slip sheet so both the limit the right and the left hand side exist they are finite and this is because we already know that phi is slip sheet ok, and then we have to prove point c ok, so c tells you that the left hand side derivative is less or equal than the right hand side ok, so take now one h positive and again we use this function here so you have that g minus h is what? is phi of x minus h minus phi of x minus h this is less or equal and this is why this is because g is increasing then phi of x plus h minus phi of x divided by h this is equal to g of h and so the inequality is preserved when you take h when you let h goes to 0 so they taking the limit 0 plus then we are done ok, finally point d ok, so both derivatives are monoton increasing ok, so take x less than y and so we consider fx plus h minus phi of x this is less than phi of y plus h minus phi of y divided by h and again you take the limit from both sides for h which converts to 0 from the left and from the side from the left or from the right and then we obtain they are both monoton increasing ok, so what remains to prove is that they coincide except to a countable number of points ok, so so we have that you observe that the left for instance the left oh no, sorry the right hand side it means the same has at most a countable number of points of discontinuities and this is why it is monoton and finite cause it is and finite will not be ok, so we can take a point of continuity of the right hand side derivative ok so we have that if you have that for instance y is less than x by the previous lemma you have two facts one is that the x plus phi of x is less or equal than phi of x minus phi of y x minus y less or equal than the x minus phi of x this is one and by step c we have another fact so we have that the x minus phi of x is less or equal than the x plus phi of x so we combine the two and what we get we get that is less or equal than the x minus phi of x equal than of x ok, but I recall you the x is a point of continuity for the phi of x plus then if we let y tends to x we get a conclusion so if y tends to x then we are done ok, then the two coincide ok, so this concludes the proof of the full lemma proposition ok and then ok, now we want to introduce the concept of supporting life of a convex function but before doing this we need a very quick result and I couldn't you need this so it was just you combine one and two and here you get these two inequalities phi of y ok so you have a minus here you have plus and plus here x and here y you let y tends to x and then you are done so if again you have we start by convex function now and let x not be a point in a b ok, we have that if we have which somehow plays the role of a slope is in belongs to this interval to close at 1 for dx minus of phi x not dx ok, this may reduce to just one point in view of it can also be open then we have that phi of x is larger equal then what is called the supporting line of phi ok, the proof so we divide two cases is less than x not ok, then for what we saw before we have that incremental quotient of phi x phi x not x minus x not is lesser equal than dx minus phi x not which is lesser equal than m ok and so we prove because is less than x not ok, so the sign are reversed and the other, if you prove the other way around you have that analogously is larger equal than and so we are done ok, just a brief definition so the line y phi x not plus m times x minus x not is called the supporting line of phi at x not if it always lies below the graph of phi so namely you have that phi of x is larger equal than f phi of x x not plus mx minus x not this is for any x in AP now we prove the so called the inequality so do you notice inequality ok, so you have phi in a convex function and then you have ok, actually this time we have to let a equal to minus infinity and b equal to to plus infinity so f is defined in the whole real line and we assume that f is an integrable function on 0,1 ok, then the convex inequality tells you that if you have think that the integral between 0,1 of phi computed in f of t in dt this is lesser equal than phi of the integral provided phi is convex we prove this ok, so we introduce alpha to be equal to the integral between 0,1 of f of t in dt ok, this is of course is finite because f is integrable and we indeed we consider the supporting line at x not at alpha actually and let y to be equal to m x minus alpha plus and so in equation of the supporting line we have that and phi f of t is larger or equal then m f of t minus alpha plus phi alpha and then from both side with respect to t so we have that 0,1 ok, so this is 0 because of our choice of alpha and so we are done ok, so we have that indeed so and this concludes the proof of the Jensenian quality ok, we can consider now this exercise ok, so we consider a sequence finite sequence of positive number and then we have the following so we have that the geometric mean is always less or equal than the arithmetic one ok, this is a consequence of the Jensenian quality so the geometric mean is the following, you take the product with this phi I denote the product between this positive number and then I write this too ok, i n is less or equal to 1 over n times the sum of bi so this is the arithmetic mean and this is the geometric one so do you have any idea on how to prove this ok, so the convex function that is convenient to to consider to prove this is the exponential function so so we consider phi of x equal to e to the x ok, which of course is convex ok and then you have the interval 0,1 and you divide it into n sub interval of the same length ok so 1 up to n, so you have call them so i1 i denote them with ii and you have that the measure of ii is equal to 1 over n of course these are disjoint ok, now so somehow we fix our choice for the convex function and now we have also to fix a choice for the integral function and this would be done in terms of this interval so we will we will define the function f has the sum the linear combination of the characteristic function of ii where the coefficients will be fixed later the coefficients are alpha i and I will choose them later ok, so we want to compute the two parts which appear in the Jensen inequality so we have that so let me go down here ok, you have that exponential of the sum of alpha i t is equal to e the sum of alpha i times 1 over n this is dot ok, this is the first part the other one is the integral of the exponential of this 0,1 of course ok, so this is equal to what, this is equal to the sum over j j goes from 1 to n oh, you can just consider the integral you can split this all this integral into the integral over this the integral over this interval ij of the sum of alpha i and moreover what you have if you observe that you observe that we have that the integral over ij of this exponential of the sum dT this is equal to just the integral over ij or e alpha ij because the other one the other terms are equal to 0 outside ij and ok, so this is equal finally what you get is this is equal to the sum over j of the integral of ij e alpha j equal to and this is to dot so you combine one dot with two dot and then which what remains to fix are the choice of alpha j so ok, if you want to take the same this is not important ok, so if we set if we set alpha j to be the logarithm of beta j this is possible because our beta j was positive strictly positive then we have that oh, you have that e and this is equal to and so we are done ok let's conclude this is for the Jensen inequality you have to combine this dot and two dot are the two part are the two side of the Jensen inequality for the special choice of phi and f if you check I mean this is for the Jensen inequality ok so we start by saying that calculating the left hand side of the Jensen inequality for our special choice of phi and f and we saw that this is equal to this then we come to the second the right hand side of the Jensen inequality for our special choice of phi and f and we saw that these are equal to this so we combine the two and we use the Jensen inequality and you have this inequality ok you're not convinced it is I mean it's simple, it's just because of the Jensen inequality and so we are done ok, just let me introduce a brief result concerning a new argument actually and then we can stop ok, this concerns somehow the LP spaces this would be so the last argument of the course somehow are the spaces which has a function that has the p-power integrable, ok so the first theorem would be the Young inequality so you start from a and b to positive number no negative and you take some lambda and lambda is in between 0 and 1 ok, then we have the following a lambda times b1 minus lambda is less or equal than the sum of lambda a plus 1 minus lambda b, ok, somehow I mean, either we prove it but I will prove it anyway but of course this is you can see it as a special instance of the of the Jensen inequality, ok or what we just prove here ok, so we define the function f of t equal to lambda t plus 1 minus lambda minus t lambda for t positive ok, we do the derivative of t is equal to lambda plus minus lambda t minus 1 and ok I collect lambda 1 minus so we have that prime of t is equal to 0 if and only if t is equal to 1 and moreover we have that it is f is increasing so the derivative is positive for t larger than 1 and is negative for t less than 1 so we have that f as a minimum and yes, f has a minimum in t equal to 1 and so we have that since we have t equal to 1 is a minimizer and since f1 there is equal to 0 then we have then for any t, which is different from 1 we have that f of t is positive so what we want to prove is that lambda a plus 1 minus lambda b is larger equal than a lambda b1 minus lambda so in particular what this means that lambda a plus 1 minus lambda b minus a lambda b1 minus lambda is larger equal than 0 so call this star and now we have that if b is equal to 0 this is trivial otherwise we can divide for b if b is equal to 0 then star is trivial if b is positive so we have that this becomes lambda a plus 1 minus lambda b minus a lambda b1 minus lambda is equal to so you collect b here and then you have lambda times a divided by b plus 1 minus lambda minus a lambda b minus lambda a b ok so if you put let t equal to a over b then somehow we are done so we have that ok so we have that for what we saw before is different from 1 so when a and b are different we have that b f of t is positive ok then we are done for t different from 1 and moreover we have that f of t is equal to 0 if and only if t is equal to 1 if and only if a is equal to b and this tells you that for the young inequality you have the equality in the if and only if a is equal to lambda in the young inequality the quality sign holds if and only if a is equal to b let's prove the theorem ok so on Tuesday we will go on with the TLP spaces for today I think we are done