Sobolevology. 1. Definitions and Notation. When α = 1 this seminorm is the same as the Lipschitz constant of the function f. 2.

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1 Sobolevology 1. Definitions and Notation 1.1. The domain. is an open subset of R n Hölder seminorm. For α (, 1] the Hölder seminorm of exponent α of a function is given by f(x) f(y) [f] α = sup x y x y α. When α = 1 this seminorm is the same as the Lipschitz constant of the function f. 2. Inequalities 2.1. Poincaré. If ϕ Cc () and if has width L (i.e. (, L) R n 1 ) then ϕ(x) 2 dx C P () ϕ 2 dx where the Poincaré constant C P () is bounded by C P () 1 L Sobolev (p = 1). For any f Cc 1 (R n ) one has ( ) (n 1)/n f n/(n 1) dx Cn f dx. R n R n The constant C n Sobolev (1 p < n). For any f Cc 1 (R n ) and any 1 p < n one has ( ) (n p)/np ( ) 1/p. f np/(n p) dx Cn,p f p dx R n R n (See problems for the constant C n,p ) 2.4. Morrey. For any f Cc 1 (R n ) and any p > n one has ( 1/p. [f] 1 n/p C n,p f dx) p R n 1

2 3. Spaces 3.1. The space W 1,p (). For 1 p the space W 1,p () consists of all f L p () for which the distributional partial derivatives f x i belong to L p (). The norm on W 1,p () is ( ) 1/p f W 1,p = { f p + f p } dx The space W 1,p () is by definition the closure in W 1,p () of C c () H 1 () and H 1 (). When p = 2 the following notation is commonly used W 1,2 () = H 1 (), W 1,2 () = H 1 (). These spaces are Hilbert spaces, and there are various inner products on them which define equivalent norms. On H 1 () one defines { } (u, v) H 1 = uv + u v dx. This expression also defines an inner product on the subspace H 1 (). The following expression defines an inner product on H 1() but not on H1 (): (u, v) H 1 = u v dx. The corresponding norm on H 1 () is u 2 H 1 () = u(x) 2 dx. It follows from the Poincaré inequality that this norm is equivalent with the H 1 norm defined above H 1 (). The space of all distributions on which can be written as with f 1,..., f n L 2 () is defined to be H 1 (). The quantity defines a norm on H 1 (). g = f = f 1 x f n x n g 2 H 1 = inf g= f f 2 dx 3.4. H 1 () is the dual of H 1 (). Let g be any distribution on. Then the functional ϕ g, ϕ extends to a bounded linear functional on H 1 () if and only if g H 1 (). Furthermore, the H 1 norm of g defined in 3.3 is also given by g H 1 = sup g, v. v H 1 1 2

3 3.5. Theorem. If f L 2 () then v f(x)v(x) dx defines a bounded linear functional on H 1 (). Identifying the function f with the linear functional it defines, we may think of f as an element of H 1 (). One has where C P () is the Poincaré constant of. f H 1 C P () f L 2, Proof. Use the Poincaré inequality, which says that for any v H 1 () one has v L 2 C P v H 1, to bound fvdx as follows: fvdx f L 2 v L 2 C P f L 2 v H Solving the Poisson equation 4.1. The Poisson equation and boundary value problem. Poisson s equation is u = f, u =. A function u H 1 () is a weak solution, or a solution in the sense of distributions if v Cc () u v dx = fv dx Theorem. Let be bounded. Then for any f L 2 () there is a unique solution of u = f, This is a special case of the following theorem. u H 1 () Theorem. Let be bounded. Then for any g H 1 () there is a unique solution of u = g, u H 1 (). Proof. The definitions imply that u H 1 () is a weak solution iff (1) v H 1 (u, v) H 1 = g, v QED The right-hand side defines a bounded linear functional on H 1 (), because, by definition g, v g H 1 v for all v H 1 H1. Every bounded linear functional on a Hilbert space such as H1 () is of the form v (u, v) H 1 for one and only one u H 1. QED 4.4. Theorem. If u H 1 () is the solution to u = g with g H 1 (), then u H 1 g H 1. Proof. u satisfies (1) for all v H 1. Choose v = u and you get u 2 H = (u, u) 1 H 1 = g, u g H 1 u H 1. Cancel u H 1 s left and right. QED 3

4 5. Problems 5.1. Membership in Sobolev spaces. Let B be the open unit ball, and define for each a > the function f a (x) = x a. For which values of a > does one have f a W 1,p (B)? 5.2. Bad (nowhere continuous) Sobolev functions. Let 1 p < n, and let B be the open unit ball in R n. Let {x i B} i N be a dense sequence of points. Find numbers b i > (i N) and a > so that the series f(x) = i=1 b i x x i a converges in W 1,p (B). Background to the next three problems. The Sobolev inequality says for any p < n that f L p () implies f L np/(n p) (). If you let p n in this statement you would get: f L n = f L. This turns out not to be true however (for a counterexample, if you insist, try functions of the form f(x) = log x α for appropriate α >.) 5.3. Sobolev constant. Derive the Sobolev inequality with 1 < p < n from the Sobolev inequality with p = 1, and find an explicit upper bound for C n,p (assuming that C n 1) Sobolev inequality when f L n (). When R n and f n dx < neither the Sobolev inequality nor the Morrey inequality apply. Prove that for any f W 1,n () and any p < one has f L p () C( f n dx) 1/n. The constant depends on p and (you must assume that has finite volume). (Hint: if f L n and < then Hölder s inequality implies that f L r () for any r [1, n). Pick the right r for the given p, and use C n,p from the previous problem.) 5.5. Sobolev inequality when f L n (), the sequel. Use the result from the previous problem to show that if f L n () for some bounded domain R n, then there is a constant c > such that e c f(x) dx <. Hint: use the Taylor expansion e c f = k= ck f k /k! and estimate the integral of the terms in this expansion using the previous problem Membership in H 1 (). In 3.5 we showed that L 2 () H 1 () for any bounded open domain R n. A. For which q [1, ] does one have L q () H 1 ()? (The answer will depend on n; you need the Sobolev inequalities to find the appropriate range of q; the cases n = 1 and n = 2 are a bit different from the case n > 3.) B. Suppose the domain contains the origin. For which a > does the boundary value problem u = x a, u H 1 () have a solution? (hint: if x a belongs to H 1 () then Theorem 4.3 applies. Use problem 5.1). 4

5 5.7. Poincaré from Sobolev. Let R n be an open subset with finite volume. Use the Sobolev inequality to prove that Poincaré s inequality holds for : u Cc () u 2 dx C 2/n u 2 dx where is the volume of. 5