Sobolev embeddings and interpolations

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1 embed2.tex, January, 2007 Sobolev embeddings and interpolations Pavel Krejčí This is a second iteration of a text, which is intended to be an introduction into Sobolev embeddings and interpolations. The goal is to show the main ideas of the proofs, so that the reader can derive himself/herself particular formulas in cases that are not explicitly treated in textbooks. Hence, emphasis is put on methods rather than on a collection of results. Some additional information can also be found in [, 2, 3, 4]. Note that in fact, the only tool behind all the estimates is the elementary Hölder inequality. Contents Preliminaries 2 Admissible domains 2 3 Spaces L p Ω) and W,p Ω) 4 4 Some inequalities 5 5 Hardy-Littlewood inequality 8 6 Smooth approximation of L p functions 4 7 Sobolev embeddings 5 8 Limit cases and counterexamples 8 9 Anisotropic embeddings 20 0 Interpolations 23 Preliminaries The word embedding is used in the situation of two Banach spaces U and V, endowed with respective norms U and V, and such that } V U,.) C > 0 v V : v U C v V. If.) holds, then we say that V is embedded in U. The embedding is said to be compact, if every bounded set A V is precompact in U, that is, ε > 0 a,..., a n A a A k {,..., n} : a a k U < ε..2) The following theorem represents a basic tool in the theory of compact embeddings in function spaces.

2 embed2.tex, January, Theorem. Arzelà-Ascoli) Let, be Banach spaces and let A, B be compact sets. Let CA; B) be the Banach space of all continuous mappings from A into B. Let K CA; B) be an equicontinuous set, that is, ε > 0 δ > 0 f K x, y A : x y < δ fx) fy) < ε. Then K is compact in CA; B). equicontinuous. Conversely, every relatively compact set in CA; B) is Proof. Let K CA; B) be equicontinuous, and let ε > 0 be given. We find δ > 0 such that for all f K we have fx) fy) < ε/4 whenever x y < δ. Since A is compact, there exist x,..., x p A such that for every x A there exists i I := {,..., p} such that x x i < δ. Furthermore, B is compact, hence there exist y,..., y q B such that for every y B there exists j J := {,..., q} such that y y j < ε/4. For z J p, z = {z,..., z p }, we now denote { K z = f K ; i I : fx i ) y zi < ε }. 4 Set M := {z J p : K z }. The set M is indeed finite and we have K = z J K z, hence we may fix one representative f z K z for each z J. For any f K z and x A we find x i such that x x i < δ, and estimate fx) f z x) fx) fx i ) + fx i ) y zi + f z x i ) y zi + f z x) f z x i ) < ε, which we wanted to prove. Since every finite set of mappings in CA; B) is equicontinuous, the fact that that relatively compact sets are equicontinuous follows easily. 2 Admissible domains We fix an open connected bounded set Ω N, where N N is an integer, and denote by Ω its closure and by Ω its boundary. We assume that the following condition holds see Fig. ) L) There exist δ > 0 and m N, and for each k =,..., m there exists an open convex sets k N, a Lipschitz continuous function a k : k, and a rotation A k in N represented by an N N matrix, still denoted by A k, such that A k = A T k and deta k = ), such that i) Ω m j= A kg k ), ii) G k = {y N ; y = y, y N ), y k, y N a k y ) δ, a k y ) + δ)}, iii) G k = {y G k ; y N a k y ) δ, a k y ))}, iv) G 0 k = {y G k ; y N = a k y )}. v) Ω A k G k ) = A k G k ), vi) Ω A k G k ) = A k G 0 k ),

3 embed2.tex, January, y k G k G+ k A k Ω y N Figure. A domain with Lipschitzian boundary. If L) i) vi) hold, then we say that Ω has Lipschitzian boundary. As an example, consider the spaces C Ω) of continuous real functions defined on Ω, endowed with the norm f C,0 = sup{ fx) ; x Ω}, and C Ω) of continuously differentiable real functions on Ω, endowed with the norm { N } f C, = sup fx) + f x) x i ; x Ω. Proposition 2. If Ω has Lipschitzian boundary, then the space C Ω) is compactly embedded in C Ω). k= Proof. Condition.) is automatically satisfied. Furthermore, let K C Ω) be bounded. Hence, there exists M > 0 such that f K x Ω : fx) + N f x) x i M. We are thus in the situation of Theorem. with = N, =, A = Ω, B = [ M, M], provided we check that K is equicontinuous. Let x, y Ω be arbitrarily chosen. We find a Lipschitz continuous function ξ : [0, ] Ω and a constant C > 0 such that ξ0) = x, ξ) = y, ξ σ) C x y a. e. this is possible by the hypotheses on Ω), and use the chain rule to estimate fx) fy) = = 0 0 k= d fξσ)) dσ dσ fξσ)), ξ σ) dσ MC x y, where we denote by, the canonical scalar product in N. The relative compactness now follows from Theorem..

4 embed2.tex, January, Spaces L p Ω) and W,p Ω) Let Ω N be any open set. We denote as usual by L p Ω) the space of measurable functions u : Ω, for which the norm u p,ω is finite, where u p,ω = /p dx) Ω ux) if p =. sup ess x Ω The spaces L p Ω) with the above norms are Banach spaces. We say that v L p Ω) is a generalized partial derivative of u L p Ω) with respect to x i, i {,..., N}, if for every Lipschitz continuous function ϕ : Ω with compact support in Ω, that is, 3.) K = K Ω x Ω \ K : ϕx) = 0, 3.2) we have ux) ϕ x) dx = vx) ϕx) dx. 3.3) Ω x i Ω By [4, Chap. 2, Sect. 2.2], condition 3.3) is fulfilled if and only if u is absolutely continuous along almost all lines parallel to the x i -axis and v coincides with u/ x i almost everywhere. The Sobolev space W,p Ω) is defined as the subspace of L p Ω) of all functions u, which together with all generalized partial derivatives u/ x i belong to L p Ω). With the norm W,p Ω) is also a Banach space. u ;p,ω = u p,ω + N u x i, 3.4) p,ω The following result is crucial for the proof of embedding theorems, and its proof can be found in [4, Chap. 2, Sect. 3.6]. We fix an open bounded connected set Ω N with Lipschitzian boundary, and an open ball B N such that Ω B. We define W,p B to be the subset of W,p N ) consisting of all functions vanishing outside B. The norms in L p N ), W,p N ) will simply be denoted by p, ;p, respectively. Theorem 3. There exists a linear prolongation operator E p : W,p Ω) W,p B every u W,p Ω) we have i) E p ux) = ux) for a. e. x Ω; ii) There exists a constant c p > 0 such that for every u W,p Ω) we have iii) For every u W,p Ω) we have i= E p u ;p c p u ;p,ω ; E p u p c p u p,ω ; such that for

5 embed2.tex, January, Some inequalities This section collects some auxiliary inequalities that are needed in the sequel. Proposition 4. oung s inequality) Let f : [0, ) [0, ) be an absolutely continuous increasing function, f0) = 0. Then for every x, y 0 we have see Fig. 2) xy x fu) du + y 0 0 f v) dv, 4.) where f is the inverse function to f. Proof. Substituting v = fu) we have, with the convention f y) x that x 0 fu) du + y 0 f v) dv = x 0 fu) + uf u)) du + = x f y) if f y) < x, f y) xfx) + xy fx)) = xy. For < p <, we denote by p the conjugate exponent p = eciprocally, p is the conjugate of p and we have x uf u) du p p. 4.2) p + p =, p = p. 4.3) As an immediate consequence of Proposition 4. we obtain, putting fx) = x p, that xy p xp + p yp 4.4) for every x, y 0 and < p <. v y 0 u v = fu) x Figure 2. oung s inequality.

6 embed2.tex, January, Proposition 4.2 Hölder s inequality) Let Ω N be any open set and let p be arbitrary. Then for every f L p Ω) and g L p Ω) we have fx) gx) dx f p,ω g p,ω, 4.5) with the convention =, =. Ω Proof. The case p = or p = is obvious. For < p < we set F x) = fx) f p,ω, Gx) = gx). g p,ω By 4.4) we have F x) Gx) p F x) p + p Gx) p = fx) p p f p p,ω + gx) p, p g p p,ω hence F x) Gx) dx which we wanted to prove. Ω Ω F x) Gx) dx p + p =, Proposition 4.3 Minkowski s inequality) Let n, m be open sets, and let f : [0, ) be a measurable function. Then for every < p < we have fx, y) dx) p dy) /p f p x, y) dy) /p dx. 4.6) Proof. where Then For y and > 0 set F y) = f x, y) = F p y) dy = Hölder = f x, y) dx, gy) = F p y), { min{, fx, y)} if max{ x, y } <, 0 if max{ x, y }. F y) gy) dy = f p x, y) dy ) /p ) /p f p x, y) dy dx ) f x, y) gy) dy dx ) /p g p y) dy dx ) /p F p y) dy,

7 embed2.tex, January, hence /p ) /p /p F p y) dy) f p x, y) dy dx f p x, y) dy) dx, and we obtain the result from Fatou s Lemma letting tend to +. Note that replacing by a finite set {,..., n}, fx, y) by f k y), k =,..., n, and dx by n k=, the Minkowski inequality reads n n f k f k p,, 4.7) k= p, k= which is nothing but the triangle inequality for the norm p,. The following example shows that the Minkowski inequality cannot be reversed. Example 4.4 Consider = = 0, ), and fx, y) = x y) + ) /p for some p >. Then 0 fx, y) dx) p ) /p dy = f p x, y) dy ) /p dx = 0 ) p ) /p x y y) /p dx dy = x 0 y) dx ) /p dy = +. emark 4.5 In the same way we prove that for every q < p < we have p p /p, f q x, y) dx) p/q dy) /p f p x, y) dy) q/p dx) /q. 4.8) We set in this case F y) = f q x, y) dx, gy) = F p/q) y), and estimate F p/q y) dy similarly as in the proof of Proposition 4.3. The proof of the Minkowski inequality is related to the so-called reverse Hölder inequality: fx) gx) dx C g p,ω g L p Ω) = f p,ω C. 4.9) Ω To prove this statement, it suffices to choose gx) = signf x)) f x) p with f defined analogously as in the proof of Proposition 4.3, use the fact that f x) p dx fx) gx) dx C g p,ω = C f p/p p,ω, Ω and let tend to. Ω Proposition 4.6 oung s inequality II for convolutions) Let p, q, r be given such that p + r = + q. 4.0)

8 embed2.tex, January, For u L p N ), v L r N ), and x N Then w L q N ) and wx) = set N uy) vx y) dy. w q u p v r. 4.) Proof. The case q = follows immediately from Hölder s inequality. Hence, assume that q <, and set = r/q 0, ]. To make the use of the Minkowski inequality more transparent, we write dx, dy instead of dx, dy. Then, using the fact that = r/p and N N that vx y) r dy = vy) r dy for a. e. x, we obtain w q = Hölder Minkowski = = u p v r. uy) vx y) dy q ) /q dx q ) /q uy) vx y) vx y) dy) dx uy) p vx y) p dy) q/p dx uy) q vx y) q dx) p/q dy ) /q ) /p v ) /p ) /q uy) p dy vx) q dx v r r ) /p vy) p ) dy We devote the next section to the Hardy-Littlewood inequality, the proof of which is quite involved and requires a certain number of auxiliary steps. The proof we give here is a modification of the one from [2]. 5 Hardy-Littlewood inequality We state the Hardy-Littlewood inequality in the following form. Proposition 5. Let < p, q, r < be such that + + = 2. Then there exists a constant p q r H pr > 0 such that for every f L p ), g L q ) we have fx) gy) x y /r dx dy H pr f p g q. 5.) 2

9 embed2.tex, January, An explicit estimate for H pr will be given in 5.8) below. We first fix an even locally integrable function h : [0, ), which is non-decreasing in, 0) and non-increasing in 0, + ), and establish the following easy result. Lemma 5.2 For a, b > 0 and r, s set ϕ ab r, s) = a+r b+s a+r b+s hx y) dy dx. 5.2) Then for all r, s we have ϕ ab / r 0, ϕ ab / s 0 for r < s, ϕ ab / r 0, ϕ ab / s 0 for r > s, ϕ ab r, r) = ϕ ab 0, 0). Proof. We obviously have ϕ ab r, s) = ϕ ab r s, 0) = ϕ ab 0, s r) for all r, s, hence it suffices to prove that ϕ ab / rr, 0) 0 for r > 0, ϕ ab / s0, s) 0 for s > 0. We have ϕ ab r, 0) = r = = b b b b a+b a b ha + r y) h a + r y)) dy ha + r + y) h a + r y)) dy hr + z) hr z)) dz = a+b a b hr + z) hr z)) dz. For a. e. z > 0 we have hr + z) hr z), and the assertion follows. The argument for ϕ ab / s0, s) is identical. The idea of the proof of Proposition 5. is based on approximations of the functions f and g by step functions, and for each step function we use a rearrangement formula which will be proved by induction see Fig. 3). The induction step is carried out in the following way. Lemma 5.3 Let h be as in Lemma 5.2, and let m, n N {0} be given. Let a 0,..., a n, b 0,..., b m, r 0,..., r n, s 0,..., s m be sequences such that a i > 0, b j > 0 for all i = 0,..., n, j = 0,..., m, and r i r i a i +a i, s j s j b j +b j for all i =,..., n, j =,..., m. i) If r i0 r i0 = a i0 + a i0 for some i 0 {,..., n}, then there exist a i > 0 and ri for i = 0,..., n such that ri ri a i + a i for all i =,..., n, and n a i = i=0 n a i, i=0 n i=0 m ϕ a i b j ri, s j ) = n i=0 m ϕ ai b j r i, s j ). 5.3) ii) If s j0 s j0 = b j0 + b j0 for some j 0 {,..., m}, then there exist b j > 0 and s j for j = 0,..., m such that s j s j b j + b j for all j =,..., m, and m b j = m b j, n i=0 m ϕ ai b j r i, s j) = n m ϕ ai b j r i, s j ). 5.4) i=0

10 embed2.tex, January, Proof. We prove only part i), the rest is similar. If r i0 r i0 = a i0 +a i0, then r i0 +a i0 = r i0 a i0, hence for every j we have ϕ ai0 b j r i0, s j ) + ϕ ai0 b j r i0, s j ) = ϕ a i0 b j r i 0, s j ), where r i 0 = 2 a i 0 a i0 + r i0 + r i0 ), a i 0 = a i0 + a i0. We now set r i = r i, a i = a i for i = 0,..., i 0 2, r i = r i, a i = a i for i = i 0,..., n. Then 5.3) is automatically fulfilled by construction. It remains to check that r i 0 r i 0 a i 0 a i 0 = r i0 + 2 a i 0 a i0 + r i0 + r i0 ) a i0 + a i0 a i0 = r i0 + r i0 a i0 + a i0 + 2 r i 0 r i0 a i0 a i0 ) 0, r i 0 r i 0 2 a i 0 a i 0 2 = 2 a i 0 a i0 + r i0 + r i0 ) r i0 2 a i0 a i0 a i0 2 = r i0 r i0 2 a i0 a i0 2 2 r i 0 r i0 a i0 a i0 ) 0, and the proof is complete. Lemma 5.4 Let h be as in Lemma 5.2, and let a 0,..., a n, b 0,..., b m, r 0,..., r n, s 0,..., s m be as in Lemma 5.3. Set n m A = a i, B = b j. i=0 i=0 Then n m S := ϕ ai b j r i, s j ) ϕ AB 0, 0). Proof. We proceed by induction over N = n + m. For N = 0 we have ϕ a0 b 0 r 0, s 0 ) = ϕ a0 b 0 r 0 s 0, 0) ϕ a0 b 0 0, 0) by Lemma 5.2. Suppose now that the statement is proven for some N 0, and consider n, m such that n + m = N +. We will assume for definiteness that r n s m the opposite case is fully analogous). We distinguish two cases: i) n = 0. Then we set ŝ j = s j + s m s m b m b m s j for j = 0,..., m, ŝ m = s m. Then ŝ j ŝ j = s j s j for j =,..., m, ŝ m ŝ m = b m + b m. By Lemma

11 embed2.tex, January, we have ϕ a0 b j r 0, s j ) ϕ a0 b j r 0, ŝ j ) for all j = 0,..., m. By Lemma 5.3 there exist s 0,..., s m and b 0,..., b m such that m b j = m b j, m ϕ a0 b j r 0, s j) = m ϕ a0 b j r 0, ŝ j ). We have m = N, and the induction hypothesis yields m ϕ a0 b j r 0, s j ) m ϕ a0 b j r 0, s j) ϕ AB 0, 0). ii) n > 0. Set ˆr n = max{s m, r n + a n + a n } r n ˆr i = r i for i = 0,..., n. By Lemma 5.2, we have Ŝ := n m ϕ ai b j ˆr i, s j ) S. i=0 If ˆr n = r n + a n + a n, then Ŝ ϕ AB0, 0) by Lemma 5.3 and by the induction hypothesis similarly as in case i). If ˆr n = s m > r n + a n + a n, then set s m = max{s m + b m + b m, r n + a n + a n } s m, s j = s j for j = 0,..., m, r n = s m, r i = r i for i = 0,..., n, with the convention s =, b = b 0 if m = 0. We have Ŝ = Lemma 5.2 = n i=0 n i=0 n i=0 m m n ϕ ai b j ˆr i, s j ) + ϕ ai b m ˆr i, s m ) + i=0 n ϕ ai b j r i, s j ) + ϕ ai b m r i, s m ) + m ϕ ai b j r i, s j ). i=0 m m ϕ anb j s m, s j ) + ϕ anb m s m, s m ) ϕ anb j s m, s j ) + ϕ anb m s m, s m ) By construction, we have either s m = s m + b m + b m or r n = r n + a n + a n, and the assertion follows again from Lemma 5.3 and the induction hypothesis.

12 embed2.tex, January, y b j +s j b j +s j b j +s j b j +s j x a i +r i a i +r i a a i +r i a i +r i+ +r i+ a i+ +r i+ i Figure 3. Illustration to Lemmas 5.3, 5.4. Corollary 5.5 Let h be as in Lemma 5.2, and let { i, β i ) ; i = 0,..., n}, {γ j, δ j ) ; j = 0,..., m}, be two systems of intervals such that i, β i ) i2, β i2 ) =, γ j, δ j ) γ j2, δ j2 ) = for all i i 2 {0,..., n}, j j 2 {0,..., m}. Set A = 2 n β i i ), B = 2 i=0 m δ j γ j ). Then n m i=0 i βi δj γ j hx y) dy dx ϕ AB 0, 0). Proof. We change the ordering of the intervals i, β i ), γ j, δ j ) in such a way that β i i, δ j γ j for all i =,..., n, j =,..., m, and set r i = 2 i + β i ), a i = 2 β i i ), s j = 2 γ j + δ j ), b j = 2 δ j γ j ) for i = 0,..., n, j = 0,..., m. We have r i r i a i a i = i β i 0, s j s j b j b j = γ j δ j 0, i = a i +r i, β i = a i +r i, γ j = b j +s j, δ j = b j +s j, and Lemma 5.4 completes the proof. The next step consists in a rearrangement formula we summarize in Lemma 5.6 below. We fix K, L N and for sequences < a 0 < a < < a k < +, 0 = f 0 f f K, < b 0 < b < < b L < +, 0 = g 0 g g L,

13 embed2.tex, January, consider step functions f, g of the form fx) = gy) = K f i χ aϱi),a ϱi) )x) = i= L g j χ bσj),b σj) )y) = j= K f ϱ I) χ ai,a I )x), I= L g σ J) χ bj,b J )y) J= 5.5) for x, y, where χ M is the characteristic function of a set M, and ϱ : {,..., K} {,..., K}, σ : {,..., L} {,..., L} are some permutations of indices. We now define Then f i = i k= F k for all i, and we have F k = f k f k for k =,..., K. 5.6) fx) = K i F k χ aϱi),a ϱi) )x) = i= k= K k= F k K i=k χ aϱi),a ϱi) )x). 5.7) We further introduce for k =,..., K the numbers a k = 2 K a ϱi) a ϱi) ), a K+ = 0, 5.8) i=k and for x put f x) = K F k χ a k,a k ) x) = k= K f i χ a i, a i+ ] [a i+,a i ) x). 5.9) i= Similarly, we put G l = g l g l for l =,..., L. 5.0) Then L j gy) = G l χ bσj),b σj) )y) = L G l L j= l= l= j=l As before, we introduce for l =,..., L the numbers χ bσj),b σj) )y). 5.) b l = 2 L b σj) b σj) ), b L+ = 0, 5.2) j=l and for y put g y) = L G l χ b l,b l ) y) = l= L g j χ b j, b j+ ] [b j+,b j ) y). 5.3) j= We now prove the following crucial inequality.

14 embed2.tex, January, Lemma 5.6 Let f, g be as in 5.5), and let f, g be given by 5.9), 5.3), respectively. Let h be as in Lemma 5.2. Then we have f p = f p, g q = g q for all p, q, and fx) gy) hx y) dy dx 2 f x) g y) hx y) dy dx ) Proof. The fact that the L p norms of f and f coincide, follows immediately from 5.5) and 5.9), taking into account the fact that for all i we have a ϱi) a ϱi) = 2a i a i+). The same argument works for g and g, indeed. We further have by 5.7), 5.) that 2 fx) gy) hx y) dy dx = K k= l= L F k G l By Corollary 5.5, we have for every k and l that K i=k L j=l aϱi) a ϱi) bσj) b σj) hx y) dy dx. 5.5) K L aϱi) i=k j=l a ϱi) bσj) b σj) hx y) dy dx a k b l a k b l hx y) dy dx, and 5.4) follows from 5.9), 5.3), and 5.5). We are now ready to pass to the proof of Proposition 5.. Proof of Proposition 5.. We restrict ourselves to the case that f and g are non-negative step functions of the form 5.5). The general case then follows from the density of step functions in L p ), L q ). By Lemma 5.6 we have fx) gy) x y /r dx dy 2 f x) g y) x y /r dx dy ) For y set F y) = f x) x y /r dx. The function f is even, nondecreasing in, 0) and nonincreasing in 0, + ), hence f p p x x f ξ)) p dξ 2 x f x)) p x. 5.7) Choosing = p/q, we thus obtain for every y that F y) f x)) 2x )/p f x y /r dx = 2 +/r f p/q p = 2 +/r f p/q p p f x)) p/q x +/r x y /r dx f yt)) p/q t +/r t /r dt.

15 embed2.tex, January, We now use the Minkowski inequality 4.6) to estimate the L q F q 2 +/r f p/q p Minkowski 2 +/r f p/q p = 2 +/r f p/q p = 2 +/r f p norm of F. We have f yt)) p/q t +/r t /r dt ) q ) /q f yt)) p t q +q /r t q /r dy dt ) /q t +/r t /r f yt)) p dy dt t /p t /r dt. By Hölder s inequality, Lemma 5.6, and inequality 5.6), the left-hand side of 5.) is estimated from above by g q F q. Hence, 5.) holds with = 2 +/r t /p t /r dt. 5.8) H pr dy ) /q 6 Smooth approximation of L p functions We fix a smooth C is enough for our purposes) function ϕ : N [0, ) such that ϕx) = 0 outside the set B) := {x N ; x }, and ϕx) dx =. 6.) B) For u L p N ), x N, and a parameter σ 0, ] we set ) x y u σ x) = σ N ϕ uy) dy. 6.2) σ N For all σ 0, ], the function u σ is continuously differentiable, and we have p u σ u p x) dx = ϕz)ux σz) ux)) dz dx N N B) ) p/p Hölder ϕ p x) dx ux σz) ux) p dx dz, 6.3) B) B) N hence u σ u strongly in L p N ) as σ ) as a consequence of the Mean Continuity Theorem, see [4, Chap. 2, Sect..2]. In the sequel, we will use the following relation between the L q norm of u σ and L p norm of u, which follows directly from Proposition 4.6: where r is as in 4.0). u σ q σ N/p /q) ϕ r u p q p, 6.5)

16 embed2.tex, January, Sobolev embeddings We now state and prove the main result of this text. Theorem 7. Let p, q, ) be such that and set κ := N p q > p N, p ) q 0, ). Then there exists C pq > 0 such that for every u W,p N ) and every σ 0, ] we have u σ u q C pq σ κ u p. 7.) Proof. Notice first that for every x N and σ 0, ) we obtain, integrating by parts, that N )) σ uσ x) = σ N xi y i x y ϕ uy) dy y N i= i σ σ ) x y = σ N Φ, uy) dy, 7.2) σ N where we set Φξ) = ξϕξ). This yields in particular, β ) x y u β x) u x) σ N Φ, uy) dy σ dσ 7.3) N for every 0 < < β. To estimate the difference u β u in 7.3) in the space L q N ), we make use of the Minkowski and oung II inequalities with r as in 4.0), using the notation, for as in Proposition 4.3. More specifically, we have N u β u q Minkowski oung II β β β σ N σ N σ N u p We have N/r ) = κ, hence B) ) ) x y q /q Φ uy) dy dσ dx) σ ) ) x y q ) /q Φ uy) dy dx dσ σ y ) ) r /r ) /p Φ dy ux) p dx dσ σ N ) /r β Φ x) r dx σ N/r ) dσ. 7.4) N u β u q C pq β κ κ ) u p 7.5) with C pq = Φ r /κ. Hence, for every sequence σ i 0+, u σ i is a Cauchy sequence in L q N ). By 6.4), u σ i converge to u in L p N ), hence u L q N ) and u σ i converge to u strongly) in L q N ). Letting tend to 0 and replacing β by σ, we thus obtain 7.).

17 embed2.tex, January, Corollary 7.2 Let Ω N be an open bounded connected set with Lipschitzian boundary, and let q > p N. Then the space W,p Ω) is compactly embedded in L q Ω). Proof. Assume first q p, and set u = E p u, where E p : W,p Ω) W,p B is the prolongation operator from Theorem 3.. By Theorems 3. and 7., there exist constants C and C 2 such that for every σ 0, ] we have By 6.5) and Theorem 3. we have u σ u q C σ κ u p C 2 σ κ u ;p,ω. 7.6) u σ q C 3 σ κ u p C 3 c p σ κ u p,ω, 7.7) with C 3 = ϕ r. Consequently, there exists a constant C 4 > 0 such that u q,ω u q C 4 σ κ u p,ω + σ κ u ;p,ω ) 7.8) for all σ 0, ]. According to.), W,p Ω) is thus embedded in L q Ω). To see that the embedding is compact, consider a bounded set M W,p Ω) and an arbitrary ε > 0. We fix σ > 0 such that, with the notation of Theorem 7., we have C pq σ κ u p < ε 4 u M. 7.9) With this fixed σ, every element u σ of the set M σ = {u σ ; u M} vanishes outside of the set + σ)b) =: B + σ). Moreover, M σ is bounded in C B + σ)), hence, by Proposition 2., there exist u,..., u n M such that u M k {,..., n} x B + σ) : u σ x) u σ kx) < We then have, by 7.9), 7.0), and Theorem 7., that ε 4measB + σ)). 7.0) u u σ k q u σ u σ k q + ε 4 < ε 2. 7.) For k =,..., n set M k = {u M ; u u σ k q < ε/2, and J = {k {,..., n} ; M k }. For every k J we fix one representative û k M k, so that for every u M k we have u û k q,ω < ε and M = k J M k. The proof is thus complete for q p. Let now q < p. Hölder s inequality yields u σ u q measb + σ))) /q /p u σ u p, hence the above argument remains valid. Corollary 7.3 Let Ω N be an open bounded connected set with Lipschitzian boundary, and let p > N. Then the space W,p Ω) is compactly embedded in C Ω).

18 embed2.tex, January, Proof. We repeat the argument of the proof of Theorem 7. and Corollary 7.2, putting κ := N p 0, ). A computation analogous to 7.4) yields for every x N that β ) u β x) u x) σ N x y Φ u y) dy dσ σ Hölder β σ N y ) p /p ) /p Φ dy) u x) p dx dσ σ N N ) /p β u p Φ x) p dx σ N/p dσ, 7.2) and we proceed as above. B) 8 Limit cases and counterexamples Proposition 8. Let Ω N, N 2, be an open bounded connected set with Lipschitzian boundary, and let q = p N. 8.) Then the space W,p Ω) is embedded in L q Ω). Proof. We proceed in principle as in the proof of Theorem 7.. The main difference is that the number κ is zero here and we have to proceed more carefully. We represent x N as x = x, x N ), x N, and rewrite inequality 7.3) as β u β x, x N ) u x, x N ) σ N x Φ y, x ) N y N uy, y N ) dy dy N dσ. σ σ N 8.2) With r = N/N ), we now repeat the computation from 7.4), restricted to the component x, to obtain u β, x N ) u, x N ) q β σ N Minkowski oung II N β β β N x Φ y, x N y N σ σ ) ) q /q uy, y N ) dy dy N dσ dx) σ N x Φ y, x ) ) N y N q ) /q uy, y N ) dy dx dσ dy N N σ σ N σ N y Φ σ, x ) N y N r ) /r ) /p dy ux, y N ) p dx dσ dy N σ N N σ N+N /r u, y N ) p Φ, x ) N y N dσ dy N. 8.3) σ r

19 embed2.tex, January, The function Φ, x N y N ) r vanishes if σ < x σ N y N. Moreover, Φ is bounded by a constant Φ 0 > 0. Hence, using the fact that N + N /r = 2 + /N, we have β σ N+N /r Φ, x ) N y N σ r dσ Φ 0 x N y N σ 2+/N dσ = Φ 0 r x N y N /r. We thus have u β, x N ) u, x N ) q Φ 0 r u, y N ) p x N y N /r dy N. At this point, we use the Hardy-Littlewood inequality 5.), with q replaced by q. Indeed, /q + /p + /r = 2. Hence, for every function g L q ) we have by Proposition 5. that u β, x N ) u, x N ) q gx N ) dx N C g q u p with some constant C > 0, hence, by the reverse Hölder inequality 4.9), we have u β u q C u p. 8.4) Since u σ converge strongly to u in L p N ) and their L q norms are bounded, we conclude that they converge strongly in L q N ) as well and the embedding formula follows. We now show a few examples to illustrate that the embedding inequalities are at least qualitatively) optimal. i) To see that the embedding in Proposition 8. is not compact, and that W,p Ω) is not embedded in L q Ω) if q < p N, 8.5) it suffices to fix any open set Ω, some x 0 Ω, find s 0 > 0 such that x 0 + s 0 B) Ω, and consider the family of functions ) x u s x) = s N/p x0 ϕ, s 0, s 0 ), 8.6) s with ϕ as in 6.2). We have u s p,ω = s ϕ p, u s x i = ϕ p,ω x i, u s q,ω = s ϕ q s 0, s 0 ), p where = N/p /q). In the case 8.), we have = 0. Using the fact that u s converge to 0 in L p Ω) as s 0+, we conclude that the family {u s }, having constant nonzero norm in L q Ω), does not contain any convergent subsequence in L q Ω), hence the embedding is not compact. In the case 8.5), we have < 0, hence the family {u s } is unbounded in L q Ω) and no embedding takes place.

20 embed2.tex, January, ii) In another limit case p = N, 8.7) the space W,p Ω) is embedded in L Ω) if and only if p = N =, and the embedding is not compact. For N 2, it suffices to consider Ω = B), and )) x ux) = log, 2 for any 0 < < /N. Then u is unbounded, but belongs to W,N B)). For N =, the embedding of W, Ω) into C Ω) hence L Ω)) for every bounded interval Ω is obvious. To see that it is not compact, we may consider for n N the sequence u n x) = { [ sin for x x 0 otherwise., n+)π nπ It is bounded in W, 0, /π), but sup u n x) u m x) = for all m n, hence it is not precompact in L 0, /π). iii) The assumption on the Lipschitzian boundary is substantial. We show that there exists an open simply connected set Ω 2 such that W,p Ω) is not embedded in L q Ω) for any q > p. This set can be defined as see Fig. 4) For any q > p we set Ω = { x = x, x 2 ) 2 ; 0 < x <, 0 < x 2 < e /x }. Then u pq W,p Ω), but u pq / L q Ω). u pq x) = e 2/p+q)x. ] x 2 Ω x 0 Figure 4. Non-Lipschitzian boundary. 9 Anisotropic embeddings In evolution problems, one deals with functions which depend on a space variable x Ω and time t ω, where ω is an open interval corresponding to the time of the process. For p, q <, we introduce the spaces L p ω ; L q Ω)) = { u L Ω ω) ; u p,q,ω,ω := ω u, t) p q,ω dt ) /p < }, 9.)

21 embed2.tex, January, with obvious modifications for p = or q =. We state explicitly one possible embedding result for such spaces, without going into much detail in the proof, which is fully analogous to the above ones. Theorem 9. Let Ω N be an open bounded connected set with Lipschitzian boundary, let ω be a bounded open interval, and let W p 0,q 0 ;p,q ω; Ω) be the space If W p 0,q 0 ;p,q ω; Ω) = { u L Ω ω) ; u t Lp 0 ω ; L q 0 Ω)), u } L p ω ; L q Ω)) for i =,..., N. x i p 0 p q 0 + q < N, 9.2) then the space W p 0,q 0 ;p,q ω; Ω) is compactly embedded in C Ω ω). If q 2 max{q 0, q }, p 2 max{p 0, p }, and ) ) p0 + p2 N q + q2 > ) ), 9.3) p p 2 q 0 q 2 then W p 0,q 0 ;p,q ω; Ω) is compactly embedded in L p 2 ω ; L q 2 Ω)). Hint for the proof. Consider as before the extensions to the space W p 0,q 0 ;p,q ; N ), where the norms pi,q i,ω,ω are denoted again for simplicity as pi,q i, i = 0,. For σ 0, ] and u W p 0,q 0 ;p,q ; N ), we define regularizations analogous to 6.2) in the form x y u σ x, t) = σ N λ σ, t s ) uy, s) dy ds, 9.4) σ λ N ϕ where ϕ is a smooth nonnegative function on N+, which vanishes outside B), ), and ϕx, t) dx dt =. The number λ is to be chosen as λ = B) ) + N q 0 q ) p 0 p Note that λ > 0 by 9.3). A computation similar to 7.2) 7.4) yields σ uσ x, t) = x y λσ N Φ 0 σ, t s ) u σ N σ N λ N x y Φ σ, t s σ λ y, s) dy ds s ), y uy, s) dy ds, 9.6)

22 embed2.tex, January, where Φ 0 ξ, τ) = τϕξ, τ), Φ ξ, τ) = ξϕξ, τ), hence u β x, t) u x, t) λi 0 x, t) + I x, t) 9.7) for 0 < < β, where I 0 x, t) = β σ N I x, t) = β σ N λ N Φ 0 x y σ, t s σ λ ) u s y, s) dy ds dσ, Φ x y N, ) t s σ σ λ y uy, s) dy ds dσ. 9.8) Let 9.3) hold. With the intention to use oung s inequality for convolutions again, we introduce the numbers r 0, s 0, r, s by the identities r 0 = q 0 + q 2, s 0 = p 0 + p 2, r = q + q 2, s = p + p ) We use again the notation dx, dy for N dx, N dy, and T dt, S ds for dt, ds. For t, we have Minkowski I 0, t) q2 hence oung II = Minkowski I 0 p2,q 2 oung II β β σ N σ N S S β σ N +N/r 0 β σ N +N/r 0 β σ N +N/r 0 x y Φ 0 σ, t s ) ) u q2 /q2 y, s) σ λ s dx) dy ds dσ Φ 0 σ, t s ) u σ λ, s) r 0 s ds dσ q0 Φ 0, t s ) u σ λ, s) s ds dσ, 9.0) q0 S T S Φ 0 r 0, t s ) σ λ r 0 p2 /p2 u, s) s ds) dt) dσ q0 ) Φ 0, u σ λ s 0,r 0 s dσ p0,q 0 u β = Φ 0 s0,r 0 s σ N +N/r 0+λ/s 0 dσ. 9.) p0,q 0 Similarly, Minkowski I, t) q2 oung II = β β σ N λ σ N λ S Φ S β σ N λ+n/r σ, t s σ λ S Φ x y Φ σ ) y u, s) q ds dσ r, t s ) σ λ, t s ) ) q2 ) /q2 σ λ y uy, s) dy dx ds dσ r y u, s) q ds dσ, 9.2)

23 embed2.tex, January, hence Set Minkowski I p2,q 2 oung II κ = N β σ N λ+n/r β σ N λ+n/r T Φ, S Φ, t s ) σ λ y u, s) q r σ λ ) s,r y u p,q dσ ds ) p2 dt ) /p2 dσ β = Φ s,r y u p,q σ N λ+n/r +λ/s dσ. 9.3) p 0 Then κ > 0 by 9.3), and we have + ) + ) p p 0 p 2 N + ) ) )). q q 2 q 0 q 2 p p 2 N + N r 0 + λ s 0 = N λ + N r + λ s = κ. Combining 9.7) with 9.8), 9.), and 9.3) yields ) u β u p2,q 2 C p0,p,p 2,q 0,q,q 2 β κ κ u ) t + x u p,q, 9.4) p0,q 0 and we obtain the result similarly as in Theorem 7.. Note that the order of integration in 9.) cannot be reversed. For p q we have by emark 4.5 that L q Ω ; L p ω)) is embedded into L p ω ; L q Ω)), but the opposite inclusion does not hold, see Example 4.4. On the other hand, denoting W q 0,p 0 ;q,p Ω; ω) = { u L Ω ω) ; u t Lq 0 Ω ; L p 0 ω)), u } L q Ω ; L p ω)) for i =,..., N, x i we may repeat the computations in 9.0) 9.3) with reversed order of integration, to check that conditions 9.2) and 9.3) remain valid for the compact embedding of W q 0,p 0 ;q,p Ω; ω) into C Ω ω) and L q 2 Ω ; L p 2 ω)), respectively. Let us mention one important particular case which frequently occurs in applications. We omit the proof which is the same as for the other cases. Corollary 9.2 If q 2 max{q 0, q }, and N + ) ) > p q q 2 q0 q2, 9.5) p 0 then the space W q 0,p 0 ;q,p Ω; ω) is compactly embedded in L q 2 Ω ; C ω)).

24 embed2.tex, January, Embeddings of function spaces that are anisotropic also in the space variables, for example u x i L p i ω ; L q i Ω)), i =,..., N, can be treated in the same way. The regularizations then have to be chosen in the form u σ x, t) = σ P µ x i y,..., x N y N, t s ) uy, s) dy ds, 9.6) σ µ σ µ N σ N ϕ with suitably chosen exponents µ,..., µ N. 0 Interpolations We first recall the following classical interpolation result in L p spaces. Proposition 0. Let Ω N be an open set bounded or unbounded), and let p 0 < p be given. If u L p 0 Ω) L p Ω), then u L p Ω) for all p [p 0, p ], and we have for all u L p 0 Ω) L p Ω), where u p,ω = u p 0,Ω u p,ω p 0 p p 0. p Proof. Set q = p /p. Then q = p 0 / )p, and we may use Hölder s inequality to obtain ) /p u p,ω = ux) )p ux) p dx Ω ) /pq ) /pq ux) )pq dx ux) pq dx Ω Ω = u p 0,Ω u p,ω. We now establish an interpolation formula between L p spaces and Sobolev spaces. Theorem 0.2 Let p, q, s, ) be such that and set κ := N s > q > p N, p ), γ = N q s ). q Then there exists C pqs > 0 such that for every u W,p N ) L s N ) and every σ 0, ] we have u q C pqs σ γ u s + σ κ u p ). 0.)

25 embed2.tex, January, Proof. The assertion follows from 6.5) and 7.) provided q p. In particular, for q = p we have κ =, γ = γ 0 := N/s /p), and Let now q < p. By Proposition 0. we have where This yields u p C pps σ γ 0 u s + σ u p ). 0.2) u q u s u p, = s q. s p u p Cpps σ γ 0 ) u s + σ u s u p. We now use inequality 4.4) with p replaced by /, and with x = µσ u p, y = u s /µ, where we set µ = σ )γ 0, and obtain Hence, which is precisely 0.). σ u s u p σ + )γ 0 u p + )σ γ 0 u s. u p 2C pps σ γ 0 u s + σ + )γ 0 u p ), We conclude this text with the famous Gagliardo-Nirenberg inequality. be an open bounded con- Corollary 0.3 Gagliardo-Nirenberg inequality) Let Ω N nected set with Lipschitzian boundary, and let s > q > p N. Set ϱ = s q + N s p Then there exists a constant K pqs > 0 such that for every u W,p Ω) we have Proof. u q,ω. K pqs u s,ω + u ϱ s,ω u ϱ ;p,ω). 0.3) As in the proof of Corollary 7.2, we set u = E p u. By Theorem 0.2, we have u q C pqs σ γ u s + σ κ u p ). 0.4) If u p > u s, then we set ) /γ+κ) u s σ =, u p otherwise we choose σ =. In both cases we obtain u q 2C pqs u s + u κ/γ+κ) s ) u γ/γ+κ) p. 0.5) We have κ/γ + κ) = ϱ, γ/γ + κ) = ϱ, and the desired result follows from Theorem 3.. It is in principle possible to derive from 9.4) the corresponding interpolation inequalities also for anisotropic spaces. The general formulas then become, however, rather complicated and we omit them here.

26 embed2.tex, January, eferences []. A. Adams, Sobolev spaces, Academic Press, New ork, 975. [2] O. V. Besov, V. P. Il in, S. M. Nikol skiĭ, Integral representations of functions and imbedding theorems Vol. I), Scripta Series in Mathematics, Halsted Press [John Wiley & Sons], New ork-toronto, Ont.-London, 978. [3] O. V. Besov, V. P. Il in, S. M. Nikol skiĭ, Integral representations of functions and imbedding theorems Vol. II), Scripta Series in Mathematics, Halsted Press [John Wiley & Sons], New ork-toronto, Ont.-London, 979. [4] J. Nečas, Les méthodes directes en théorie des équations elliptiques, Academia, Prague, 967.