(d). Why does this imply that there is no bounded extension operator E : W 1,1 (U) W 1,1 (R n )? Proof. 2 k 1. a k 1 a k

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1 Exercise For k 0,,... let k be the rectangle in the plane (2 k, 0 + ((0, (0, and for k, 2,... let [3, 2 k ] (0, ε k. Thus is a passage connecting the room k to the room k. Let ( k0 k (. We assume ε k < 2 k so is open and connected. Given a sequence of constants {a k } k0, define a function f so that f(x, y a k in k, f is continuous, and f is linear in the passages. (a. Find the weak derivative of f. (b. Fix an exponent p < and show that for any sequence {a k } we may find a sequence {ε k } so that the weak gradient Df lies in L p (. (c. Given < q, find a domain and a function f so that f, Df L (, but f is not in L q (. (d. Why does this imply that there is no bounded extension operator E : W, ( W, ( n? Proof. (a. On the function f is given by f(x, y a k a k (x 3 + a k. Let F y (x, y 0 and let F x be defined as { ak a k, (x, y P 2 F x (x, y k k 0, (x, y k. Now, assume {a k } and {ε k } are chosen so that f, F x L loc (. We will show that the weak derivatives of f are D x f F x and D y f F y. First, the function f is independent of the variable y, so its standard derivative (which coincides with its weak derivative in the y-direction is D y f 0 F y. To show D x f F x, we turn to the definition of weak derivative. We want to demonstrate that for any test function ϕ Cc (, ϕf x dy D x ϕfdy. ( Since ϕ has compact support in, there is a largest k such that supp ϕ, call it K. Then we can write the left-hand side of ( as ϕf x dy ϕdy. Page of 5

2 Similarly, there is a largest k such that supp ϕ k, call it K 2, so the right-hand side of ( can be written D x ϕfdy K 2 D x ϕ (x 3 dy + a k D x ϕdy. k Now, (x 3 is smooth on, so we integrate the terms in the first sum by parts to obtain D x ϕ (x 3 dy k0 ϕdy. By another round of integration by parts, the terms in the second sum evaluate to D x ϕdy k D x (D y ϕydy k D y ϕd x ydy 0. k Hence, the left- and right-hand sides of ( are equal, so D x f F x in the weak sense. (b. Since D y f 0, it suffices to show that a sequence {ε k } exists so that D x f L p (. By the work done in part (a, we know Define {ε k } { D x f p dy p m(p 2 p( k k p ( (ε 2 p( k k p 2 ( p( k ε k. } k 2 a k a k p 2 ( p( k π 2 6 <, putting D xf L p (.. Then the above sum reduces to k 2 (c. Let {x 2 x < } be the open disk in the plane, and let f(x x q. According to the proof given on Page 260 of Evans, the function u(x x α is in W,p (, i.e., u and its gradient Du are in L p (, if and only if α < n p. Given p α, n 2 and p, we have < 2, so f and Df are in q q L (. However, f x, so f / L q (. (d. I don t think this implies that no such bounded extension operator exists. In fact, since the boundary of in (c is C, there is a bounded extension operator E : W, ( W, ( n. Page 2 of 5

3 Exercise 2 Consider functions u(x, t on n+. For which exponents p and q can the following inequality hold? u L q ( n+ C Dx α u L p ( n+ + D t u L p ( n+. α 2 Proof. Suppose the inequality holds for some p and q. Choose a function u(x, t C 2, c ( n+ such that the inequality holds, and for λ > 0 let u λ (x, t : u(λx, λ 2 t. It follows. Now, on the left-hand ( that u λ L q ( n+ C α 2 Dα x u λ L p ( n+ + D t u λ L p ( n+ side of this inequality we have u λ q L q ( n+ u λ q dt n+ ( u(λx, λ 2 t q dt n ( u(x, t q dt λ 2 λ n n λ n+2 u q L q ( n+. On the right-hand side we have two rescales. First, for any subindex α such that α 2, Similarly, D α x u λ p L p ( n+ n+ D α x u λ p dt λ 2p n+ D α x u(λx, λ 2 t p dt λ 2p (n+2 D α x u p L p ( n+. D t u λ p L p ( n+ n+ D t u λ p dt λ 2p n+ D t u(λx, λ 2 t p dt λ 2p (n+2 D t u p L p ( n+. Inserting all of these equalities into the inequality above gives us λ n+2 q u L q ( n+ Cλ 2p (n+2 p Dx α u L p ( n+ + D t u L p ( n+, or equivalently, If 2 n+2 + n+2 p q Hence, 2 n+2 + n+2 p q Page 3 of 5 n+2 2 u L q ( n+ Cλ p + n+2 q α 2 Dx α u L p ( n+ + D t u L p ( n+. α 2 0, we can let λ be close enough to 0 or to obtain a contradiction. 0, i.e., q p(n+2 n+2 2p.

4 Evans, Page 307 Exercise 9 Integrate by parts to prove the interpolation inequality: Du L 2 C u /2 L D 2 u /2 2 L 2 for all u C c (. Assume is bounded, is smooth, and prove this inequality if u H 2 ( H 0(. Proof. For u Cc (, we have Du and D 2 u defined a.e., so we may assert Du 2 L Du 2 Du Du 2 n n u u n u u n u u ( n ( u 2 2 u 2 2 n n ( ( u 2 2 n max u xi x i 2 2 n i n ( n ( n u 2 2 max u xi x j 2 2 n i,j n ( n ( n u 2 2 D 2 u 2 2 n n n u L 2 D 2 u L 2. Taking the square root of both sides gives the desired result, where C n. Now, suppose u H 2 ( H 0( for some bounded open set with smooth boundary. Then we can choose a sequence {v k } C c ( converging to u in H0( and a sequence {w k } C ( converging to u in H 2 (. I am not sure why we need both sequences here; since {w k }, {Dw k } and {D 2 w k } converge to u, Du and D 2 u respectively in L 2, and the inequality holds for each w k, as k we should end up with which is precisely what we wanted to show. Du L 2 C u 2 L 2 D 2 u 2 L 2, Page 4 of 5

5 Exercise Suppose is connected and u W,p ( satisfies Du 0 a.e. in. Prove that u is constant a.e. in. Proof. The condition Du 0 a.e. in implies ud x i ϕ 0 for each i,..., n and any ϕ C c (. Fix η C c ( such that η. Then we can write ϕ(x Φη + D x i ψ(x, where Φ ϕ and ψ(x xi a [ϕ(x,..., x i, t, x i+,..., x n Φη(x,..., t,..., x n ] dt for some a such that, WLOG, the segment {x }... {x i } [a, x i ] {x i+ }... {x n }. Such a point a is guaranteed to exist since is connected. Note that ψ C c this we may deduce ϕf (Φη + D xi ψf Φ ηf + D xi ψf ( ηf ϕ. (. From Since ηf is constant (call it C, we have ϕ(f C 0 for any ϕ C c can conclude that f C almost everywhere in. (. Thus, we Page 5 of 5