Homework I, Solutions

Transcription

1 Homework I, Solutions I: (15 points) Exercise on lower semi-continuity: Let X be a normed space and f : X R be a function. We say that f is lower semi - continuous at x 0 if for every ε > 0 there exists δ > 0 so that f(x) f(x 0 ) > ε (1) whenever x 0 x < δ. We also say that f is lower semi-continuous if f is lower semicontinuous at every point of X. a) Prove that f is lower semi-continuous at x 0 if and only if for every sequence x n with lim x n = x 0, it follows that lim inf f(x n ) f(x 0 ). Assume that f is lower semi-continuous. If x n x 0 0 as n then for any given ε there exists N so that x n x 0 < δ for all n > N. Hence if f is lower semi-continuous, f(x n ) > f(x 0 ) ε for all n > N and hence lim inf f(x n) f(x 0 ) ε. Since ε is arbitrary the result follows. For the converse, assume that there is an x 0 at which f is not lower semi-continuous. This means that there exists ε > 0 so that for any δ > 0, there exists x with x x 0 < δ such that f(x) f(x 0 ) ε. This means that there is a sequence x n converging to x 0 so that for all n = 1, 2,.... In particular A contradiction. f(x n ) f(x 0 ) ε f(x 0 lim inf f(x n) f(x 0 ε. b) Prove that f is lower semi-continuous if and only if the set {x X : f(x) > t} is open for every value of t. Assume that f is lower semi-continuous. Pick any x 0 {x X : f(x) > t}. Since f(x 0 ) > t, we also have that f(x 0 ) > t + ε for ε sufficiently small. By assumption there exists δ > 0 so that f(x) > f(x 0 ) ε > t for all x with x x 0 < δ. Hence {x X : f(x) > t} is open. To see the converse, assume that {x X : f(x) > t} is open for all t R. Suppose that there exists x 0 at which f is not lower semi-continuous. Then there exists ε > 0 so that for all δ > 0 there exists x with x x 0 < δ such that f(x) f(x 0 ) ε. Pick t = f(x 0 ) ε/2. Then we there exists a sequence x n converging to x 0 so that f(x n ) f(x 0 ) ε = t ε/2 1

2 2 which contradicts the fact that {x X : f(x) > t} is open. II: (20 points) The space H 1 (): Let R n be an open set and consider L 2 (, dx). Denote by Cc () the set of infinitely differentiable functions on that have compact support. Note that, by definition, a continuous function has compact support if the closure of the set where f does not vanish is compact. Here are some facts about L 2 (, dx) and Cc (): As you know L 2 (, dx) is a Hilbert space with inner product (f, h) = f(x)h(x)dx. Another useful fact is that Cc () is dense in L 2 (, dx). We define the space H 1 () to consist of all functions f L 2 (, dx) with the property that there exist functions gf i L2 (, dx), i = 1... n such that f(x) φ (x)dx = g x f(x)φ(x)dx i, i = 1...., n. i The expression (f, h) 1 = R f(x)h(x)dx + defines obviously an inner product on H 1 (). n i=1 R g i f(x)g i h (x)dx a) Prove that the gradient gf i, i = 1,..., n is unique. Suppose that there exists another gradient h i f, i = 1,..., n. Then we have that (gf i h i f)φdx = 0, i = 1,..., n for all φ C c (). Fix the index i and set u i := g i f hi f. The fact that C c () is dense in L 2 (, dx) means that for any ε > 0 there exists φ i so that u i φ i < ε. We shall use the sign to denote the norm in L 2 (). Hence 0 = u i φ i dx = u i (φ i u i )dx + u i 2 so that u i 2 = u i (φ i u i )dx u i u i φ i by Schwarz s inequality. Thus, if u i 0 we have that which proves the claim since ε is arbitrary. u i u i φ i < ε b) Prove that H 1 () is a Hilbert space.

3 Let f n be a Cauchy sequence in H 1 (). This means that n f n f m 2 + gf i n gf i m 2 0 i=1 as n, m. This means that f n converges to f in L 2 () and gf i n converges to some g i in L 2 (). It remains to show that f φ dx = g i φdx. x i To see this note that φ f n dx f φ dx x i x i = [f n f] φ dx x i f f n φ 0 x i as n. Hence f φ φ dx = lim f n dx = lim gf x i x i i n φdx = where we used that gf i n φdx g i φdx gf i n g i φ. g i φdx, 3 III: (10 points) On L 2 (R) consider the sequences f j (x) = f(x j) and g j (x) = j 1/2 g(jx) where f, g are fixed functions in L 2 (R). Show that both sequences converge weakly to zero as j. The idea is that for a function h L 2 (R) the sequence (f j, h) tends to zero since the overlap between f j and h tends to disappear. The problem is that the functions under consideration are only in L 2 (R) and they do not have compact support. To really prove this one argues via approximations. Pick any ε > 0. There exists φ, ψ Cc (R) so that f ψ < ε and h φ < ε. If we set ψ j (x) = ψ(x j) we have that f j ψ j = f ψ by changing variables in the integration. Hence (f j, h) = (f j ψ j, h) + (ψ j, h φ) + (ψ j, φ) f j ψ j h + ψ j h φ + (ψ j, φ) ε( h + ψ ) + (ψ j, φ) For j large enough (ψ j, φ) = 0 since the two functions φ and ψ have compact support and then the supports of φ and ψ(x j) do not overlap. Thus for j sufficiently large (f j, h) ε( h + ψ ) and since ε is arbitrary this means that lim j (f j, h) = 0. For the second problem we proceed exactly the same way except that we set ψ j (x) = j 1/2 ψ(jx) so that g j ψ j = g ψ. Hence Now (ψ j, φ) j 1/2 (g j, h) ε( h + ψ ) + (ψ j, φ) ψ(jx) φ(x) dx Cj 1/2 where C = max x φ(x). The rest follows as before. ψ(jx) dx = Cj 1/2 ψ(x) dx

4 4 IV (15 points) : Let f j, g j be any two strongly convergent sequences in an arbitrary infinite dimensional Hilbert space H and h j, k j any two weakly convergent sequences in H. Prove or find a counterexample: a) The sequence (f j, g j ) is always convergent. Yes. Assume that f j resp. g j converge strongly to f resp. g. Then (f j, g j ) (f, g) = (f j f, g j ) + (f, g j g) f f j g + g g j f j which converges to zero since f j stays bounded. b) The sequence (f j, h j ) is always convergent. Yes. Suppose that h j converges weakly to h. Then (f j, h j ) (f, h) = (f j f, h j ) + (f, h j h) f f j h j + (f, h j h) 0 as n since h j is bounded by the uniform boundedness principle. c) The sequence (h j, k j ) is always convergent. No. Take h j to be an orthonormal sequence, (h j, h k ) = δ j,k and define k j = h j for j even and k j = 0 for j odd. Both sequences converge weakly to zero. But (h j, k j ) is a sequence that alternates between 0 and 1 and hence does not converge. Here (, ) denotes the inner product in H. V: Extra credit: This exercise is difficult. Let X be a complete normed space (which we assume for simplicity to be real) and assume that the norm satisfies the parallelogram identity, i.e., x + y 2 + x y 2 = 2 x y 2 for all x, y X. Prove that X is a Hilbert space, i.e., there exists an inner product (x, y) such that x = (x, x). Solution: We define f(x, y) = 1 2 ( x + y 2 x 2 y 2 ) First we show that f(x, y) is additive in each variable. Clearly f is symmetric and hence it suffices to do this for the first variable. We have to show that 0 = f(x + y, z) f(x, z) f(y, z) = 1 2 ( x + y + z 2 x + y 2 x + z 2 y + z 2 + x 2 + y 2 + z 2 Now set a = x + y, b = x + z 2 2 Using this notation we have to show that, c = y + z 2 a + b + c 2 + a + b c 2 + a + c b 2 + b + c a 2 = 4 a b c 2..

5 Now using the parallelogram identity and a + b + c 2 + a + b c 2 = 2 a + b c 2 a + c b 2 + b + c a 2 = c + (a b) 2 + c (a b) 2 = 2 c a b 2, and 2 a + b c c a b 2 = 4 a b c 2. It remains to show that f(x, y) is homogeneous in each variable. By symmetry it suffices to show this for the first variable. If p is any integer we have by induction that and for q 0 an integer f(px, y) = pf(x, y) qf( x q, y) = f(x, y) or f(x q, y) = 1 f(x, y) q so that f(rx, y) = rf(x, y) for any rational number r. The function x f(x, y) is continuous. To see this note that f(x 1, y) f(x 2, y) = 1 x1 + y 2 x 1 2 x 2 + y 2 + x = 1 2 ( x 1 + y + x 2 + y )( x 1 + y x 2 + y ) 1 2 ( x 1 + x 2 )( x 2 x 1 ) ( x 1 + x 2 + y ) x 2 x 1 by the triangle inequality. For c real, pick a sequence r n of rational numbers that converge to c. Clearly cf(x, y) = lim r n f(x, y) = lim f(r n x, y) = f(cx, y). Hence, f(x, y) satisfies the conditions of an inner product. 5