JUHA KINNUNEN. Sobolev spaces

Transcription

1 JUHA KINNUNEN Sobolev saces Deartment of Mathematics and Systems Analysis, Aalto University 217

2 Contents 1 SOBOLEV SPACES Weak derivatives Sobolev saces Proerties of weak derivatives Comleteness of Sobolev saces Hilbert sace structure Aroximation by smooth functions Local aroximation in Sobolev saces Global aroximation in Sobolev saces Sobolev saces with zero boundary values Chain rule Truncation Sequential weak comactness of Sobolev saces Difference quotients Absolute continuity on lines SOBOLEV INEQUALITIES Gagliardo-Nirenberg-Sobolev inequality Sobolev-Poincaré inequalities Morrey s inequality Lischitz functions and W 1, Summary of the Sobolev embeddings Direct methods in the calculus of variations MAXIMAL FUNCTION APPROACH TO SOBOLEV SPACES Reresentation formulas and Riesz otentials Sobolev-Poincaré inequalities Sobolev inequalities on domains A maximal function characterization of Sobolev saces Pointwise estimates Aroximation by Lischitz functions Maximal oerator on Sobolev saces

3 CONTENTS ii 4 POINTWISE BEHAVIOUR OF SOBOLEV FUNCTIONS Sobolev caacity Caacity and measure Quasicontinuity Lebesgue oints of Sobolev functions Sobolev saces with zero boundary values

4 1 Sobolev saces In this chater we begin our study of Sobolev saces. The Sobolev sace is a vector sace of functions that have weak derivatives. Motivation for studying these saces is that solutions of artial differential equations, when they exist, belong naturally to Sobolev saces. 1.1 Weak derivatives Notation. Let R n be oen, f : R and k = 1,2,... Then we use the following notations: C() = {f : f continuous in } su f = {x : f (x) } = the suort of f C () = {f C() : su f is a comact subset of } C k () = {f C() : f is k times continuously diferentiable} C k () = Ck () C () C = C k () = smooth functions k=1 C () = C () C () = comactly suorted smooth functions = test functions W A R N I N G : In general, su f. Examles 1.1: (1) Let u : B(,1) R, u(x) = 1 x. Then su u = B(,1). 1

5 CHAPTER 1. SOBOLEV SPACES 2 (2) Let f : R R be x 2, x, f (x) = x 2, x <. Now f C 1 (R) \ C 2 (R) although the grah looks smooth. (3) Let us define ϕ : R n R, 1 e x 2 1, x B(,1), ϕ(x) =, x R n \ B(,1). Now ϕ C (Rn ) and suϕ = B(,1) (exercise). Let us start with a motivation for definition of weak derivatives. Let R n be oen, u C 1 () and ϕ C (). Integration by arts gives u ϕ u dx = ϕ dx. x j x j There is no boundary term, since ϕ has a comact suort in and thus vanishes near. Let then u C k (), k = 1,2,..., and let α = (α 1,α 2,...,α n ) N n (we use the convention that N) be a multi-index such that the order of multi-index α = α α n is at most k. We denote D α u = α u x α xα 1 1 = α 1 x 1... αn x n u. T H E M O R A L : A coordinate of a multi-index indicates how many times a function is differentiated with resect to the corresonding variable. The order of a multi-index tells the total number of differentiations. Successive integration by arts gives ud α ϕ dx = ( 1) α D α uϕ dx. Notice that the left-hand side makes sense even under the assumtion u L 1 loc (). Definition 1.2. Assume that u L 1 loc () and let α Nn be a multi-index. Then v L 1 loc () is the αth weak artial derivative if u, written Dα u = v, if u D α ϕ dx = ( 1) α vϕ dx for every test function ϕ C (). We denote D u = D (,...,) = u. If α = 1, then is the weak gradient of u. Here (the jth comonent is 1). Du = (D 1 u, D 2 u..., D n u) D j u = u x j = D (,...,1,...,) u, j = 1,..., n,

6 CHAPTER 1. SOBOLEV SPACES 3 T H E M O R A L : Classical derivatives are defined as ointwise limits of difference quotients, but the weak derivatives are defined as a functions satisfying the integration by arts formula. Observe, that changing the function on a set of measure zero does not affect its weak derivatives. W A R N I N G We use the same notation for the weak and classical derivatives. It should be clear from the context which interretation is used. Remarks 1.3: (1) If u C k (), then the classical weak derivatives u to order k are also the corresonding weak derivatives of u. In this sense, weak derivatives generalize classical derivatives. (2) If u = almost everywhere in an oen set, then D α u = almost everywhere in the same set. Lemma 1.4. A weak αth artial derivative of u, if it exists, is uniquely defined u to a set of measure zero. Proof. Assume that v, ṽ L 1 () are both weak αth artial derivatives of u, that loc is, ud α ϕ dx = ( 1) α vϕ dx = ( 1) α ṽϕ dx for every ϕ C (). This imlies that Claim: v = ṽ almost everywhere in. (v ṽ)ϕ dx = for every ϕ C (). (1.1) Reason. Let (i.e. is oen and is a comact subset of ). The sace C ( ) is dense in L ( ) (we shall return to this later). There exists a sequence of functions ϕ i C ( ) such that ϕ i 2 in and ϕ i sgn(v ṽ) almost everywhere in as i. Here sgn is the signum function. Identity (1.1) and the dominated convergence theorem, with the majorant (v ṽ)ϕ i 2( v + ṽ ) L 1 ( ), give = lim i = (v ṽ)ϕ i dx = (v ṽ)sgn(v ṽ) dx = lim (v ṽ)ϕ i dx i v ṽ dx This imlies that v = ṽ almost everywhere in for every. Thus v = ṽ almost everywhere in. From the roof we obtain a very useful corollary.

7 CHAPTER 1. SOBOLEV SPACES 4 Corollary 1.5 (Fundamental lemma of the calculus of variations). If f L 1 loc () satisfies f ϕ dx = for every ϕ C (), then f = almost everywhere in. T H E everywhere. M O R A L : This is an integral way to say that a function is zero almost Examle 1.6. Let n = 1 and = (,2). Consider x, < x 1, u(x) = 1, 1 x < 2, and 1, < x 1, v(x) =, 1 x < 2. We claim that u = v in the weak sense. To see this, we show that for every ϕ C ((,2)). 2 uϕ dx = 2 vϕ dx Reason. An integration by arts and the fundamental theorem of calculus give 2 u(x)ϕ (x) dx = 1 xϕ (x) dx + 1 = xϕ(x) }{{} = =ϕ(1) ϕ(x) dx = 1 ϕ (x) dx ϕ(x) dx + ϕ(2) ϕ(1) }{{} = 2 vϕ(x) dx for every ϕ C ((,2)). 1.2 Sobolev saces Definition 1.7. Assume that is an oen subset of R n. The Sobolev sace W k, () consists of functions u L () such that for every multi-index α with α k, the weak derivative D α u exists and D α u L (). Thus W k, () = {u L () : D α u L (), α k}.

8 CHAPTER 1. SOBOLEV SPACES 5 If u W k, (), we define its norm and u W k, () = ( α k u W k, () = D α u dx α k esssu D α u., 1 <, Notice that D u = D (,...,) u = u. Assume that is an oen subsets of. We say that is comactly contained in, denoted, if is a comact subset of. A function u W k, loc (), if u W k, ( ) for every. T H E M O R A L : Thus Sobolev sace W k, () consists of functions in L () that have weak artial derivatives u to order k and they belong to L (). Remarks 1.8: (1) As in L saces we identify W k, functions which are equal almost everywhere. (2) There are several ways to define a norm on W k, (). The norm W k, () is equivalent, for examle, with the norm D α u L (), 1. α k and W k, () is also equivalent with max α k Dα u L (). (3) For k = 1 we use the norm ( ) u W 1, () = u 1 L () + Du L () and ( = u dx + u W 1, () = esssu We may also consider equivalent norms and when 1 < and Du dx u + esssu Du. ( n u W 1, () = u L () + D j u L () u W 1, () = u L () + j=1, 1 <, n D j u L (), j=1 u W 1, () = u L () + Du L () u W 1, () = max{ u L (), D 1 u L (),..., D n u L ()}.,

9 CHAPTER 1. SOBOLEV SPACES 6 Examle 1.9. Let u : B(,1) [, ], u(x) = x α, α >. Clearly u C (B(,1) \ {}), but u is unbounded in any neighbourhood of the origin. We start by showing that u has a weak derivative in the entire unit ball. When x, we have Thus Gauss theorem gives u (x) = α x α 1 x j x j x = α, j = 1,..., n. x α+2 x j Du(x) = α x x α+2. D j (uϕ) dx = B(,1)\B(,ε) uϕν j ds, (B(,1)\B(,ε)) where ν = (ν 1,...,ν n ) is the outward ointing unit ( ν = 1) normal of the boundary and ϕ C (B(,1)). As ϕ = on B(,1), this can be written as D j uϕ dx + B(,1)\B(,ε) By rearranging terms, we obtain ud j ϕ dx = B(,1)\B(,ε) ud j ϕ dx = B(,1)\B(,ε) D j uϕ dx + B(,1)\B(,ε) uϕν j ds. B(,ε) uϕν j ds. (1.2) B(,ε) Let us estimate the last term on the right-hand side. Since ν(x) = x x, we have ν j (x) = x j x, when x B(,ε). Thus uϕν j ds B(,ε) ϕ L (B(,1)) ε α ds (B(,ε)) = ϕ L (B(,1))ω n 1 ε n 1 α as ε, if n 1 α >. Here ω n 1 = H n 1 (B(,1)) is the (n 1)-dimensional measure of the shere B(,1). Next we study integrability of D j u. We need this information in order to be able to use the dominated convergence theorem. A straightforward comutation gives B(,1) if n 1 α >. D j u dx Du dx = α B(,1) x α 1 dx B(,1) 1 1 = α x α 1 ds dr = αω n 1 B(,r) 1 = αω n 1 r n α 2 dr = αω n 1 1 n α 1 rn α 1 r α 1+n 1 dr <,

10 CHAPTER 1. SOBOLEV SPACES 7 The following argument shows that D j u is a weak derivative of u also in a neighbourhood of the origin. By the dominated convergence theorem ) lim (ud j ϕχ B(,1)\B(,ε) dx ud j ϕ dx = B(,1) B(,1) ε = lim ε = lim ε = = ud j ϕ dx B(,1)\B(,ε) D j uϕ dx + lim B(,1)\B(,ε) ε lim B(,1) ε D juϕχ B(,1)\B(,ε) dx D j uϕ dx. B(,1) uϕν j ds B(,ε) Here we used the dominated convergence theorem twice: First to the function ud j ϕχ B(,1)\B(,ε), which is dominated by u Dϕ L 1 (B(,1)), and then to the function D j uϕχ B(,1)\B(,ε), which is dominated by Du ϕ L 1 (B(,1)). We also used (1.2) and the fact that the last term there converges to zero as ε. Now we have roved that u has a weak derivative in the unit ball. We note that u L (B(,1)) if and only if α + n >, or equivalently, α < n. On the other hand, Du L (B(,1), if (α + 1) + n >, or equivalently, α < n. Thus u W 1, (B(,1)) if and only if α < n. Let (r i ) be a countable and dense subset of B(,1) and define u : B(,1) [, ], Then u W 1, (B(,1)) if α < n. Reason. 1 u(x) = 2 i x r i α. i=1 x ri α W 1, (B(,1)) 1 x ri i=1 2 i α W 1, (B(,1)) 1 = x α 2 i W 1, (B(,1)) i=1 = x α W 1, (B(,1)) <. Note that if α >, then u is unbounded in every oen subset of B(,1) and not differentiable in the classical sense in a dense subset. T H E oen subset. M O R A L : Functions in W 1,, 1 < n, n 2, may be unbounded in every

11 CHAPTER 1. SOBOLEV SPACES 8 Examle 1.1. Observe, that u(x) = x α, α >, does not belong to W 1,n (B(,1). However, there are unbounded functions in W 1,n, n 2. Let u : B(,1) R, ( ( )) log log x, x, u(x) =, x =. Then u W 1,n (B(,1)) when n 2, but u L (B(,1)). This can be used to construct a function in W 1,n (B(,1) that is unbounded in every oen subset of B(, 1) (exercise). T H E M O R A L : Functions in W 1,, 1 n, n 2, are not continuous. Later we shall see, that every W 1, function with > n coincides with a continuous function almost everywhere. Examle The function u : B(,1) R, 1, x n >, u(x) = u(x 1,..., x n ) =, x n <, does not belong to W 1, (B(,1) for any 1 (exercise). 1.3 Proerties of weak derivatives The following general roerties of weak derivatives follow rather directly from the definition. Lemma Assume that u, v W k, () and α k. Then (1) D α u W k α, (), (2) D β (D α u) = D α (D β u) for all multi-indices α,β with α + β k, (3) for every λ,µ R, λu + µv W k, () and (4) if is oen, then u W k, ( ), D α (λu + µv) = λd α u + µd α v, (5) (Leibniz s formula) if η C (), then ηu W k, () and D α (ηu) = ( ) α D β η D α β u, β α β where ( ) α α! = β β!(α β)!, α! = α 1!...α n! and β α means that β j α j for every j = 1,..., n.

12 CHAPTER 1. SOBOLEV SPACES 9 T H E M O R A L : Weak derivatives have the same roerties as classical derivatives of smooth functions. Proof. (1) Follows directly from the definition of weak derivatives. See also (2). (2) Let ϕ C (). Then Dβ ϕ C (). Therefore ( 1) β D β (D α u)ϕ dx = D α ud β ϕ dx = ( 1) α for all test functions ϕ C (). Notice that ud α+β ϕ dx = ( 1) α ( 1) α+β D α+β uϕ dx α + α + β =α α n + (α 1 + β 1 ) (α n + β n ) =2(α α n ) + β β n =2 α + β. As 2 α is an even number, the estimate above, together with the uniqueness results Lemma 1.4 and Corollary 1.5, imlies that D β (D α u) = D α+β u. (3) and (4) Clear. (5) First we consider the case α = 1. Let ϕ C (). By Leibniz s rule for differentiable functions and the definition of weak derivative ηud α ϕ dx = = (ud α (ηϕ) u(d α η)ϕ) dx (ηd α u + ud α η)ϕ dx for all ϕ C (). The case α > 1 follows by induction (exercise). 1.4 Comleteness of Sobolev saces One of the most useful roerties of Sobolev saces is that they are comlete. Thus Sobolev saces are closed under limits of Cauchy sequences. A sequence (u i ) of functions u i W k, (), i = 1,2,..., converges in W k, () to a function u W k, (), if for every ε > there exists i ε such that Equivalently, u i u W k, () < ε when i i ε. lim u i u W i k, () =. A sequence (u i ) is a Cauchy sequence in W k, (), if for every ε > there exists i ε such that u i u j W k, () < ε when i, j i ε.

13 CHAPTER 1. SOBOLEV SPACES 1 W A R N I N G : This is not the same condition as u i+1 u i W k, () < ε when i i ε. Indeed, the Cauchy sequence condition imlies this, but the converse is not true (exercise). Theorem 1.13 (Comleteness). The Sobolev sace W k, (), 1, k = 1,2,..., is a Banach sace. T H E M O R A L : The saces C k (), k = 1,2,..., are not comlete with resect to the Sobolev norm, but Sobolev saces are. This is imortant in existence arguments for PDEs. Proof. Ste 1: W k, () is a norm. Reason. (1) u W k, () = u = almost everywhere in. = u W k, () = imlies u L () =, which imlies that u = almost everywhere in. = u = almost everywhere in imlies D α uϕ dx = ( 1) α ud α ϕ dx = for all ϕ C (). This together with Corollary 1.5 imlies that Dα u = almost everywhere in for all α, α k. (2) λu W k, () = λ u W k, (), λ R. Clear. (3) The triangle inequality for 1 < follows from the elementary inequality (a + b) α a α + b α, a, b, < α 1, and Minkowski s inequality, since u + v W k, () = ( α k ( α k ( D α u + D α v L () ( D α u L () + D α ) v L () D α u L () α k + ( D α v L () α k = u W k, () + v W k, (). Ste 2: Let (u i ) be a Cauchy sequence in W k, (). As D α u i D α u j L () u i u j W k, (), α k, it follows that (D α u i ) is a Cauchy sequence in L (), α k. The comleteness of L () imlies that there exists u α L () such that D α u i u α in L (). In articular, u i u (,...,) = u in L ().

14 CHAPTER 1. SOBOLEV SPACES 11 Ste 3: We show that D α u = u α, α k. We would like to argue ud α ϕ dx = lim i = lim ( 1) α i = ( 1) α u i D α ϕ dx D α u i ϕ dx D α uϕ dx for every ϕ C (). On the second line we used the definition of the weak derivative. Next we show how to conclude the fist and last inequalities above. 1 < < Let ϕ C (). By Hölder s inequality we have u i D α ϕ dx ud α ϕ dx = (u i u)d α ϕ dx u i u L () D α ϕ L () and consequently we obtain the first inequality above. The last inequality follows in the same way, since D α u i ϕ dx = 1, = D α uϕ dx Dα u i u α L () ϕ L (). A similar argument as above (exercise). This means that the weak derivatives D α u exist and D α u = u α, α k. As we also know that D α u i u α = D α u, α k, we conclude that u i u W k, (). Thus u i u in W k, (). Remark W k, (), 1 < is searable. In the case k = 1 consider the maing u (u, Du) from W 1, () to L () L ()) n and recall that a subset of a searable sace is searable (exercise). However, W 1, () is not searable. 1.5 Hilbert sace structure The sace W k,2 () is a Hilbert sace with the inner roduct u, v W k,2 () = D α u, D α v L 2 (), α k where Observe that D α u, D α v L 2 () = D α ud α v dx. u W k,2 () = u, u 1 2 W k,2 ().

15 CHAPTER 1. SOBOLEV SPACES Aroximation by smooth functions This section deals with the question whether every function in a Sobolev sace can be aroximated by a smooth function. Define φ C (Rn ) by 1 c e x 2 1, x < 1, φ(x) =, x 1, where c > is chosen so that For ε >, set R n φ(x) dx = 1. φ ε (x) = 1 ε n φ ( x ε ). The function φ is called the standard mollifier. Observe that φ ε, suφ ε = B(,ε) and R n φ ε (x) dx = 1 ε n R n φ( x ε ) dx = 1 ε n R n φ(y)ε n d y = R n φ(x) dx = 1 for all ε >. Here we used the change of variable y = x ε, dx = εn d y. Notation. If R n is oen with, we write ε = {x : dist(x,) > ε}, ε >. If f L 1 loc (), we obtain its standard convolution mollification f ε : ε [, ], f ε (x) = (f φ ε )(x) = f (y)φ ε (x y) d y. T H E M O R A L : Since the convolution is a weighted integral average of f over the ball B(x,ε) for every x, instead of it is well defined only in ε. If = R n, we do not have this roblem. Remarks 1.15: (1) For every x ε, f ε (x) = f (y)φ ε (x y) d y = (2) By a change of variables z = x y we have f (y)φ ε (x y) d y = f (y)φ ε (x y) d y. B(x,ε) f (x z)φ ε (z) dz

16 CHAPTER 1. SOBOLEV SPACES 13 (3) For every x ε, f ε (x) B(x,ε) (4) If f C (), then f ε C ( ε ), whenever f (y)φ ε (x y) d y φ ε f (y) d y <. B(x,ε) < ε < ε = 1 dist(su f,). 2 Reason. If x ε s.t. dist(x,su f ) > ε (in articular, for every x ε \ ε ) then B(x,ε) su f =, which imlies that f ε (x) =. Lemma 1.16 (Proerties of mollifiers). (1) f ε C ( ε ). (2) f ε f almost everywhere as ε. (3) If f C(), then f ε f uniformly in every. (4) If f L loc (), 1 <, then f ε f in L ( ) for every. W A R N I N G : (4) does not hold for =, since there are functions in L () that are not continuous. Proof. (1) Let x ε, j = 1,..., n, e j = (,...,1,...,) (the jth comonent is 1). Choose h > such that B(x, h ) ε and let h R, h < h. Then f ε (x + he j ) f ε (x) = 1 [ ( ) 1 x + he j y ( x y ) ] h ε n φ φ f (y) d y h ε ε B(x+he j,ε) B(x,ε) Let us set = B(x, h + ε). Now and B(x + he j,ε) B(x,ε). Claim: [ ( ) 1 x + he j y ( x y φ φ h ε ε ) ] 1 ε φ ( x y ) x j ε for all y as h. Reason. Let ψ(x) = φ ( x y ) ε. Then ψ (x) = 1 φ ( x y ), j = 1,..., n x j ε x j ε and ψ(x + he j ) ψ(x) = h t (ψ(x + te j)) dt = h Dψ(x + te j ) e j dt. Thus ψ(x + he j ) ψ(x) 1 ε h h Dψ(x + te j ) e j dt ( ) x + te Dφ j y dt ε h ε Dφ L (R n ).

17 CHAPTER 1. SOBOLEV SPACES 14 This estimate shows that we can use the Lebesgue dominated convergence theorem (on the third row) to obtain f ε f ε (x + he j ) f ε (x) (x) = lim x j h h 1 = lim h ε n = 1 1 ε n ε = = [ ( 1 x + he j y φ h ε φ ( x y ) f (y) d y x j ε φ ε (x y) f (y) d y x j ( ) φε f (x). x j A similar argument shows that D α f ε exists and ) φ ( x y ) ] f (y) d y ε D α f ε = D α φ ε f in ε for every multi-index α. (2) Recall that B(x,ε) φ ε(x y) d y = 1. Therefore we have f ε (x) f (x) = φ ε (x y)f (y) d y f (x) φ ε (x y) d y B(x,ε) B(x,ε) = φ ε (x y)(f (y) f (x) d y B(x,ε) 1 ( x y ) ε n φ f (y) f (x) d y B(x,ε) ε 1 n φ L (R n ) f (y) f (x) d y B(x, ε) B(x,ε) for almost everywhere x. Here n = B(,1) and the last convergence follows from the Lebesgue s differentiation theorem. (3) Let, < ε < dist(, ), and x. Because is comact and f C(), f is uniformly continuous in, that is, for every ε > there exists δ > such that f (x) f (y) < ε for all x, y with x y < δ. By combining this with an estimate from the roof of (ii), we conclude that 1 f ε (x) f (x) n φ L (R n ) f (y) f (x) d y < n φ L B(x, ε) (R n ) ε B(x,ε) for all x if ε < δ. Claim: (4) Let. f ε dx f dx whenever < ε < dist(, ) and < ε < dist(,).

18 CHAPTER 1. SOBOLEV SPACES 15 Reason. Take x. By Hölder s inequality imlies f ε (x) = φ ε (x y)f (y) d y B(x,ε) ( φ ε (x y 1 φ ε (x y f (y) d y B(x,ε) φ ε (x y) d y B(x,ε) (B(x,ε) φ ε (x y) f (y) d y By raising the revious estimate to ower and by integrating over, we obtain f ε (x) dx φ ε (x y) f (y) d y dx B(x,ε) = φ ε (x y) f (y) dx d y = f (y) φ ε (x y) dx d y = f (y) d y. Here we used Fubini s theorem and once more the fact that the integral of φ ε is one. Since C( ) is dense in L ( ). Therefore for every ε > there exists g C( ) such that ( f g dx ε 3. By (2), we have g ε g uniformly in as ε. Thus ( g ε g dx su g ε g 1 < ε when ε > is small enough. Now we use the Minkowski s inequality and the revious claim to conclude that ( f ε f dx ( f ε g ε dx ( ( + g ε g dx + ( ( 2 g f dx + 2 ε 3 + ε 3 = ε. 3, g ε g dx g f dx Thus f ε f in L ( ) as ε.

19 CHAPTER 1. SOBOLEV SPACES Local aroximation in Sobolev saces Next we show that the convolution aroximation converges locally in Sobolev saces. Theorem Let u W k, (), 1 <. then (1) D α u ε = D α u φ ε in ε and (2) u ε u in W k, ( ) for every. T H E M O R A L : Smooth functions are dense in local Sobolev saces. Thus every Sobolev function can be locally aroximated with a smooth function in the Sobolev norm. Proof. (1) Fix x ε. Then D α u ε (x) = D α (u φ ε )(x) = (u D α φ ε )(x) = D α x φ ε(x y)u(y) d y = ( 1) α D α y (φ ε(x y))u(y) d y. Here we first used the roof of Lemma 1.16 (1) and then the fact that ( φ x j ( x y ε )) = x j ( φ ( y x ε )) = y j ( φ ( x y For every x ε, the function ϕ(y) = φ ε (x y) belongs to C (). Therefore D α y (φ ε(x y))u(y) d y = ( 1) α D α u(y)φ ε (x y) d y. By combining the above facts, we see that D α u ε (x) = ( 1) α + α D α u(y)φ ε (x y) d y = (D α u φ ε )(x). Notice that ( 1) α + α = 1. (2) Let, and choose ε > s.t. ε. By (i) we know that D α u ε = D α u φ ε in, α k. By Lemma 1.16, we have D α u ε D α u in L ( ) as ε, α k. Consequently ε )). u ε u W k, ( ) = ( α k D α u ε D α u L ( ).

20 CHAPTER 1. SOBOLEV SPACES Global aroximation in Sobolev saces The next result shows that the convolution aroximation converges also globally in Sobolev saces. Theorem 1.18 (Meyers-Serrin). If u W k, (), 1 <, then there exist functions u i C () W k, () such that u i u in W k, (). T H E M O R A L : Smooth functions are dense in Sobolev saces. Thus every Sobolev function can be aroximated with a smooth function in the Sobolev norm. In articular, this holds true for the function with a dense infinity set in Examle 1.9. Proof. Let = and { i = x : dist(x,) > 1 } B(, i), i = 1,2,... i Then = i and i=1 Claim: There exist η i C ( i+2 \ i 1 ), i = 1,2,..., such that η i 1 and η i (x) = 1 for every x. i=1 This is a artition of unity subordinate to the covering { i }. Reason. By using the distance function and convolution aroximation we can construct η i C ( i+2\ i 1 ) such that η i 1 and η i = 1 in i+1 \ i (exercise). Then we define η i (x) = η i (x) j=1 η j(x), i = 1,2,... Observe that the sum is only over four indices in a neighbourhood of a given oint. Now by Lemma 1.12 (5), η i u W k, () and su(η i u) i+2 \ i 1. Let ε >. Choose ε i > so small that su(φ εi (η i u)) i+2 \ i 1 (see Remark 1.15 (4)) and φ εi (η i u) η i u W k, () < ε 2 i, i = 1,2,...

21 CHAPTER 1. SOBOLEV SPACES 18 By Theorem 1.17 (2), this is ossible. Define v = φ εi (η i u). i=1 This function belongs to C (), since in a neighbourhood of any oint x, there are at most finitely many nonzero terms in the sum. Moreover, v u W k, () = φ εi (η i u) η i u i=1 i=1 W k, () φεi (η i u) η i u W k, () i=1 ε i=1 2 i = ε. Remarks 1.19: (1) The Meyers-Serrin theorem 1.18 gives the following characterization for the Sobolev saces W 1, (), 1 < : u W 1, () if and only if there exist functions u i C () W k, (), i = 1,2,..., such that u i u in W k, () as i. In other words, W 1, () is the comletion of C () in the Sobolev norm. Reason. = Theorem = Theorem (2) The Meyers-Serrin theorem 1.18 is false for =. Indeed, if u i C () W 1, () such that u i u in W 1, (), then u C 1 () (exercise). Thus secial care is required when we consider aroximations in W 1, (). (3) Let. The roof of Theorem 1.17 and Theorem 1.18 shows that for every ε > there exists v C () such that v u W 1, ( ) < ε. (4) The roof of Theorem 1.18 shows that not only C () but also C () is dense in L (), 1 <. 1.9 Sobolev saces with zero boundary values In this section we study definitions and roerties of first order Sobolev saces with zero boundary values in an oen subset of R n. A similar theory can be develoed for higher order Sobolev saces as well. Recall that, by Theorem 1.18, the Sobolev sace W 1, () can be characterized as the comletion of C () with resect to the Sobolev norm when 1 <.

22 CHAPTER 1. SOBOLEV SPACES 19 Definition 1.2. Let 1 <. The Sobolev sace with zero boundary values W 1, () is the comletion of C () with resect to the Sobolev norm. Thus u W 1, () if and only if there exist functions u i C (), i = 1,2,..., such that u i u in W 1, () as i. The sace W 1, () is endowed with the norm of W 1, (). T H E M O R A L : The only difference comared to W 1, () is that functions in W 1, () can be aroximated by C () functions instead of C () functions, that is, W 1, () = C () and W 1, () = C (), where the comletions are taken with resect to the Sobolev norm. A function in W 1, () has zero boundary values in Sobolev s sense. We may say that u, v W 1, () have the same boundary values in Sobolev s sense, if u v W 1, (). This is useful, for examle, in Dirichlet roblems for PDEs. W A R N I N G : Roughly seaking a function in W 1, () belongs to W 1, (), if it vanishes on the boundary. This is a delicate issue, since the function does not have to be zero ointwise on the boundary. We shall return to this question later. Remark W 1, () is a closed subsace of W 1, () and thus comlete (exercise). Remarks 1.22: (1) Clearly C () W1, () W 1, () L (). (2) If u W 1, (), then the zero extension ũ : R n [, ], belongs to W 1, (R n ) (exercise). u(x), x, ũ(x) =, x R n \, Lemma If u W 1, () and su u is a comact subset of, then u W 1, (). Proof. Let η C () be a cutoff function such that η = 1 on the suort of u. Claim: If u i C (), i = 1,2,..., such that u i u in W 1, (), then ηu i C () converges to ηu = u in W 1, (). Reason. We observe that ( ) ηu i ηu W 1, () = ηu i ηu L () + D(ηu i ηu) 1 L () ηu i ηu L () + D(ηu i ηu) L (),

23 CHAPTER 1. SOBOLEV SPACES 2 where ( ηu i ηu L () = ( = ηu i ηu dx η u i u dx ( η L () u i u dx and by Lemma 1.12 (5) as i. ( D(ηu i ηu) L () = ( = ( D(ηu i ηu) dx (u i u)dη + (Du i Du)η dx ( + (u i u)dη dx ( Dη L () + η L () u i u dx ( (Du i Du)η dx Du i Du dx Since ηu i C (), i = 1,2,..., and ηu i u in W 1, (), we conclude that u W 1, (). Since W 1, () W 1, (), functions in these saces have similar general roerties and they will not be reeated here. Thus we shall focus on roerties that are tyical for Sobolev saces with zero boundary values. Lemma W 1, (R n ) = W 1, (R n ), 1 <. T H E M O R A L : The standard Sobolev sace and the Sobolev sace with zero boundary value coincide in the whole sace. W A R N I N G : W 1, (B(,1)) W 1, (B(,1)), 1 <. Thus the saces are not same in general. Proof. Assume that u W 1, (R n ). Let η k C (B(, k + 1)) such that η = 1 on B(, k), η k 1 and Dη k c. Lemma 1.23 imlies uη k W 1, (R n ). Claim: uη k u in W 1, (R n ) as k.

24 CHAPTER 1. SOBOLEV SPACES 21 Reason. u uη k W 1, (R n ) u uη k L (R n ) + D(u uη k ) L (R n ) ( ( = u(1 η k ) dx + D(u(1 η k )) dx R n R n ( ( = u(1 η k ) dx + (1 η k )Du udη k ) dx R n R n ( ( u(1 η k ) dx + (1 η k )Du dx R n R n ( + udη k dx. R n We note that lim k u(1 η k ) = almost everywhere and u(1 η k ) u L 1 (R n ) will do as an integrable majorant. The dominated convergence theorem gives ( u(1 η k ) dx. R n A similar argument shows that ( R n (1 η k )Du dx as k. Moreover, by the dominated convergence theorem ( R n udη k dx c ( u dx B(,k+1)\B(,k) ( = c u χ B(,k+1)\B(,k) dx R n as k. Here u χ B(,k+1)\B(,k) u L 1 (R n ) will do as an integrable majorant. Since uη k W 1, (R n ), i = 1,2,..., uη k u in W 1, (R n ) as k and W 1, () is comlete, we conclude that u W 1, (). 1.1 Chain rule We shall rove some useful results for the first order Sobolev saces W 1, (), 1 <. Lemma 1.25 (Chain rule). If u W 1, () and f C 1 (R) such that f L (R) and f () =, then f u W 1, () and D j (f u) = f (u)d j u, j = 1,2,..., n almost everywhere in.

25 CHAPTER 1. SOBOLEV SPACES 22 Proof. By Theorem 1.18, there exist a sequence of functions u i C () W 1, (), i = 1,2,..., such that u i u in W 1, () as i. Let ϕ C (). Claim: Reason. (f u)d j ϕ dx = lim i f (u i )D j ϕ dx. 1 < < By Hölder s inequality f (u)d j ϕ dx f (u i )D j ϕ dx f (u) f (u i ) Dϕ dx ( f (u) f (u i ) dx f ( u u i dx ( Dϕ dx ( Dϕ dx. On the last row, we used the fact that u f (u) f (u i ) = f (t) dt u ui f u u i. Finally, the convergence to zero follows, because the first and the last term are bounded and u i u in L (). = 1, = A similar argument as above (exercise). Next, we use the claim above, integration by arts for smooth functions and the chain rule for smooth functions to obtain (f u)d j ϕ dx = lim i f (u i )D j ϕ dx = lim D j (f (u i ))ϕ dx i = lim i f (u i )D j u i ϕ dx = f (u)d j uϕ dx = (f u)d j uϕ dx, j = 1,..., n, for every ϕ C (). We leave it as an exercise to show the fourth inequality in the dislay above. Finally, we need to show that f (u) and f (u) u x are in L (). Since j u f (u) = f (u) f () = f (t) dt f u, we have ( / ( f (u) dx f u dx <,

26 CHAPTER 1. SOBOLEV SPACES 23 and similarly, ( f (u)d j u ( dx f Du dx < Truncation The truncation roerty is an imortant roerty of first order Sobolev saces, which means that we can cut the functions at certain level and the truncated function is still in the same Sobolev sace. Higher order Sobolev saces do not enjoy this roerty, see Examle 1.6. Theorem If u W 1, (), then u + = max{u,} W 1, (), u = min{u,} W 1, (), u W 1, () and Du + Du almost everywhere in {x : u(x) > }, = almost everywhere in {x : u(x) }, Du almost everywhere in {x : u(x) }, = Du almost everywhere in {x : u(x) < }, and Du almost everywhere in {x : u(x) > }, Du = almost everywhere in {x : u(x) = }, Du almost everywhere in {x : u(x) < }. T H E M O R A L : In contrast with C 1, the Sobolev sace W 1, are closed under taking absolute values. Proof. Let ε > and let f ε : R R, f ε (t) = t 2 + ε 2 ε. The function f ε has the following roerties: f ε C 1 (R), f ε () = lim f ε(t) = t for every t R, ε (f ε ) (t) = 1 2 (t2 + ε 2 ) 1/2 2t = t t 2 + ε 2 for every t R, and (f ε ) 1 for every ε >. From Lemma 1.25, we conclude that f ε u W 1, () and (f ε u)d j ϕ dx = (f ε ) (u)d j uϕ dx, j = 1,..., n,

27 CHAPTER 1. SOBOLEV SPACES 24 for every ϕ C (). We note that 1, t >, lim (f ε) (t) =, t =, ε 1, t <, and consequently u D j ϕ dx = lim ε = lim ε = (f ε u)d j ϕ dx (f ε ) (u)d j uϕ dx D j u ϕ dx j = 1,..., n, for every ϕ C (), where D j u is as in the statement of the theorem. We leave it as an exercise to rove that the first equality in the dislay above holds. The other claims follow from formulas u + = 1 2 (u + u ) and u = 1 2 ( u u). Remarks 1.27: (1) If u, v W 1, (), then max{u, v} W 1, () and min{u, v} W 1, (). Moreover, Du almost everywhere in {x : u(x) v(x)}, D max{u, v} = Dv almost everywhere in {x : u(x) v(x)}, and Du almost everywhere in {x : u(x) v(x)}, D min{u, v} = Dv almost everywhere in {x : u(x) v(x)}. If u, v W 1, (), then max{u, v} W 1, () and min{u, v} W 1, () (exercise). Reason. max{u, v} = 1 2 (u + v + u v ) and min{u, v} = 1 2 (u + v u v ). (2) If u W 1, () and λ R, then Du = almost everywhere in {x : u(x) = λ} (exercise). (3) If u W 1, () and λ R, then min{u,λ} W 1, () and loc Du almost everywhere in {x : u(x) < λ}, D min{u,λ} = almost everywhere in {x : u(x) λ}.

28 CHAPTER 1. SOBOLEV SPACES 25 A similar claim also holds for max{u, λ}. This imlies that a function u W 1, () can be aroximated by the truncated functions u λ = max{ λ,min{u,λ}} λ almost everywhere in {x : u(x) λ}, = u almost everywhere in {x : λ < u(x) < λ}, λ almost everywhere in {x : u(x) λ}, in W 1, (). (Here λ >.) Reason. By alying the dominated convergence theorem to we have u u λ 2 ( u + u λ ) 2 +1 u L 1 (), lim λ u u λ dx = lim u u λ dx =, λ and by alying the dominated convergence theorem to Du Du λ Du L 1 (), we have lim λ Du Du λ dx = lim Du Du λ dx =. λ T H E M O R A L : Bounded W 1, functions are dense in W 1, Sequential weak comactness of Sobolev saces In this section we assume that 1 < < and that R n. We recall the definition of weak convergence in L. Definition A sequence (f i ) of functions in L () converges weakly in L () to a function f L (), if where = lim i f i g dx = 1 is the conjugate exonent of. f g dx for every g L (), If f i f weakly in L (), then (f i ) is bounded in L (), that is, f i L (), i = 1,2,..., and su f i L () <. i

29 CHAPTER 1. SOBOLEV SPACES 26 Moreover, f L () liminf i f i L (). (1.3) T H E M O R A L : The L -norm is lower semicontinuous with resect to the weak convergence. Thus a weakly converging sequence is bounded. The next result shows that the converse is true u to a subsequence. Theorem Let 1 < <. Assume that the sequence (f i ) is bounded in L (). Then there exists a subsequence (f ik ) and f L () such that f ik f weakly in L () as k. T H E M O R A L : This shows that L with 1 < < is weakly sequentially comact, that is, every bounded sequence in L has a weakly converging subsequence. One of the most useful alications of weak convergence is in comactness arguments. A bounded sequence in L does not need to have any convergent subsequence with convergence interreted in the standard L sense. However, there exists a weakly converging subsequence. Remark 1.3. Theorem 1.29 is equivalent to the fact that L saces are reflexive for 1 < <. Weak convergence is a very weak mode of convergence and sometimes we need a tool to ugrade it to strong convergence. Theorem 1.31 (Mazur s lemma). ] Assume that X is a normed sace and that a sequence (x i ) converges weakly to x as i in X. Then for every ε >, there exists k N and a convex combination k i=1 a ix i such that k x a i x i < ε. i=1 Recall, that in a convex combination k i=1 a ix i we have a i and k i=1 a i = 1. Observe that some of the coefficients a i may be zero so that the convex combination is essentially for a subsequence. T H E M O R A L : For every weakly converging sequence, there is a subsequence of convex combinations that converges strongly. Thus weak convergence is ugraded to strong convergence for a subsequence of convex combinations. Remark Mazur s lemma can be used to give a roof for (1.3) (exercise). Theorem Let 1 < <. W 1, (). Assume that (u i ) is a bounded sequence in Then there exists a subsequence (u ik ) and u W 1, () such that u ik u weakly in L () and Du ik Du weakly in L () as k. Moreover, if u i W 1, (), i = 1,2..., then u W 1, ().

30 CHAPTER 1. SOBOLEV SPACES 27 Proof. (1) Assume that u W 1, () Since (u i ) is a bounded sequence in L (), by Theorem 1.29 there exists a subsequence (u ik ) and u L () such that u ik u weakly in L () as k. Since (D j u i ), j = 1,..., n, is a bounded sequence in L (), by assing to subsequences successively, we obtain a subsequence (D j u ik ) and g j L () such that D j u ik g j weakly in L () as k for every j = 1,..., n. Claim: g j = D j u, j = 1,..., n. Reason. For every ϕ C (), we have ud j ϕ dx = lim k = lim = k u ik D j ϕ dx D j u ik ϕ dx g j ϕ dx. Here the first equality follows from the definition of weak convergence, the second equality follows form the definition of weak derivative and the last equality follows from weak convergence. Thus g j = D j u, j = 1,..., n. This shows that the weak artial derivatives D j u, j = 1,..., n, exist and belong to L (). It follows that u W 1, (). (2) Then assume that u W 1, (). We use the notation above. By Mazur s lemma, see Theorem 1.31, there there exists a subsequence of (u ik ), denoted by (u i ), such that for convex combinations k a i u i u i=1 and k a i Du i Du in L () as k. i=1 This shows that Since k j=1 k a i u i u in W 1, () as k. i=1 a i u i W 1, () and k j=1 a i Du i W 1, (), comleteness of W 1, () imlies u W 1, () Remarks 1.34: (1) Theorem 1.33 is equivalent to the fact that W 1, saces are reflexive for 1 < <. (2) Another way to see that W 1, saces are reflexive for 1 < < is to recall that a closed subsace of a reflexive sace is reflexive. Thus it is enough to find an isomorhism between W 1, () and a closed subsace of L (,R n+1 ) = L (,R n ) L (,R n ). The maing u (u, Du) will do

31 CHAPTER 1. SOBOLEV SPACES 28 for this urose. This holds true for W 1, () as well. This aroach can be used to characterize elements in the dual sace by the Riesz reresentation theorem. Theorem Let 1 < <. Assume that (u i ) is a bounded sequence in W 1, () and u i u almost everywhere in. Then u W 1, (), u i u weakly in L () and Du i Du weakly in L (). Moreover, if u i W 1, (), i = 1,2..., then u W 1, (). T H E M O R A L : In order to show that u W 1, () it is enough to construct functions u i W 1, (), i = 1,2,..., such that u i u almost everywhere in and su i u i W 1, ()) <. Proof. We ass to subsequences several times in the argument and we denote all subsequences again by (u i ). By Theorem 1.33 there exists a subsequence (u i ) and a function ũ W 1, () such that u i ũ weakly in L () and Du i Dũ weakly in L () as i. By Mazur s lemma, see Theorem 1.31, there exists a subsequence (u i ) such that the convex combinations k a i u i ũ i=1 and k a i Du i Dũ in L () as k. i=1 Since convergence in L () imlies that there is a subsequence that converges almost everywhere in and u i u almost everywhere in imlies that k a i u i u almost everywhere in as k, j=1 we conclude that ũ = u and Dũ = Du almost everywhere in. This show that the weak limit is indeendent of the choice of the subsequences, which imlies that u i u weakly in L () and Du i Du weakly in L (). Remark Theorem 1.33 and Theorem 1.35 do not hold when = 1 (exercise) Difference quotients In this section we give a characterization of W 1,, 1 < <, in terms of difference quotients. This aroach is useful in regularilty theory for PDEs. Moreover, this characterization does not involve derivatives. Definition Let u L 1 loc () and. The j th difference quotient is D h j u(x) = u(x + he j) u(x), j = 1,..., n, h for x and h R such that < h < dist(,). We denote D h u = (D h 1 u,..., Dh n u).

32 CHAPTER 1. SOBOLEV SPACES 29 T H E M O R A L : Note that the definition of the difference quotient makes sense at every x whenever < h < dist(x,). If = R n, then the definition makes sense for every h. Theorem (1) Assume u W 1, (), 1 <. Then for every, we have D h u L ( ) c Du L () for some constant c = c(n, ) and all < h < dist(,). (2) If u L ( ), 1 < <, and there is a constant c such that D h u L ( ) c for all < h < dist(,), then u W 1, ( ) and Du L ( ) c. T H E M O R A L : Pointwise derivatives are defined as limit of difference quotients and Sobolev saces can be characterized by integrated difference quotients. W A R N I N G : Claim (2) does not hold for = 1 (exercise). Proof. (1) First assume that u C () W 1, (). Then u(x + he j ) u(x) = = = h h h t (u(x + te j)) dt Du(x + te j ) e j dt u x j (x + te j ) dt, j = 1,..., n, for all x, < h < dist(,). By Hölder s inequality D h j u(x) = u(x + he j ) u(x) h 1 h u h (x + te j ) x dt j ( 1 h ) u 1/ h (x + te j ) x dt h 1 1 j, which imlies D h j u(x) 1 h h u (x + te j ) x dt j

33 CHAPTER 1. SOBOLEV SPACES 3 Next we integrate over and switch the order of integration by Fubini s theorem to conclude D h j u(x) dx 1 h u h (x + te j ) x j = 1 h u h (x + te j ) x j u (x) x dx. j dt dx dx dt The last inequality follows from the fact that, for < h < dist(,), we have u (x + te j ) x dx u (x) j x dx. j Using the elementary inequality (a a n ) α n α (a α aα n ), a i, α >, we obtain ( ) n D h u(x) dx = D h 2 n j u(x) 2 dx n 2 D h j=1 j u(x) dx j=1 n n = c D h j=1 j u(x) dx c u (x) j=1 x dx j c Du(x) dx The general case u W 1, () follows by an aroximation, see Theorem 1.18 (exercise). (2) Let ϕ C ( ). Then by a change of variables we see that, for < h < dist(suϕ, ), we have u(x) ϕ(x + he j) ϕ(x) dx = h u(x) + u(x he j ) ϕ(x) dx, j = 1,..., n. h This shows that ud h j ϕ dx = (D h j u)ϕ dx, j = 1,..., n. (1.4) By assumtion su D h < h <dist( j u L ( ) <.,) From reflexivity of L ( ), 1 < <, it follows that there exists g j L ( ), j = 1,..., n and a subsequence h i such that D h i j u g j weakly in L ( ).

34 CHAPTER 1. SOBOLEV SPACES 31 Therefore u ϕ dx = x j u = lim hi = lim = hi ( lim h i Dh i ϕ j ) dx ud h i ϕ dx j g j ϕ dx (D h i u)ϕ dx j for every ϕ C ( ). Here the second equality follows from the dominated convergence theorem and the last equality is the weak convergence. Thus u x = g j j, j = 1,..., n, in the weak sense and u W 1, ( ). This is essentially the same argument as in the roof of Theorem Absolute continuity on lines In this section we relate weak derivatives to classical derivatives and give a characterization W 1, in terms of absolute continuity on lines. Recall that a function u : [a, b] R is absolutely continuous, if for every ε >, there exists δ > such that if a = x 1 < y 1 x 2 < y 2... x m < y m = b is a artition of [a, b] with then m (y i x i ) < δ, i=1 m u(y i ) u(x i ) < ε. i=1 Absolute continuity can be characterized in terms of the fundamental theorem of calculus. Theorem A function u : [a, b] R is absolutely continuous if and only if there exists a function g L 1 ((a, b)) such that u(x) = u(a) + x a g(t) dt. By the Lebesgue differentiation theorem g = u almost everywhere in (a, b). T H E M O R A L : Absolutely continuous functions are recisely those functions for which the fundamental theorem of calculus holds true. Examles 1.4: (1) Every Lichitz continuous function u : [a, b] R is absolutely continuous. (2) The Cantor function u is continuous in [, 1] and differentiable almost everywhere in (, 1), but not absolutely continuous in [, 1].

35 CHAPTER 1. SOBOLEV SPACES 32 Reason. u(1) = 1 = u() + 1 u (t) dt. }{{} = The next result relates weak artial derivatives with the classical artial derivatives. Theorem 1.41 (Nikodym, ACL characterization). Assume that u W 1, loc (), 1 and let. Then there exists u : [, ] such that u = u almost everywhere in and u is absolutely continuous on (n 1)-dimensional Lebesgue measure almost every line segments in that are arallel to the coordinate axes and the classical artial derivatives of u coincide with the weak artial derivatives of u almost everywhere in. Conversely, if u L () and loc there exists u as above such that D i u L (), i = 1,..., n, then u W1, loc loc (). T H E M O R A L : This is a very useful characterization of W 1,, since many claims for weak derivatives can be reduced to the one-dimensional claims for absolute continuous functions. In addition, this gives a ractical tool to show that a function belongs to a Sobolev sace. Remarks 1.42: (1) The ACL characterization can be used to give a simle roof of Examle 1.9 (exercise). (2) In the one-dimensional case we obtain the following characterization: u W 1, ((a, b)), 1, if u can be redefined on a set of measure zero in such a way that u L ((a, b)) and u is absolutely continuous on every comact subinterval of (a, b) and the classical derivative exists and belongs to u L ((a, b)). Moreover, the classical derivative equals to the weak derivative almost everywhere. (3) A function u W 1, () has a reresentative that has classical artial derivatives almost everywhere. However, this does not give any information concerning the total differentiability of the function. See Theorem (4) The ACL characterization can be used to give a simle roof of the Leibniz rule. If u W 1, () L () and v W 1, () L (), then uv W 1, () and D j (uv) = vd j u + ud j v, j = 1,..., n, almost everywhere in (exercise), comare to Lemma 1.12 (5). (5) The ACL characterization can be used to give a simle roof for Lemma 1.25 and Theorem The claim that if u, v W 1, (), then max{u, v} W 1, () and min{u, v} W 1, () follows also in a similar way (exercise).

36 CHAPTER 1. SOBOLEV SPACES 33 (6) The ACL characterization can be used to show that if is connected, u W 1, () and Du = almost everywhere in, then u is a constant loc almost everywhere in (exercise). Proof. Since the claims are local, we may assume that = R n and that u has a comact suort. = Let u i = u εi, i = 1,2,..., be a sequence of standard convolution aroximations of u such that su u i B(, R) for every i = 1,2,... and u i u W 1,1 (R n ) < 1 2 i, i = 1,2,... By Lemma 1.16 (2), the sequence of convolution aroximations converges ointwise almost everywhere and thus the limit lim i u i (x) exists for every x R n \ E for some E R n with E =. We define lim u (x) = u i(x), x R n \ E, i, x E. We fix a standard base vector in R n and, without loss of generality, we may assume that it is (,...,,1). Let and f i (x 1,..., x n 1 ) = R ( u i+1 u i + n u i+1 x j j=1 f (x 1,..., x n 1 ) = f i (x 1,..., x n 1 ). i=1 u ) i x (x 1,..., x n ) dx n j By the monotone convergence theorem and Fubini s theorem f dx 1... dx n 1 = f i dx 1... dx n 1 R n 1 R n 1 i=1 = f i dx 1... dx n 1 i=1 R n 1 ( = u i+1 u i + u i+1 u ) i i=1 R n x j x dx j 1 < 2 i <. i=1 This shows that f L 1 (R n 1 ) and thus f < (n 1)-almost everywhere in R n 1. Let x = (x 1,..., x n 1 ) R n 1 such that f ( x) <. Denote g i (t) = u i ( x, t) and g(t) = u ( x, t). Claim: (g i ) is a Cauchy sequence in C(R).

37 CHAPTER 1. SOBOLEV SPACES 34 Reason. Note that where Thus i 1 g i = g 1 + (g k+1 g k ), i = 1,2,..., g k+1 (t) g k (t) = k=1 t (g k+1 g k )(s) ds g k+1 (s) g k (s) ds R u k+1 ( x, s) u k ( x, s) x n x ds f k( x). n R i 1 i 1 (g k+1 (t) g k (t)) g k+1 (t) g k (t) k=1 k=1 f k ( x) = f ( x) < for every t R. This imlies that (g i ) is a Cauchy sequence in C(R). Since C(R) is comlete, there exists g C(R) such that g i g uniformly in R. It follows that { x} R R n \ E. Claim: (g i ) is a Cauchy sequence in L1 (R). k=1 Reason. R i 1 i 1 (g k+1 (t) g k (t)) dt g k+1 (t) g k (t) dt R k=1 k=1 i 1 k=1 f k ( x) f ( x) <. This imlies that (g i ) is a Cauchy sequence in L 1 (R). Since L 1 (R) is comlete, there exists g L 1 (R) such that g i g in L1 (R) as i. Claim: g is absolutely continuous in R. Reason. t t g(t) = lim g i (t) = lim g i i i (s) ds = g (s) ds This imlies that g is absolutely continuous in R and g = g almost everywhere in R. Claim: g is the weak derivative of g. Reason. Let ϕ C (R). Then gϕ dt = lim R i g i ϕ dt = lim R i g iϕ dt = gϕ dt. R R

38 CHAPTER 1. SOBOLEV SPACES 35 Thus for every ϕ C (Rn ) we have and by Fubini s theorem u ( x, x n ) ϕ ( x, x n ) dx n = R x n R u u ϕ Rn u dx = ϕ dx. R n x n x n x n ( x, x n )ϕ( x, x n ) dx n This shows that u has the classical artial derivatives almost everywhere in R n and that they coincide with the weak artial derivatives of u almost everywhere in R n. = Assume that u has a reresentative u as in the statement of the theorem. For every ϕ C (Rn ), the function u ϕ has the same absolute continuity roerties as u. By the fundamental theorem of calculus (u ϕ) ( x, t) dt = x n for (n 1)-almost every x R n 1. Thus u ( x, t) ϕ ( x, t) dt = R x n and by Fubini s theorem R R n u ϕ x n dx = R u x n ( x, t)ϕ( x, t) dt Rn u x n ϕ dx. Since u = u almost everywhere in R n, we see that u x n is the nth weak artial derivative of u. The same argument alies to all other artial derivatives u x, j j = 1,..., n as well. Examle The radial rojection u : B(,1) B(,1), u(x) = x x is discontinuous at the origin. However, the coordinate functions x j x, j = 1,..., n, are absolutely continuous on almost every lines. Moreover, ( ) x j D i = δ i j x x i x j x x x 2 L (B(,1)) whenever 1 < n. Here 1, i = j, δ i j =, i j, is the Kronecker symbol. By the ACL characterization the coordinate functions of u belong to W 1, (B(,1)) whenever 1 < n. Remark We say that a closed set E to be removable for W 1, (), if E = and W 1, ( \ E) = W 1, () in the sense that every function in W 1, ( \ E)

39 CHAPTER 1. SOBOLEV SPACES 36 can be aroximated by the restrictions of functions in C (). Theorem 1.41 imlies the following removability theorem for W 1, ): if H n 1 (E) =, then E is removable for W 1, (). Observe, that if H n 1 (E) =, then E is contained in a measure zero set of lines in a fixed direction (equivalently the rojection of E onto a hyerlane also has H n 1 -measure zero). This result is quite shar. For examle, let = B(,1) and E = {x B(,1) : x 2 = }. Then < H n 1 (E) <, but E is not removable since, using Theorem 1.41 again, it is easy to see that the function which is 1 on the uer half-lane and on the lower half-lane does not belong to W 1, ). With a little more work we can show that E = E B(, 1 2 ) is not removable for W1, (B(,1)).

40 2 Sobolev inequalities The term Sobolev inequalities refers to a variety of inequalities involving functions and their derivatives. As an examle, we consider an inequality of the form ( R n u q dx q c ( R n Du dx (2.1) for every u C (Rn ), where constant < c < and exonent 1 q < are indeendent of u. By density of smooth functions in Sobolev saces, see Theorem 1.18, we may conclude that (2.1) holds for functions in W 1, (R n ) as well. Let u C (Rn ), u, 1 < n and consider u λ (x) = u(λx) with λ >. Since u C (Rn ), it follows that (2.1) holds true for every u λ with λ > with c and q indeendent of λ. Thus ( R n u λ q dx q c ( R n Du λ dx for every λ >. By a change of variables y = λx, dx = 1 λ n d y, we see that R n u λ (x) q dx = R n u(λx) q dx = R n u(y) q 1 λ n d y = 1 λ n R n u(x) q dx and Du λ (x) dx = λ Du(λx) dx R n R n = λ λ n Du(y) d y R n = λ λ n Du(x) dx. R n Thus 1 λ n q ( R n u q dx q c λ λ n ( R n Du dx 37

41 CHAPTER 2. SOBOLEV INEQUALITIES 38 for every λ >, and equivalently, u L q (R n ) cλ 1 n + n q Du L (R n ). Since this inequality has to be indeendent of λ, we have 1 n + n q = q = n n. T H E M O R A L : There is only one ossible exonent q for which inequality (2.1) may hold true for all comactly suorted smooth functions. For 1 < n, the Sobolev conjugate exonent of is Observe that (1) >, (2) If n, then and (3) If = 1, then = n n 1. = n n. 2.1 Gagliardo-Nirenberg-Sobolev inequality The following generalized Hölder s inequality will be useful for us. Lemma 2.1. Let 1 1,..., k with k = 1 and assume f i L i (), i = 1,..., k. Then f 1... f k dx k f i L i (). i=1 Proof. Induction and Hölder s inequality (exercise). Sobolev roved the following theorem in the case > 1 and Nirenberg and Gagliardo in the case = 1. Theorem 2.2 (Gagliardo-Nirenberg-Sobolev). Let 1 < n. c = c(n, ) such that There exists for every u W 1, (R n ). ( R n u dx ( c R n Du dx