Mat 1501 lecture notes, penultimate installment

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1 Mat 1501 ecture notes, penutimate instament 1. bounded variation: functions of a singe variabe optiona) I beieve that we wi not actuay use the materia in this section the point is mainy to motivate the definition we wi ater introduce of BV functions and to reca some resuts, perhaps famiiar, that we wi see extend in a natura way to BV functions of severa variabes. so if you have imited amounts of time, then I recommend that you read other parts of these notes first.) Definition 1. We wi say that a function f : is has bounded variation if 1) sup fx i ) fx i 1 ) <, i= where the supremum is taken over a sequences... < x i < x i+1 <... such that x i ± as i ±. The quantity appearing on the eft in 1) is caed the tota variation of f and is denoted T V f). Here are some facts about functions of bounded variation that may be famiiar from anaysis casses: Any function of bounded variation may be written as the difference of bounded monotone functions. That is, if f has bounded variation, there exist bounded nondecreasing functions f 1, f 2 such that 2) f = f 1 f 2. Moreover, this can be done in such a way that 3) T V f) = T V f 1 ) + T V f 2 ). As a resut 4) fx ) := im y x fy) and fx + ) := im y x fy) both exist for every x. If f ) is a sequence of functions such that sup T V f ) = M <, and sup f x) <,,x then there is a subsequence and a function f such that f f a.e. and in L 1, and T V f) M. Exercise 1. Prove this. One way to do this is to first prove it is true if each f is nondecreasing, and then deduce the genera case from 2), 3). if f C 1 then 5) T V f) = f x) dx, so that a C 1 function has bounded variation if and ony if this integra is finite. Exercise 2. Prove 5). In fact it can be deduced in about one sentence from the fundamenta theorem of cacuus.) 1

2 2 Very cosey reated to the above definition, we have: Definition 2. We wi say that f : is a BV function if there exists a signed adon measure µ such that 6) fx)φ x) dx = φx) dµx) for a φ Cc 1 ) and µ ) <. When 6) hods, we wi write f BV ), and we wi write f to denote the signed measure µ in 6). Note that if f is C 1, then fx)φ x) dx = φx)f x) dx for a φ C 1 c ). Motivated by this, we interpret 6) as asserting that the weak) derivative of f is a signed measure. Hence the notation f for the measure in 6). The reationship between the two definitions is given by Lemma 1. If f : has bounded variation, then f BV ). Moreover, for any interva a, b), the measure f satisfies 7) u a, b)) = fb ) fa + ). Conversey, if f BV ), then there exists a function f of bounded variation such that f = f a.e.. Proof. 1. First, we wi prove 6) under the assumption that f is a bounded, nondecreasing function. Assume that this hods, and define λ : Cc 1 ) λφ) := fx)φ x) dx By the dominated convergence theorem and a change of variabes, φx h) φx) fx) fx h) λφ) = im fx) dx = im φx) dx. h 0 h h 0 h The fact that f is nondecreasing impies that 8) λφ) 0 if φ 0. Hence λφ) λψ) if φx) ψx) for a x. In particuar, if φ sup 1 and supp φ) a, b), then fx) fx h) λφ) im 1 a,b) x) dx h 0 h 1 a ) b = im fx) dx + fx) dx h 0 h a h b h = fb ) fa ). It foows that λ is continuous with respect to the topoogy of C c ). Since its domain of definition is a dense subset of C c ), we concude that λ has a unique extension to a continuous inear functiona λ : C c ). Such a functiona can

3 1. BOUNDED VAIATION: FUNCTIONS OF A SINGLE VAIABLE OPTIONAL) 3 be represented by a adon measure µ. since λ is positive in the sense of 8), and ceary λ inherits this property.) In particuar, 6) hods for this µ. 2. If f has bounded variation, we can write it as a difference of nondecreasing functions f 1 f 2 and appy Step 1 to both of these, to obtain a measure µ satisfying 6). 3. We remark that if f has bounded variation, then 6) continues to hod if φ is merey Lipschitz continuous. This can be deduced by approximating φ by smooth functions in a standard way and appying 6) in the smooth case) to each approximant. Now for a < b and 0 < ε < 1 2 b a), et 0 if x a, b) φ ε x) := 1 if x a + ε, b ε) inear if x [a, a + ε] [b ε, b]. Then using 6), µa, b)) = im φ ε x) dµx) ε 0 = im fx)φ ε 0 εx) = im ε 0 1 ε a+ε a = fb ) fa + ). fx) dx + 1 ε b b ε fx) dx Hence 7) hods. 4 Now assume that f BV ). Without giving a the detais, there are a coupe of ways to show that f coincides a.e. with a function f on bounded variation. Let us define gx) by gx) := µ, x)) One can then check that g has bounded variation, and that f g) = 0 in the sense that f g)φ dx = 0 for a φ C 1 c ). One can further check that, as a consequence, there exists some constant c such that f g = c a.e.. Then the concusions foow, with f = g + c. Aternativey, one can write f ε := ψ ε f, where ψ ε is a standard moifier. Then for every e > 0, one can check that T V f ε ) = f εx) dx µ ). Then it foows from Exercise 1 that there exists a subsequence that converges a.e. to a function f of bounded variation. On the other hand, we know that f ε f in L 1, and hence upon passing to a subsequence) amost everywhere. Thus f = f a.e.. )

4 4 We wi be interested in generaizations of the above definitions to spaces of functions N, for exampe, or more generay N E, where E is a metric space. 2. bounded variation : functions of severa variabes We say that a function u : is a BV function if u L 1 ), and there exists some C > 0 such that uy) ζy) dy C ζ for a ζ C 1 c, ). If this hods, we wi write u BV N ). For such a function u, the map ζ Cc 1 ; ) uy) ζy) dy is a inear functiona that is continuous with respect to the topoogy of C 0, ) and hence extends to a bounded inear functiona λ u C 0, ). Then a version of the iesz epresentation Theorem asserts that there exists a adon measure on, which we wi denote Du, and a Du -measurabe function σ :, such that λ u ζ) = ζ σ d Du for a ζ C 0, ) We wi aso sometimes write u xi to denote the signed measure defined by u xi A) = σ i d Du, A Bore A and Du to denote the vector-vaued measure u x1,..., u xn ) = Du σ. Then we can identify Du as the gradient of u in the weak sense. For this reason, one often says that u BV ) if and ony if Du is a measure. We reca that the iesz epresentation Theorem aso guarantees that { } Du O) = sup u ζ dl n : ζ Cc 1 ; ), suppζ) O, ζ 1 for any open O. Note that Du, σ and Du are characterized by the identity 9) u ζ dl n = ζ σ d Du = ζ ddu We have just combined and earier identities, with partiay new notation.) We wi use the notation u BV := u + Du ). We coect some basic facts about BV functions. Lemma 2. If u C 1 ) L 1 ) and u dx <, then u BV ) and Du = L n u.

5 2. BOUNDED VAIATION : FUNCTIONS OF SEVEAL VAIABLES 5 Proof. This foows directy by conparing 9) with the formua u ζ dl n = ζ u dl n for ζ Cc 1, ), which foows for u C 1 from the divergence theorem. The previous emma impies that if u C 1, then Du = L n u and that σx) = u u x). Athough σ is undefined where u = 0, this does not concern us, since the set {x : ux) = 0} has Du -measure 0, and σ : is ony required to be Du -measurabe. The next resut is rather straightforward exercise especiay if you have ever seen any simiar arguments, which you shoud have seen by now...) Lemma 3. weak ower semicontinuity: Assume that u ) is a sequence of BV functions, and that sup Du ) < and that u u in L 1. Then Du Du weaky as measures, and Du ) im inf Du ) Lemma 4. weak density of smooth functions. Assume that u BV ), and et u ε := ψ ε u, where ψ ε x) = 1 ε ψ x n ε ) for some fixed ψ C c ) such that suppψ) B0, 1), ψ 0, ψ = 1. Then u ε u in L 1 ), and } Du ε Du weaky as measures, and Du Du ε Du ε ) Du ). Proof. It is a standard fact that if u L 1 ) then ψ ε u u in L 1 as ε 0. This can be deduced for exampe from the density of continuous functions in L 1, which is a consequence of Lusin s Theorem.) Then in view of the above emma, we aready know that Du ε Du weaky as measures, and aso that 10) Du O) im inf ε 0 Du ε O) for a open O. So to prove that Du ε Du weaky as measures, we ony need to show that Du F ) im sup Du ε F ) for a cosed F. ε 0 This wi foow once we verify that 11) Du ) = im ε 0 Du ε ), since then if F is cosed, writing O := \ F, we have Du F ) = Du ) Du O) im Du ε ) im inf Du ε O) ε ε = im sup Du ε F ). ε So to compete the proof, we must ony check 11).

6 6 To do this, et ζ Cc 1 ; ) be any vector fied with compact support in, and such that ζ 1. Then for any ε > 0, using basic properties of convoutions, u ε ζ dl n = ψ ε u ζ dl n = uψ ε ζ) dl n = u ψ ε ζ) dl n. It foows from basic properties of convoutions that ψ ε ζ ζ 1, and hence that the right-hand side is bounded by Du ). Since this is true for a ζ, we concude that Du ε ) Du ) for every ε > 0. By combining this with 10) for O = ) we obtain 11). Lemma 5. if u BV ) and v, then 12) τ v u u L1 ) v Du ), for τ v ux) := ux v) As a resut, if ψ ε ) is a standard moifier, then 13) ψ ε u u L 1 ε Du ). Proof. In view of Lemma 4, it suffices to prove 12) for u C 1 ) BV ). This is an exercise see beow.) And if u C 1 BV, then ux v) ux) = = d ux sv) ds dx ds ux sv) v ds dx e v ux sv) dx ds 0 n = v u = v Du ). It foows rather directy that ψ ε u u L 1 ψ ε y) ux y) ux) dx dy ε Du ). B0,ε) Exercise 3. Show that if 12) hods for every u C 1 ) BV ), then in fact it hods for every u BV ). Lemma 6. compactness. Assume that u ) is a sequence of BV functions and that sup u BV <. Then there exists a subsequence that converges to a imit u in L 1 oc n ). That is, for every bounded open W, u u L W ) 0 as..

7 2. BOUNDED VAIATION : FUNCTIONS OF SEVEAL VAIABLES 7 Proof. For every, et u,ε := ψ ε u. It is a standard fact that u,ε is smooth, and u,ε = ψ ε u ψ ε u 1 Cε n+1) sup u BV. Simiary u,ε = ψ ε u ψ ε u 1 Cε n sup u BV. Thus for every fixed ε > 0, the sequence {u,ε } =1 is bounded and equicontinuous. By the Arzea-Ascoi Theorem and a diagonaization argument, it foows that there is a subsequence {u,ε } that converges uniformy on compact sets, and and hence aso in L 1 oc, to a imit u,ε. Note aso that for every bounded open set W, using 13), u,ε u,δ L1 W ) im inf u,ε u,ε L1 W ) + u,ε u L1 W ) ε + δ) sup Du ). + u u,δ L1 W ) + u,ε u,δ L1 W ) Since L 1 is compete, it foows that u := L 1 im ε 0 u,ε exists and that u u,ε L 1 ) = sup u u,ε L 1 W ) Cε. W bounded, open Then for every bounded open W and every ε > 0, im sup u u L 1 W ) im sup u u,ε L 1 W ) + u,ε u,ε L 1 W ) ) + u,ε u L 1 W ) where we have again used 13). Since ε is arbitrary, this competes the proof. For u BV ), we wi use the notation Du Bx, r)) MDux) := sup r>0 α n r n Cε where α n := L n B0, 1)). Thus, MDu is the maxima function associated to the tota variation measure Du. It foows from rather standard arguments, using the Vitai covering emma, that L n {x : MDux) < λ} ) c n λ Du n ). In particuar, MDu < L n amost everywhere. We wi need the foowing fact. Exercise 4. Assume that u BV ), and define u ε := ψ ε u as in Lemma 4. Prove that MDu ε x) MDux) as ε 0, for every x. ) Lemma 7. Assume that u BV ). If x, y are Lebesgue points of u then 14) ux) uy) C n x y MDux) MDuy))

8 8 Proof. 1. We caim that if x is a Lebesgue point of u, then ux) uy) 15) dy MDux). x y Bx,r) We first prove this for u BV C 1 ). It ceary suffices to prove it for x = 0. Thus we compute, using Fubini s Theorem and a change of variabes, u0) uy) dy = d B0,r) y α n r n usy) ds B0,r) 0 y ds dy 1 1 α n r n usy) ds dy B0,r) = α n r n s n uz) dz ds = MDu0). B0,rs) Du B0, rs)) α n rs) n ds This is 15). For arbitrary u BV, we define u ε = ψ ε u as in Lemma 4. Then 15) appies to every u ε for every ε. Aso, it is rather standard, and not hard to check, that u ε u at every Lebesgue point of u, so 15) foows from Exercise Next, we define θ = θn) 0, 1) by requiring that if x 1, x 2 are any distinct points in, and r := x 2 x 1, then L n Bx 1, r) Bx 2, r)) α n r n = 3θ The point is that the numerator depends ony on r := x 2 x 1 and scaes ike r n, so that such a number exists. Now fix any two Lebesgue points x 1, x 2 of u, and et A := Bx 1, r) Bx 2, r), for r := x 2 x 1. We caim that for i = 1, 2, 16) L n {z A : ux i) uz) > 1 ) x i z θ MDux i)} < 1 3 Ln A). Indeed, for any k > 0, Chebyshev s inequaity impies that k L n {z A : ux ) i) uz) > k} < x i z A ux i ) uz) dz x i z Bx i,r) α n r n MDux i ) ux i ) uz) dz x i z = Ln A) MDux i ) 3θ Setting k = 1 θ MDux i), we deduce 16). 3. It foows from 16) that there exists z A such that if we define C n := 1 θ, then ux i ) uz) C n MDux i ) for i = 1, 2 x i z

9 2. BOUNDED VAIATION : FUNCTIONS OF SEVEAL VAIABLES 9 Since x i z r := x 1 x 2 for z A, it foows that ux 1 ) ux 2 ) ux 1 ) uz) + uz) ux 2 ) C n x 2 x 1 MDux 1 )+MDux 2 )). Coroary 1. If u BV ), then the set {x, ux)) : x } +1 that is, the graph of u) is countaby n-rectifiabe. Exercise 5. Suppy the short) proof. Finay we prove Lemma 8. Assume that µ is a signed adon measure on with finite tota variation, and that ζ dµ C ζ for a ζ Cc 1, ). Then µ L n, and dµ dl is a BV function. n As a consequence, if L is a bounded inear functiona on Cc ) such that Lφ) C φ for a φ C c ), L ζ) C ζ for a ζ Cc 1 ), then there is a function u BV ) such that Lφ) = uφ dx for a φ C c ). Proof. Let u ε := ψ ε µ, where ψ ε is a standard moifier, so that u ε x) := ψ ε x y) dµy). Then it is straightforward to check that u ε is a BV function, and moreover arguing as in the proof of 11) in Lemma 4) that u ε L 1 µ ), Du ε ) sup{ ζ dµ : ζ Cc 1, ), ζ 1} for a ε > 0. Thus there exists a subseqnece ε and a function u BV such that u ε u in L 1 oc n ). Then for any φ C c ), u φ dl n = im u ε φ dl n ε 0 n = im ψ ε ε φd mu 0 n = φd mu. This impies that µ L n, and then it foows that u BV. The fina concusion of the emma then is a consequence of the iesz epresentation Theorem.

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