MAT201C Lecture Notes: Introduction to Sobolev Spaces

Transcription

1 MAT2C Lecture Notes: Introduction to Sobolev Spaces Steve Shkoller Department of Mathematics University of California at Davis Davis, CA 9566 USA May 26, 2 These notes, intended for the third quarter of the graduate Analysis sequence at UC Davis, should be viewed as a very short introduction to Sobolev space theory, and the rather large collection of topics which are foundational for its development. This includes the theory of L p spaces, the Fourier series and the Fourier transform, the notion of weak derivatives and distributions, and a fair amount of differential analysis (the theory of differential operators). Sobolev spaces and other very closely related functional frameworks have proved to be indispensable topologies for answering very basic questions in the fields of partial differential equations, mathematical physics, differential geometry, harmonic analysis, scientific computation, and a host of other mathematical specialities. These notes provide only a brief introduction to the material, essentially just enough to get going with the basics of Sobolev spaces. As the course progresses, I will add some additional topics and/or details to these notes. In the meantime, a good reference is Analysis by Lieb and Loss, and of course Applied Analysis by Hunter and Nachtergaele, particularly Chapter 2, which serves as a nice compendium of the material to be presented. If only I had the theorems! Then I should find the proofs easily enough. Bernhard Riemann ( ) Facts are many, but the truth is one. Rabindranath Tagore (86-94)

2 Shkoller CONTENTS Contents L p spaces 4. Notation Definitions and basic properties Basic inequalities The space (L p (), L p() is complete Convergence criteria for L p functions The space L () Approximation of L p () by simple functions Approximation of L p () by continuous functions Approximation of L p () by smooth functions Continuous linear functionals on L p () A theorem of F. Riesz Weak convergence Integral operators Appendix : The monotone and dominated convergence theorems and Fatou s lemma Appendix 2: The Fubini and Tonelli Theorems Exercises The Sobolev spaces H k () for integers k Weak derivatives Definition of Sobolev Spaces A simple version of the Sobolev embedding theorem Approximation of W k,p () by smooth functions Hölder Spaces Morrey s inequality The Gagliardo-Nirenberg-Sobolev inequality Local coordinates near Sobolev extensions and traces The subspace W,p () Weak solutions to Dirichlet s problem Strong compactness Exercises The Fourier Transform Fourier transform on L (R n ) and the space S(R n ) The topology on S(R n ) and tempered distributions Fourier transform on S (R n )

3 Shkoller CONTENTS 3.4 The Fourier transform on L 2 (R n ) Bounds for the Fourier transform on L p (R n ) Convolution and the Fourier transform An explicit computation with the Fourier Transform Applications to the Poisson, Heat, and Wave equations Exercises The Sobolev Spaces H s (R n ), s R H s (R n ) via the Fourier Transform Fractional-order Sobolev spaces via difference quotient norms Fractional-order Sobolev spaces on domains with boundary The space H s (R n +) The Sobolev space H s () The Sobolev Spaces H s (T n ), s R The Fourier Series: Revisited The Poisson Integral Formula and the Laplace operator Exercises Regularity of the Laplacian on 94 8 Inequalities for the normal and tangential decomposition of vector fields on 8. The regularity of Tangential and normal derivatives Some useful inequalities Elliptic estimates for vector fields The div-curl lemma 6 9. Exercises

4 Shkoller L P SPACES L p spaces. Notation We will usually use to denote an open and smooth domain in R d, for d =, 2, 3,... In this chapter on L p spaces, we will sometimes use to denote a more general measure space, but the reader can usually think of a subset of Euclidean space. C k () is the space of functions which are k times differentiable in for integers k. C () then coincides with C(), the space of continuous functions on. C () = k C k (). spt f denotes the support of a function f, and is the closure of the set {x f(x) }. C () = {u C() spt u compact in }. C k () = Ck () C (). C () = C () C (). We will also use D() to denote this space, which is known as the space of test functions in the theory of distributions..2 Definitions and basic properties Definition.. Let < p < and let (, M, µ) denote a measure space. If f : R is a measurable function, then we define ( f L p () := ) f p p dx and f L () := ess sup x f(x). Note that f L p () may take the value. Unless stated otherwise, we will usually consider to be a smooth, open subset of R d, and we will assume that all functions under consideration are measurable. Definition.2. The space L p () is the set L p () = {f : R f L p () < }. The space L p () satisfies the following vector space properties:. For each α R, if f L p () then αf L p (); 2. If f, g L p (), then so that f + g L p (). f + g p 2 p ( f p + g p ), 4

5 Shkoller L P SPACES 3. The triangle inequality is valid if p. The most interesting cases are p =, 2,, while all of the L p arise often in nonlinear estimates. Definition.3. The space l p, called little L p, will be useful when we introduce Sobolev spaces on the torus and the Fourier series. For p <, we set { } l p = {x n } n Z x n p <, where Z denotes the integers..3 Basic inequalities n= Convexity is fundamental to L p spaces for p [, ). Lemma.4. For λ (, ), x λ ( λ) + λx. Proof. Set f(x) = ( λ)+λx x λ ; hence, f (x) = λ λx λ = if and only if λ( x λ ) = so that x = is the critical point of f. In particular, the minimum occurs at x = with value f() = ( λ) + λx x λ. Lemma.5. For a, b and λ (, ), a λ b λ λa + ( λ)b with equality if a = b. Proof. If either a = or b =, then this is trivially true, so assume that a, b >. Set x = a/b, and apply Lemma to obtain the desired inequality. Theorem.6 (Hölder s inequality). Suppose that p and < q < with p + q =. If f Lp and g L q, then fg L. Moreover, fg L f L p g L q. Note that if p = q = 2, then this is the Cauchy-Schwarz inequality since fg L = (f, g) L 2. Proof. We use Lemma.5. Let λ = /p and set a = f p f p, and b = g q L g q p L p 5

6 Shkoller L P SPACES for all x. Then a λ b λ = a /p b /p = a /p b /q so that f g f L p g L q f p p f p L p + q g q g q L q. Integrating this inequality yields f g dx f L p g L q ( f p p f p L p + g q q g q L q ) dx = p + q =. Definition.7. The exponent q = p. p p (or q = p ) is called the conjugate exponent of Theorem.8 (Minkowski s inequality). If p and f, g L p then f + g L p f L p + g L p. Proof. If f + g = a.e., then the statement is trivial. Assume that f + g a.e. Consider the equality f + g p = f + g f + g p ( f + g ) f + g p, and integrate over to find that f + g p dx Since q = p p, from which it follows that [ ( f + g ) f + g p ] dx Hölder s ( f L p + g L p) f + g p L q. f + g p ( ) = f + g p q dx L, q ( which completes the proof, since p = q. ) f + g p q dx f L p + g L q, Corollary.9. For p, L p () is a normed linear space. 6

7 Shkoller L P SPACES Example.. Let denote a subset of R n whose Lebesgue measure is equal to one. If f L () satisfies f(x) M > for almost all x, then log(f) L () and satisfies log fdx log( fdx). To see this, consider the function g(t) = t log t for t >. Compute g (t) = t = so t = is a minimum (since g () > ). Thus, log t t and letting t t we see that t log t t. (.) Since log x is continuous and f is measurable, then log f is measurable for f >. t = f(x) f in (.) to find that L Let f L f(x) log f(x) log f L f(x) f L. (.2) Since g(x) log f(x) h(x) for two integrable functions g and h, it follows that log f(x) is integrable. Next, integrate (.2) to finish the proof, as ( ) f(x) f dx =. L.4 The space (L p (), L p() is complete Recall the a normed linear space is a Banach space if every Cauchy sequence has a limit in that space; furthermore, recall that a sequence x n x in if lim n x n x =. The proof of completeness makes use of the following two lemmas which are restatements of the Monotone Convergence Theorem and the Dominated Convergence Theorem, respectively (see the Appendix for this chapter). Lemma. (MCT). If f n L (), f (x) f 2 (x), and f n L () C <, then lim n f n (x) = f(x) with f L () and f n f L as n. Lemma.2 (DCT). If f n L (), lim n f n (x) = f(x) a.e., and if g L () such that f n (x) g(x) a.e. for all n, then f L () and f n f L. Proof. Apply the Dominated Convergene Theorem to the sequence h n = f n f a.e., and note that h n 2g. Theorem.3. If p < then L p () is a Banach space. Proof. Step. The Cauchy sequence. Let {f n } n= denote a Cauchy sequence in L p, and assume without loss of generality (by extracting a subsequence if necessary) that f n+ f n L p 2 n. 7

8 Shkoller L P SPACES Step 2. Conversion to a convergent monotone sequence. Define the sequence {g n } n= as It follows that g =, g n = f + f 2 f + + f n f n for n 2. g g 2 g n so that g n is a monotonically increasing sequence. Furthermore, {g n } is uniformly bounded in L p as ( p gndx p = g n p L f p L p + f i f i L p) ( f L p + ) p ; thus, by the Monotone Convergence Theorem, g p n g p a.e., g L p, and g n g a.e. Step 3. Pointwise convergence of {f n }. For all k, i=2 f n+k f n = f n+k f n+k + f n+k + f n+ + f n+ f n Therefore, f n f a.e. Since n+k+ i=n+ f n f + f i f i = g n+k g n a.e. n f i f i g n g for all n N, i=2 it follows that f g a.e. Hence, f n p g p, f p g p, and f f n p 2g p, and by the Dominated Convergence Theorem, lim f f n p dx = lim f f n p dx =. n n.5 Convergence criteria for L p functions If {f n } is a sequence in L p () which converges to f in L p (), then there exists a subsequence {f nk } such that f nk (x) f(x) for almost every x (denoted by a.e.), but it is in general not true that the entire sequence itself will converge pointwise a.e. to the limit f, without some further conditions holding. 8

9 Shkoller L P SPACES Example.4. Let = [, ], and consider the subintervals [, ] [ ] [, 2 2,,, ] [, 3 3, 2 ] [ ] [ 2, 3 3,,, ] [, 4 4, 2 ] [ 2, 4 4, 3 ] 4 [ ] [ 3, 4,,, ], 5 Let f n denote the indicator function of the n th interval of the above sequence. Then f n L p, but f n (x) does not converge for any x [, ]. Example.5. Set = R, and for n N, set f n = [n,n+]. Then f n (x) as n, but f n L p = for p [, ); thus, f n pointwise, but not in L p. Example.6. Set = [, ], and for n N, set f n = n [, n ]. Then f n(x) a.e. as n, but f n L = ; thus, f n pointwise, but not in L. Theorem.7. For p <, suppose that {f n } L p () and that f n (x) f(x) a.e. If lim n f n L p () = f L p (), then f n f in L p (). Proof. Given a, b, convexity implies that ( ) a+b p 2 2 (ap + b p ) so that (a + b) p 2 p (a p + b p ), and hence a b p 2 p ( a p + b p ). Set a = f n and b = f to obtain the inequality 2 p ( f n p + f p ) f n f p. Since f n (x) f(x) a.e., 2 p f p dx = ( lim 2 p ( f n p + f p ) f n f p) dx. n Thus, Fatou s lemma asserts that 2 p f p ( dx lim inf 2 p ( f n p + f p ) f n f p) dx n = 2 p f p dx + 2 p lim f n p + lim inf n n = 2 p f p dx lim sup f n f p dx. n ( ) f n f p dx As f p dx <, the last inequality shows that lim sup n f n f p dx. It follows that lim sup n f n f p dx = lim inf n f n f p dx =, so that lim n f n f p dx =. 9

10 Shkoller L P SPACES.6 The space L () Definition.8. With f L () = inf{m f(x) M a.e.}, we set L () = {f : R f L () < }. Theorem.9. (L (), L ()) is a Banach space. Proof. Let f n be a Cauchy sequence in L (). It follows that f n f m f n f m L () a.e. and hence f n (x) f(x) a.e., where f is measurable and essentially bounded. Choose ɛ > and N(ɛ) such that f n f m L () < ɛ for all n, m N(ɛ). Since f(x) f n (x) = lim m f m (x) f n (x) ɛ holds a.e. x, it follows that f f n L () ɛ for n N(ɛ), so that f n f L (). Remark.2. In general, there is no relation of the type L p L q. For example, suppose that = (, ) and set f(x) = x 2. Then f L (, ), but f L 2 (, ). On the other hand, if = (, ) and f(x) = x, then f L 2 (, ), but f L (, ). Lemma.2 (L p comparisons). If p < q < r, then (a) L p L r L q, and (b) L q L p + L r. Proof. We begin with (b). Suppose that f L q, define the set E = {x : f(x) }, and write f as f = f E + f E c = g + h. Our goal is to show that g L p and h L r. Since g p = f p E f q E and h r = f r E c f q E c, assertion (b) is proven. For (a), let λ [, ] and for f L q, ( f L q = ) ( f q q dx = ) f λq f ( λ)q q dµ ( ) f λq L f ( λ)q q p L r = f λ L p f ( λ) L r. Theorem.22. If µ() and q > p, then L q L p. Proof. Consider the case that q = 2 and p =. Then by the Cauchy-Schwarz inequality, f dx = f dx f L () 2 µ().

11 Shkoller L P SPACES.7 Approximation of L p () by simple functions Lemma.23. If p [, ), then the set of simple functions f = n i= a i Ei, where each E i is an element of the σ-algebra A and µ(e i ) <, is dense in L p (, A, µ). Proof. If f L p, then f is measurable; thus, there exists a sequence {φ n } n= functions, such that φ n f a.e. with of simple φ φ 2 f, i.e., φ n approximates f from below. Recall that φ n f p a.e. and φ n f p 2 p f p L, so by the Dominated Convergence Theorem, φ n f L p. Now, suppose that the set E i are disjoint; then by the definition of the Lebesgue integral, If a i, then µ(e i ) <. φ p ndx = n a i p µ(e i ) <..8 Approximation of L p () by continuous functions i= Lemma.24. Suppose that R n is bounded. Then C () is dense in L p () for p [, ). Proof. Let K be any compact subset of. The functions F K, n (x) = + n dist(x, K) C () satisfy F K, n, and decrease monotonically to the characteristic function K. The Monotone Convergence Theorem gives f K, n K in L p (), p <. Next, let A be any measurable set, and let λ denote the Lebesgue measure. Then λ(a) = sup{µ(k) : K A, K compact}. It follows that there exists an increasing sequence of K j of compact subsets of A such that λ(a\ j K j ) =. By the Monotone Convergence Theorem, Kj A in L p () for p [, ). According to Lemma.23, each function in L p () is a norm limit of simple functions, so the lemma is proved.

12 Shkoller L P SPACES.9 Approximation of L p () by smooth functions For R n open, for ɛ > taken sufficiently small, define the open subset of by ɛ := {x dist(x, ) > ɛ}. Definition.25 (Mollifiers). Define η C (R n ) by η(x) := { Ce ( x 2 ) if x < if x, with constant C > chosen such that R n η(x)dx =. For ɛ >, the standard sequence of mollifiers on R n is defined by η ɛ (x) = ɛ n η(x/ɛ), and satisfy R n η ɛ (x)dx = and spt(η ɛ ) B(, ɛ). Definition.26. For R n open, set L p loc () = {u : R u Lp ( ) }, where means that there exists K compact such that K. We say that is compactly contained in. Definition.27 (Mollification of L ). If f L loc (), define its mollification f ɛ = η ɛ f in ɛ, so that f ɛ (x) = η ɛ (x y)f(y)dy = B(,ɛ) η ɛ (y)f(x y)dy x ɛ. Theorem.28 (Mollification of L p ()). (A) f ɛ C ( ɛ ). (B) f ɛ f a.e. as ɛ. (C) If f C (), then f ɛ f uniformly on compact subsets of. (D) If p [, ) and f L p loc (), then f ɛ f in L p loc (). 2

13 Shkoller L P SPACES Proof. Part (A). We rely on the difference quotient approximation of the partial derivative. Fix x ɛ, and choose h sufficiently small so that x+he i ɛ for i =,..., n, and compute the difference quotient of f ɛ : f ɛ (x + he i ) f(x) h = ɛ n = ɛ n [ h h ( ) ( )] x + hei y x y η η f(y)dy ɛ ɛ ( ) ( )] x + hei y x y η f(y)dy for some open set. On, [ ( ) ( )] x + hei y x y lim η η = ( ) η x y, h h ɛ ɛ ɛ x i ɛ so by the Dominated Convergence Theorem, f ɛ η ɛ (x) = (x y)f(y)dy. x i x i A similar argument for higher-order partial derivatives proves (A). Step 2. Part (B). By the Lebesgue differentiation theorem, lim f(y) f(x) dy = for a.e. x. ɛ B(x, ɛ) B(x,ɛ) Choose x for which this limit holds. Then f ɛ (x) f(x) η ɛ (x y) f(y) f(x) dy = ɛ n B(x,ɛ) B(x,ɛ) C B(x, ɛ) [ η ɛ η((x y)/ɛ) f(y) f(x) dy B(x,ɛ) f(x) f(y) dy as ɛ. ɛ Step 3. Part (C). For, the above inequality shows that if f C () and hence uniformly continuous on, then f ɛ (x) f(x) uniformly on. Step 4. Part (D). For f L p loc (), p [, ), choose open sets U D ; then, for ɛ > small enough, f ɛ L p (U) f L p (D). 3

14 Shkoller L P SPACES To see this, note that f ɛ (x) = B(x,ɛ) B(x,ɛ) η ɛ (x y) f(y) dy η ɛ (x y) (p )/p η ɛ (x y) /p f(y) dy ( ) (p )/p ( /p η ɛ (x y)dy η ɛ (x y) f(y) dy) p, B(x,ɛ) B(x,ɛ) so that for ɛ > sufficiently small f ɛ (x) p dx U U D B(x,ɛ) η ɛ (x y) f(y) p dydx ( ) f(y) p η ɛ (x y)dx dy B(y,ɛ) D f(y) p dy. Since C (D) is dense in L p (D), choose g C (D) such that f g L p (D) < δ; thus f ɛ f L p (U) f ɛ g ɛ L p (U) + g ɛ g L p (U) + g f L p (U) 2 f g L p (D) + g ɛ g L p (U) 2δ + g ɛ g L p (U).. Continuous linear functionals on L p () Let L p () denote the dual space of L p (). For φ L p (), the operator norm of φ is defined by φ op = sup L p ()= φ(f). p Theorem.29. Let p (, ], q = p. For g Lq (), define F g : L p () R as F g (f) = fgdx. Then F g is a continuous linear functional on L p () with operator norm F g op = g L q (). Proof. The linearity of F g again follows from the linearity of the Lebesgue integral. Since F g (f) = fgdx fg dx f L p g L q, with the last inequality following from Hölder s inequality, we have that sup f L p = F g (f) g L q. 4

15 Shkoller L P SPACES For the reverse inequality let f = g q sgn g. f is measurable and in L p since f p = f q q = g q and since fg = g q, so that g L q = F g (f) f L p F g (f) = ( = fgdx = F g op. f p dx ) ( g q dx = ( p g q dx ) q ) g q p + q dx = f L p g L q Remark.3. Theorem.29 shows that for < p, there exists a linear isometry g F g from L q () into L p (), the dual space of L p (). When p =, g F g : L () L () is rarely onto (L () is strictly larger than L ()); on the other hand, if the measure space is σ-finite, then L () = L ().. A theorem of F. Riesz Theorem.3 (Representation theorem). Suppose that < p < and φ L p (). Then there exists g L q (), q = p p such that φ (f) = fgdx f L p (), and φ op = g L q. Corollary.32. For p (, ) the space L p (, µ) is reflexive, i.e., L p () = L p (). The proof Theorem.3 crucially relies on the Radon-Nikodym theorem, whose statement requires the following definition. Definition.33. If µ and ν are measure on (, A) then ν µ if ν(e) = for every set E for which µ(e) =. In this case, we say that ν is absolutely continuous with respect to µ. Theorem.34 (Radon-Nikodym). If µ and ν are two finite measures on, i.e., µ() <, ν() <, and ν µ, then F (x) dν(x) = F (x)h(x)dµ(x) (.3) holds for some nonnegative function h L (, µ) and every positive measurable function F. 5

16 Shkoller L P SPACES Proof. Define measures α = µ + 2ν and ω = 2µ + ν, and let H = L 2 (, α) (a Hilbert space) and suppose φ : L 2 (, α) R is defined by φ (f) = fdω. We show that φ is a bounded linear functional since φ(f) = f d(2µ + ν) f d(2µ + 4ν) = 2 f dα f L 2 (x,α) α(). Thus, by the Riesz representation theorem, there exists g L 2 (, α) such that φ(f) = f dω = fg dα, which implies that f(2g )dν = f(2 g)dµ. (.4) Given F a measurable function on, if we set f = F 2 g 2g and h = 2g then F dν = F h dx which is the desired result, if we can prove that /2 g(x) 2. Define the sets { En = x g(x) < 2 } { and En 2 = x g(x) > 2 + }. n n By substituting f = E j n, j =, 2 in (.4), we see that µ(e j n) = ν(e j n) = for j =, 2, from which the bounds /2 g(x) 2 hold. Also µ({x g(x) = /2}) = and ν({x g(x) = 2}) =. Notice that if F =, then h L (). Remark.35. The more general version of the Radon-Nikodym theorem. Suppose that µ() <, ν is a finite signed measure (by the Hahn decomposition, ν = ν + ν + ) such that ν µ; then, there exists h L (, µ) such that F dν = F h dµ. Lemma.36 (Converse to Hölder s inequality). Let µ() <. Suppose that g is measurable and fg L () for all simple functions f. If { } M(g) = sup fg dµ : f is a simple function <, (.5) f L p = then g L q (), and g L q () = M(g). 6

17 Shkoller L P SPACES Proof. Let φ n be a sequence of simple functions such that φ n g a.e. and φ n g. Set f n = φ n q sgn (φ n ) φ n q L q so that f n L p = for p = q/(q ). By Fatou s lemma, Since φ n g a.e., then g L q () lim inf φ n L n q () = lim inf n g L q () lim inf n f n φ n dµ lim inf n The reverse inequality is implied by Hölder s inequality. f n φ n dµ. f n g dµ M(g). Proof of the L p () representation theorem. We have already proven that there exists a natural inclusion ι : L q () L p () which is an isometry. It remains to show that ι is surjective. Let φ L p () and define a set function ν on measurable subsets E by ν(e) = E dν =: φ( E ). Thus, if µ(e) =, then ν(e) =. Then f dν =: φ(f) for all simple functions f, and by Lemma.23, this holds for all f L p (). Radon-Nikodym theorem, there exists g L () such that f dν = fg dµ f L p (). But φ(f) = f dν = By the fg dµ (.6) and since φ L p (), then M(g) given by (.5) is finite, and by the converse to Hölder s inequality, g L q (), and φ op = M(g) = g L q (). 7

18 Shkoller L P SPACES.2 Weak convergence The importance of the Representation Theorem.3 is in the use of the weak-* topology on the dual space L p (). Recall that for a Banach space B and for any sequence φ j in the dual space B, φ j φ in B weak-*, if φ j, f φ, f for each f B, where, denotes the duality pairing between B and B. Theorem.37 (Alaoglu s Lemma). If B is a Banach space, then the closed unit ball in B is compact in the weak -* topology. Definition.38. For p <, a sequence {f n } L p () is said to weakly converge to f L p () if f n (x)φ(x)dx f(x)φ(x)dx φ L q (), q = p p. We denote this convergence by saying that f n f in L p () weakly. Given that L p () is reflexive for p (, ), a simple corollary of Alaoglu s Lemma is the following Theorem.39 (Weak compactness for L p, < p < ). If < p < and {f n } is a bounded sequence in L p (), then there exists a subsequence {f nk } such that f nk f in L p () weakly. Definition.4. A sequence {f n } L () is said to converge weak-* to f L () if f n (x)φ(x)dx f(x)φ(x)dx φ L (). We denote this convergence by saying that f n f in L () weak-*. Theorem.4 (Weak-* compactness for L ). If {f n } is a bounded sequence in L (), then there exists a subsequence {f nk } such that f nk f in L () weak-*. Lemma.42. If f n f in L p (), then f n f in L p (). Proof. By Hölder s inequality, g(f n f)dx f n f L p g L q. Note that if f n is weakly convergent, in general, this does not imply that f n is strongly convergent. 8

19 Shkoller L P SPACES Example.43. If p = 2, let f n denote any orthonormal sequence in L 2 (). From Bessel s inequality f n gdx g 2 L 2 (), we see that f n in L 2 (). n= This example shows that the map f f L p is continuous, but not weakly continuous. It is, however, weakly lower-semicontinuous. Theorem.44. If f n f weakly in L p (), then f L p lim inf n f n L p. Proof. As a consequence of Theorem.3, f L p () = sup fgdx = g L q () = sup lim g L q () = n sup lim inf f n L p g L q. g L q () = n f n gdx Theorem.45. If f n f in L p (), then f n is bounded in L p (). Theorem.46. Suppose that R n is a bounded. Suppose that If < p <, then f n f in L p (). sup f n L p () M < and f n f a.e. n Proof. Egoroff s theorem states that for all ɛ >, there exists E such that µ(e) < ɛ and f n f uniformly on E c. By definition, f n f in L p () for p (, ) if (f n f)gdx for all g L q (), q =. We have the inequality p p (f n f)gdx E f n f g dx + f n f g dx. E c Choose n N sufficiently large, so that f n (x) f(x) δ for all x E c. By Hölder s inequality, E c f n f g dx f n f L p (E c ) g L q (E c ) δµ(e c ) g L q () Cδ for a constant C <. 9

20 Shkoller L P SPACES By the Dominated Convergence Theorem, f n f L p () 2M so by Hölder s inequality, the integral over E is bounded by 2M g L q (E). Next, we use the fact that the integral is continuous with respect to the measure of the set over which the integral is taken. In particular, if h is integrable, then for all δ >, there exists ɛ > such that if the set E ɛ has measure µ(e ɛ ) < ɛ, then E ɛ hdx δ. To see this, either approximate h by simple functions, or use the Dominated Convergence theorem for the integral E ɛ (x)h(x)dx. Remark.47. The proof of Theorem.46 does not work in the case that p =, as Hölder s inequality gives f n f g dx f n f L () g L (E), so we lose the smallness of the right-hand side. E Remark.48. Suppose that E is bounded and measurable, and let g = E. If f n f in L p (), then f n (x)dx f(x)dx; E hence, if f n f, then the average of f n converges to the average of f pointwise..3 Integral operators If u : R n R satisfies certain integrability conditions, then we can define the operator K acting on the function u as follows: Ku(x) = k(x, y)u(y)dy, R n where k(x, y) is called the integral kernel.the mollification procedure, introduced in Definition.27, is one example of the use of integral operators; the Fourier transform is another. Definition.49. Let L(L p (R n ), L p (R n )) denote the space of bounded linear operators from L p (R n ) to itself. Using the Representation Theorem.3, the natural norm on L(L p (R n ), L p (R n )) is given by K L(L p (R n ),L p (R n )) = sup sup Kf(x)g(x)dx. R n E f L p = g L q = Theorem.5. Let p <, Ku(x) = R k(x, y)u(y)dy, and suppose that n k(x, y) dx C y R n and k(x, y) dy C 2 x R n, R n R n where < C, C 2 <. Then K : L p (R n ) L p (R n ) is bounded and K L(L p (R n ),L p (R n p )) C p C p 2. 2

21 Shkoller L P SPACES In order to prove Theorem.5, we will need another well-known inequality. Lemma.5 (Cauchy-Young Inequality). If p + q =, then for all a, b, ab ap p + bq q. Proof. Suppose that a, b >, otherwise the inequality trivially holds. ab = exp(log(ab)) = exp(log a + log b) (since a, b > ) ( = exp p log ap + ) log bq q p exp(log ap ) + q exp(log bq ) (using the convexity of exp) = ap p + bq q where we have used the condition p + q =. Lemma.52 (Cauchy-Young Inequality with δ). If p + q =, then for all a, b, ab δ a p + C δ b q, δ >, with C δ = (δp) q/p q. Proof. This is a trivial consequence of Lemma.5 by setting ab = a (δp) /p b (δp) /p. Proof of Theorem.5. According to Lemma.5, f(y)g(x) f(y) p p k(x, y)f(y)g(x)dydx R n R n R n k(x, y) R n p C p f p L p + C 2 q g q L q. dx f(y) p dy + R n k(x, y) R n q + g(x) q q dy g(x) q dx so that 2

22 Shkoller L P SPACES To improve this bound, notice that k(x, y)f(y)g(x)dydx R n R n k(x, y) dx tf(y) p k(x, y) dy + dy t g(x) q dx R n R n p R n R n q C t p p f p L + C 2t q g q p L =: F (t). q q Find the value of t for which F (t) has a minimum to establish the desired bounded. Theorem.53 (Simple version of Young s inequality). Suppose that k L (R n ) and f L p (R n ). Then k f L p k L f L p. Proof. Define K k (f) = k f := k(x y)f(y)dy. R n Let C = C 2 = k L (R n ). Then according to Theorem.5, K k : L p (R n ) L p (R n ) and K k L(L p (R n ),L p (R n )) C. Theorem.5 can easily be generalized to the setting of integral operators K : L q (R n ) L r (R n ) built with kernels k L p (R n ) such that + r = p + q. Such a generalization leads to Theorem.54 (Young s inequality). Suppose that k L p (R n ) and f L q (R n ). Then k f L r k L p f L q for + r = p + q..4 Appendix : The monotone and dominated convergence theorems and Fatou s lemma Let R d denote an open and smooth subset. The domain is called smooth whenever its boundary is a smooth (d )-dimensional hypersurface. Theorem.55 (Monotone Convergence Theorem). Let f n : R {+ } denote a sequence of functions, f n, and suppose that the sequence f n is monotonically increasing, i.e., f f 2 f 3 Then lim f n (x)dx = n 22 lim f n(x)dx. n

23 Shkoller L P SPACES Lemma.56 (Fatou s Lemma). Suppose the sequence f n : R and f n. Then lim inf f n(x)dx lim inf f n (x)dx. n n Example.57. Consider = (, ) R and suppose that f n = n (,/n). Then f n(x)dx = for all n N, but lim inf n f n(x)dx =. Theorem.58 (Dominated Convergence Theorem). Suppose the sequence f n : R, f n (x) f(x) almost everywhere (with respect to Lebesgue measure), and furthermore, f n g L (). Then f L () and lim n f n (x)dx = f(x)dx. Equivalently, f n f in L () so that lim n f n f L () =. In the exercises, you will be asked to prove that the Monotone Convergence Theorem implies Fatou s Lemma which, in turn, implies the Dominated Convergence Theorem..5 Appendix 2: The Fubini and Tonelli Theorems Let (, A, µ) and (Y, B, ν) denote two fixed measure spaces. The product σ-algebra A B of subsets of Y is defined by A B = {A B : A A, B B}. The set function µ ν : A B [, ] defined by for each A B A B is a measure. (µ ν)(a B) = µ(a) ν(b) Theorem.59 (Fubini). Let f : Y R be a µ ν-integrable function. Then both iterated integrals exist and f d(µ ν) = f dµdν = f dνdµ. Y Y The existence of the iterated integrals is by no means enough to ensure that the function is integrable over the product space. As an example, let = Y = [, ] and µ = ν = λ with λ the Lebesgue measure. Set Y f(x, y) = { x 2 y 2, (x 2 +y 2 ) 2 (x, y) (, ), (x, y) = (, ). 23

24 Shkoller L P SPACES Then a standard computation shows that f(x, y)dxdy = π 4, f(x, y)dydx = π 4. Fubini s theorem shows, of course, that f is not integrable over [, ] 2 There is a converse to Fubini s theorem, however, according to which the existence of one of the iterated integrals is sufficient for the integrability of the function over the product space. The theorem is known as Tonelli s theorem, and this result is often used. Theorem.6 (Tonelli). Let (, A, µ) and (Y, B, ν) denote two σ-finite measure spaces, and let f : Y R be a µ ν-measurable function. If one of the iterated integrals Y f dνdµ or Y f dµdν exists, then the function f is µ ν-integrable and hence, the other iterated integral exists and f d(µ ν) = f dµdν = f dνdµ..6 Exercises Y Y Y Problem.. Use the Monotone Convergence Theorem to prove Fatou s Lemma. Problem.2. Use Fatou s Lemma to prove the Dominated Convergence Theorem. Problem.3. Let R d denote an open and smooth subset. Let (a, b) R be an open interval, and let f : (a, b) R be a function such that for each t (a, b), f(t, ) : R is integrable and df dt (t, x) exists for each (t, x) (a, b). Futhermore, assume that there is an integrable function g : [, ) such that sup t (a,b) df dt (t, x) g(x) for all x. Show that the function h defined by h(t) f(t, x)dx is differentiable and that the derivative is given by dh dt (t) = d df f(t, x)dx = (t, x)dx dt dt for each t (a, b). Hint: You will need to use the definition of the derivative for a real valued function function r : (a, b) R which is dr dt (t r(t ) = lim +h) r(t ) h h, as well as the Mean Value Theorem from calculus which states the following: Let (t, t 2 ) R and let q : (t, t 2 ) R be differentiable on (t, t 2 ). Then q(t 2) q(t ) t 2 t = dq dt (t ) where t is some point between t and t 2. Problem.4. Let denote an open subset of R n. If f L () L (), show that f L p () for < p <. If is bounded, then show that lim p f L p = f L. (Hint: For ɛ >, you can prove that the set E = {x : f(x) > f L ɛ} has positive Lebesgue measure, and the inequality [ f L ɛ] E f holds.) 24

25 Shkoller L P SPACES Problem.5. Theorem.7 states that if p <, f L p, {f n } L p, f n f a.e., and lim n f n L p = f L p, then lim n f n f L p. Show by an example that this theorem is false when p =. Problem.6. Show that equality holds in the inequality a λ b λ λa + ( λ)b, λ (, ), a, b if and only if a = b. Use this to show that if f L p and g L q for < p, q < and =, then p + q fg dx = f L p g L q holds if and only if there exists two constants C and C 2 (not both zero) such that C f p = C 2 g q holds. Problem.7. Use the result of Problem.6 to prove that if f, g L 3 () satisfy f L 3 = g L 3 = f 2 g dx =, then g = f a.e. Problem.8. Given f L (S ), < r <, define P r f(θ) = n= ˆf n r n e inθ, ˆfn = 2π f(θ)e inθ dθ. 2π Show that where P r f(θ) = p r f(θ) = 2π p r (θ φ)f(φ)dφ, 2π p r (θ) = 2π Show that 2π p r (θ)dθ =. n= r n e inθ = r 2 2r cos θ + r 2. 25

26 Shkoller L P SPACES Problem.9. If f L p (S ), p <, show that P r f f in L p (S ) as r. Problem.. Suppose that Y = [, ] 2 is the unit square in R 2 and let a(y) denote a Y -periodic function in L (R 2 ). For ɛ >, let a ɛ (x) = a( x ɛ ), and let ā = Y a(y)dy denote the average value of a. Prove that a ɛ ā as ɛ. Problem.. Let f n = n (, n ). Prove that f n in L 2 (, ), that f n in L (, ), but that f n does not converge strongly in L 2 (, ). 26

27 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K 2 The Sobolev spaces H k () for integers k 2. Weak derivatives Definition 2. (Test functions). For R n, set C () = {u C () spt(u) V }, the smooth functions with compact support. Traditionally D() is often used to denote (), and D() is often referred to as the space of test functions. C For u C (R), we can define du dx by the integration-by-parts formula; namely, du dx (x)φ(x)dx = u(x) dφ dx (x)dx φ C (R). R Notice, however, that the right-hand side is well-defined, whenever u L loc (R) R Definition 2.2. An element α Z n + (nonnegative integers) is called a multi-index. For such an α = (α,..., α n ), we write D α = α x α αn x αn and α = α n + α n. Example 2.3. Let n = 2. If α =, then α = (, ); if α =, then α = (, ) or α = (, ). If α = 2, then α = (, ). Definition 2.4 (Weak derivative). Suppose that u L loc (). Then vα L loc () is called the α th weak derivative of u, written v α = D α u, if u(x) D α φ(x)dx = ( ) α v α (x)φ(x)dx φ C (). Example 2.5. Let n = and set = (, 2). Define the function { x, x < u(x) =, x 2. Then the function v(x) = {, x <, x 2 is the weak derivative of u. To see this, note that for φ C 2 u(x) dφ dx (x)dx = = = x dφ dx (x)dx dφ dx (x)dx (, 2), φ(x)dx + xφ + φ 2 = φ(x)dx v(x)φ(x)dx. 27

28 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Example 2.6. Let n = and set = (, 2). Define the function { x, x < u(x) = 2, x 2. Then the weak derivative does not exist! To prove this, assume for the sake of contradiction that there exists v L loc () such that for all φ C (, 2), 2 Then 2 v(x)φ(x)dx = v(x)φ(x)dx = = = 2 u(x) dφ dx (x)dx. x dφ 2 dx (x)dx 2 dφ dx (x)dx φ(x)dx φ() + 2φ() φ(x)dx + φ(). Suppose that φ j is a sequence in C (, 2) such that φ j() = and φ j (x) for x. Then = φ j () = which provides the contradiction. 2 v(x)φ j (x)dx φ j (x)dx, Definition 2.7. For p [, ], define W,p () = {u L p () weak derivative exists, Du L p ()}, where Du is the weak derivative of u. Example 2.8. Let n = and set = (, ). Define the function f(x) = sin(/x). Then u L (, ) and du dx = cos(/x)/x2 L loc (, ), but u W,p () for any p. Definition 2.9. In the case p = 2, we set H () = W,p (). Example 2.. Let = B(, ) R 2 and set u(x) = x α. We want to determine the values of α for which u H (). Since x α = 3 j= (x jx j ) α/2, then xi x α = α x α 2 x i is well-defined away from x =. Step. We show that u L loc (). To see this, note that x α dx = 2π whenever α < 2. r α rdrdθ < 28

29 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Step 2. Set the vector v(x) = α x α 2 x (so that each component is given by v i (x) = α x α 2 x i ). We show that u(x)dφ(x)dx = v(x)φ(x)dx φ C (B(, )). B(,) B(,) To see this, let δ = B(, ) B(, δ), let n denote the unit normal to δ (pointing toward the origin). Integration by parts yields δ x α Dφ(x)dx = 2π δ α φ(x)n(x)δdθ + α x α 2 x φ(x)dx. δ Since lim δ δ α 2π φ(x)n(x)dθ = if α <, we see that lim δ x α Dφ(x)dx = lim α δ δ x α 2 x φ(x)dx δ Since 2π r α rdrdθ < if α <, the Dominated Convergence Theorem shows that v is the weak derivative of u. Step 3. v L 2 (), whenever 2π r 2α 2 rdrdθ < which holds if α <. Remark 2.. Note that if the weak derivative exists, it is unique. To see this, suppose that both v and v 2 are the weak derivative of u on. Then (v v 2 )φdx = for all φ C (), so that v = v 2 a.e. 2.2 Definition of Sobolev Spaces Definition 2.2. For integers k and p, W k,p () = {u L loc () Dα u exists and is in L p () for α k}. Definition 2.3. For u W k,p () define u W k,p () = D α u p L p () α k p for p <, and u W k, () = α k D α u L (). The function W k,p () is clearly a norm since it is a finite sum of L p norms. 29

30 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Definition 2.4. A sequence u j u in W k,p () if lim j u j u W k,p () =. Theorem 2.5. W k,p () is a Banach space. Proof. Let u j denote a Cauchy sequence in W k,p (). It follows that for all α k, D α u j is a Cauchy sequence in L p (). Since L p () is a Banach space (see Theorem.9), for each α there exists u α L p () such that D α u j u α in L p (). When α = (,..., ) we set u := u (,..., ) so that u j u in L p (). We must show that u α = D α u. For each φ C (), ud α φdx = lim u j D α φdx j = ( ) α lim D α u j φdx j = ( ) α u α φdx ; thus, u α = D α u and hence D α u j D α u in L p () for each α k, which shows that u j u in W k,p (). Definition 2.6. For integers k and p = 2, we define H k () = W k,2 (). H k () is a Hilbert space with inner-product (u, v) H k () = α k (Dα u, D α v) L 2 (). 2.3 A simple version of the Sobolev embedding theorem For two Banach spaces B and B 2, we say that B is embedded in B 2 if u B2 C u B for some constant C and for u B. We wish to determine which Sobolev spaces W k,p () can be embedded in the space of continuous functions. To motivate the type of analysis that is to be employed, we study a special case. Theorem 2.7 (Sobolev embedding in 2-D). For kp 2, max u(x) C u x R 2 W k,p (R 2 ) u C (). (2.) 3

31 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Proof. Given u C (), we prove that for all x spt(u), u(x) C D α u(x) L p () α k. By choosing a coordinate system centered about x, we can assume that x = ; thus, it suffices to prove that u() C D α u(x) L p () α k. Let g C ([, )) with g, such that g(x) = for x [, 2 ] and g(x) = for x [ 3 4, ). By the fundamental theorem of calculus, u() = = = ( )k (k )! r [g(r)u(r, θ)]dr = r 2 r [g(r)u(r, θ)]dr r k k r [g(r)u(r, θ)]dr = Integrating both sides from to 2π, we see that u() = ( )k 2π(k )! 2π r (r) r [g(r)u(r, θ)]dr ( )k (k )! r k 2 k r [g(r)u(r, θ)]rdrdθ. The change of variables from Cartesian to polar coordinates is given by By the chain-rule, x(r, θ) = r cos θ, y(r, θ) = r sin θ. r u(x(r, θ), y(r, θ)) = x u cos θ + y u sin θ, r k 2 k r [g(r)u(r, θ)]rdr 2 r u(x(r, θ), y(r, θ)) = 2 xu cos 2 θ xyu cos θ sin θ + 2 yu sin 2 θ. It follows that r k = α k a α(θ)d α, where a α consists of trigonometric polynomials of θ, so that u() = ( )k r k 2 a α (θ)d α [g(r)u(x)]dx 2π(k )! B(,) α k C r k 2 L q (B(,)) D α (gu) L p (B(,)) ( C α k Hence, we require p(k 2) p + > or kp > 2. ) p r p(k 2) p p rdr u W k,p (R 2 ). 3

32 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K 2.4 Approximation of W k,p () by smooth functions Recall that ɛ = {x dist(x, ) > ɛ}. Theorem 2.8. For integers k and p <, let u ɛ = η ɛ u in ɛ, where η ɛ is the standard mollifier defined in Definition.25. Then (A) u ɛ C ( ɛ ) for each ɛ >, and (B) u ɛ u in W k,p loc () as ɛ. Definition 2.9. A sequence u j u in W k,p loc () if u j u in W k,p ( ) for each. Proof of Theorem 2.8. Theorem.28 proves part (A). Next, let v α denote the the αth weak partial derivative of u. To prove part (B), we show that D α u ɛ = η ɛ v α in ɛ. For x ɛ, D α u ɛ (x) = D α η ɛ (x y)u(y)dy = Dx α η ɛ (x y)u(y)dy = ( ) α Dy α η ɛ (x y)u(y)dy = η ɛ (x y)v α (y)dy = (η ɛ v α )(x). By part (D) of Theorem.28, D α u ɛ v α in L p loc (). It is possible to refine the above interior approximation result all the way to the boundary of. We record the following theorem without proof. Theorem 2.2. Suppose that R n is a smooth, open, bounded subset, and that u W k,p () for some p < and integers k. Then there exists a sequence u j C () such that u j u in W k,p (). It follows that the inequality (2.) holds for all u W k,p (R 2 ). 32

33 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K 2.5 Hölder Spaces Recall that for R n open and smooth, the class of Lipschitz functions u : R satisfies the estimate u(x) u(y) C x y x, y for some constant C. Definition 2.2 (Classical derivative). A function u : R is differentiable at x if there exists f : L(R n ; R n ) such that u(x) u(y) f(x) (x y) x y. We call f(x) the classical derivative (or gradient) of u(x), and denote it by Du(x). Definition If u : R is bounded and continuous, then u C () = max x u(x). If in addition u has a continuous and bounded derivative, then u C () = u C () + Du C (). The Hölder spaces interpolate between C () and C (). Definition For < γ, the space C,γ () consists of those functions for which u C,γ () := u C () + [u] C,γ () <, where the γth Hölder semi-norm [u] C,γ () is defined as ( ) u(x) u(y) [u] C,γ () = max x,y x y γ. x y The space C,γ () is a Banach space. 2.6 Morrey s inequality We can now offer a refinement and extension of the simple version of the Sobolev Embedding Theorem 2.7. Theorem 2.24 (Morrey s inequality). For n < p, let B(x, r) R n and let y B(x, r). Then u(x) u(y) Cr n p Du L p (B(x,2r)) u C (R n ). 33

34 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K In fact, Morrey s inequality holds for all u W,p (B(x, 2r)) (see Problem 2.9 in the Exercises). Notation 2.25 (Averaging). Let B(, ) R n. The volume of B(, ) is given by α n = π n 2 Γ( n 2 +) and the surface area is Sn = nα n. We define f(y)dy = α n r n f(y)dy B(x,r) f(y)ds = B(x,r) nα n r n B(x,r) B(x,r) f(y)ds. Lemma For B(x, r) R n, y B(x, r) and u C (B(x, r)), Du(y) u(y) u(x) dy C dy. x y n B(x,r) B(x,r) Proof. For some < s < r, let y = x + sω where ω S n = B(, ). By the fundamental theorem of calculus, for < s < r, Since ω =, it follows that Thus, integrating over S n yields u(x + sω) u(x) = = u(x + sω) u(x) S n u(x + sω) u(x) dω s s s s s d u(x + tω)dt dt Du(x + tω) ωdt. Du(x + tω) dt. Du(x + tω) dωdt S n Sn Du(x + tω) tn B(x,r) Du(y) dy, x y n dωdt tn where we have set y = x + tω. Multipling the above inequality by s n and integrating s from to r shows that r u(x + sω) u(x) dωs n ds rn Du(y) dy S n n B(x,r) x y n Cα n r n Du(y) dy, x y n which proves the lemma. 34 B(x,r)

35 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Proof of Theorem Assume first that u C (B(x, 2r)). Let D = B(x, r) B(y, r) and set r = x y. Then u(x) u(y) = u(x) u(y) dz D u(x) u(z) dz + u(y) u(z) dz. D Since D equals the intersection of two balls of radius r, it is clear that can choose a constant C, depending only on the dimension n, such that It follows that u(x) u(y) D D B(x, r) = D D B(y, r) = C. u(x) u(z) dz + u(y) u(z) dz D ( ) C u(x) u(z) dz + u(y) u(z) dz D D u(x) u(z) dz + C u(y) u(z) dz. B(x, r) C Thus, by Lemma 2.26, u(x) u(z) dz C and B(x,r) B(y,r) so that u(x) u(z) dz C and by Hölder s inequality, u(x) u(y) C B(x,r) B(x,r) B(y,r) u(x) u(y) C ( B(,2r) B(y,r) x z n Du(z) dz C x z n Du(z) dz C B(x,2r) B(x,2r) B(x,2r) x z n Du(z) dz x z n Du(z) dz x z n Du(z) dz (2.2) ) p ( s p( n) p ) p p s n dsdω Du(z) p dz B(x,2r) 35

36 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Morrey s inequality implies the following embedding theorem. Theorem 2.27 (Sobolev embedding theorem for k = ). There exists a constant C = C(p, n) such that u C, np (R n ) C u W,p (R n ) u W,p (R n ). Proof. First assume that u C (Rn ). Given Morrey s inequality, it suffices to show that max u C u W,p (R n ). Using Lemma 2.26, for all x R n, u(x) B(x,) C B(x,) C u W,p (R n ), u(x) u(y) dy + u(y) dy B(x,) Du(y) x y n dy + C u L p (R n ) the last inequality following whenever p > n. Thus, u, np C u C (R n ) W,p (R n ) u C (R n ). (2.3) By the density of C (Rn ) in W,p (R n ), there is a sequence u j C (Rn ) such that By (2.3), for j, k N, u j u W,p (R n ). u j u k C, n p (R n ) C u j u k W,p (R n ). Since C, n p (R n ) is a Banach space, there exists a U C, n p (R n ) such that u j U in C, n p (R n ). It follows that U = u a.e. in. By the continuity of norms with respect to strong convergence, we see that U, np C u C (R n ) W,p (R n ) which completes the proof. In proving the above embedding theorem, we established that that for p > n, we have the inequality u L (R n ) C u W,p (R n ). (2.4) 36

37 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K We will see later that (2.4), via a scaling argument, leads to the following important interpolation inequality: for p > n, n p n u L (R n p ) C(n, p) Du L p (R n ) u p L p (R n ). Another important consequence of Morrey s inequality is the relationship between the weak and classical derivative of a function. We begin by recalling the definition of classical differentiability. A function u : R n R m is differentiable at a point x if there exists a linear operator L : R n R m such that for each ɛ >, there exists δ > with y x < δ implying that u(y) u(x) L(y x) ɛ y x. When such an L exists, we write Du(x) = L and call it the classical derivative. As a consequence of Morrey s inequality, we extract information about the classical differentiability properties of weak derivatives. Theorem 2.28 (Differentiability a.e.). If R n, n < p and u W,p loc (), then u is differentiable a.e. in, and its gradient equals its weak gradient almost everywhere. Proof. We first restrict n < p <. By a version Lebesgue s differentiation theorem, for almost every x, lim Du(x) Du(z) r B(x,r) p dz =, (2.5) where Du denotes the weak derivative of u. Thus, for r > sufficiently small, we see that Du(x) Du(z) p dz < ɛ. B(x,r) Fix a point x for which (2.5) holds, and define the function Notice that w x (x) = and that w x (y) = u(y) u(x) Du(x) (y x). D y w x (y) = Du(y) Du(x). Set r = x y. Since u(y) u(x) Du(x) (y x) = w x (y) w x (x), an application of the inequality (2.2) that we obtained in the proof of Morrey s inequality then yields the 37

38 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K estimate u(y) u(x) Du(x) (y x) C = C Cr n p B(x,2r) B(x,2r) ( Cr ( B(x,2r) C x y ɛ, D z w x (z) dz x z n Du(z) Du(x) x z n dz Du(z) Du(x) p dz B(x,2r) Du(z) Du(x) p dz from which it follows that Du(x) is the classical derivative of u at the point x. The case that p = follows from the inclusion W,,p loc () Wloc () for all p <. 2.7 The Gagliardo-Nirenberg-Sobolev inequality In the previous section, we considered the embedding for the case that p > n. Theorem 2.29 (Gagliardo-Nirenberg inequality). For p < n, set p = np n p. Then u L p (R n ) C p,n Du L p (R n ) u W,p (R n ). Proof for the case n = 2. Suppose first that p = in which case p = 2, and we must prove that u L 2 (R 2 ) C Du L (R 2 ) u C (R 2 ). (2.6) Since u has compact support, by the fundamental theorem of calculus, ) p ) p so that and u(x, x 2 ) = u(x, x 2 ) u(x, x 2 ) x u(y, x 2 )dy = u(y, x 2 ) dy 2 u(x, y 2 ) dy 2 x2 2 u(x, y 2 )dy 2 Du(y, x 2 ) dy Du(x, y 2 ) dy 2. 38

39 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Hence, it follows that u(x, x 2 ) 2 Integrating over R 2, we find that u(x, x 2 ) 2 dx dx 2 ( ( Du(y, x 2 ) dy Du(x, y 2 ) dy 2 ) Du(y, x 2 ) dy Du(x, y 2 ) dy 2 dx dx 2 ) 2 Du(x, x 2 ) dx dx 2 which is (2.6). Next, if p < 2, substitute u γ for u in (2.6) to find that ( ) u 2γ 2 dx Cγ u γ Du dx R 2 R 2 Cγ Du L p (R 2) ( Choose γ so that 2γ = p(γ ) p ; hence, γ = p 2 p, and so that ( R 2 u 2p 2 p dx ) 2 p u p(γ ) p R 2 2p Cγ Du L p (R 2), ) p p dx u L 2 p 2p C p,n Du L (R n p (R n ) (2.7) ) for all u C (R2 ). Since C (R2 ) is dense in W,p (R 2 ), there exists a sequence u j C (R2 ) such that Hence, by (2.7), for all j, k N, so there exists U L 2p 2 p (R n ) such that u j u in W,p (R 2 ). u j u k L 2p 2 p (R n ) C p,n Du j Du k L p (R n ) u j U in L 2p 2 p (R n ). Hence U = u a.e. in R 2, and by continuity of the norms, (2.7) holds for all u W,p (R 2 ). 39

40 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K Proof for the general case of dimension n. Following the proof for n = 2, we see that so that u(x) n n u(x) n n dx Π n i= ( ( ( Π n i= ( ) Du dy Du(x,..., y i,..., x n ) dy i ) n ) Du(x,..., y i,..., x n ) dy i ( n Π n i=2 ) ( n Du dy Π n i=2 ) Du dy i where the last inequality follows from Hölder s inequality. Integrating the last inequality with respect to x 2, we find that where I = ( u(x) n n dx dx 2 < ) n Du dx dy 2 Du dy, I i = Applying Hölder s inequality, we find that u(x) n n dx dx 2 ( Π n i=3 Du dx dy 2 ) n ( ( n dx n dx Du dx dy i ) n, Π n i= i 2 Du dx dy i for i = 3,..., n. n Ii dx 2, Du dy dx 2 ) n Du dx dx 2 dy i ) n. Next, continue to integrate with respect to x 3,..., x n to find that ( u n n dx Π n i= R n ( = R n Du dx ) n n. Du dx...dy i...dx n ) n This proves the case that p =. The case that < p < n follows identically as in the proof of n = 2. 4

41 Shkoller 2 THE SOBOLEV SPACES H K () FOR INTEGERS K It is common to employ the Gagliardo-Nirenberg inequality for the case that p = 2; as stated, the inequality is not well-defined in dimension two, but in fact, we have the following theorem. Theorem 2.3. Suppose that u H (R 2 ). Then for all q <, u L q (R 2 ) C q u H (R 2 ). Proof. Let x and y be points in R 2, and write r = x y. Let θ S. Introduce spherical coordinates (r, θ) with origin at x, and let g be the same cut-off function that was used in the proof of Theorem 2.7. Define U := g(r)u(r, θ). Then and u(x) = u(x) U (r, θ)dr r x y U x y DU(r, θ) rdr. (r, θ)rdr r Integrating over S, we obtain: u(x) 2π B(x,) x y DU(y) dy := K DU, R 2 where the integral kernel K(x) = 2π B(,) x. Using Young s inequality from Theorem.54, we obtain the estimate K f L q (R 2 ) K L k (R 2 ) f L 2 (R 2 ) for Using the inequality (2.8) with f = DU, we see that u L q (R 2 ) C DU L 2 (R 2 ) [ k = q 2 +. (2.8) y k dy B(,) [ C DU L 2 (R 2 ) r k dr = C u H (R 2 ) [ q ] k. ] k ] k When q, k 2, so u Lq(R2) Cq 2 u H (R 2 ). 4