l(y j ) = 0 for all y j (1)

Transcription

1 Problem 1. The closed linear span of a subset {y j } of a normed vector space is defined as the intersection of all closed subspaces containing all y j and thus the smallest such subspace. 1 Show that the closed linear span of {y j } is the closure of the linear span Y of {y j }, which consists of all finite linear combinations of the y j. Prove that a point z of a normed linear space belongs to the closed linear span S of a subset {y j } of if and only if every bounded linear functional l that vanishes on the subset vanishes at z. That is, if and only if implies that lz = 0. Hint: For, use the Hahn-Banach theorem. Proof. ly j = 0 for all y j 1 1 Since S contains all y j, we have that Y S. Since S is closed, we obtain Y S. To prove the converse, note that the closure Y of the subspace Y is again a subspace. So Y is a closed subspace containing all y j. But S is precisely the intersection of all such subspaces, and therefore S Y. Since l is linear, assumption 1 implies that ly = 0 for all y Y. For all z S there exists a sequence of y n Y with the property that y n z 0, as a consequence of 1. Since l is continuous, this implies that lz = lim ly n = 0. Conversely, suppose that z S, such that δ := inf z y > 0. y S Consider the subspace Z of all points of the form y + αz, with y S and α R assuming for simplicity that is a real vector space. Define a linear functional l 0 : Z R by l 0 y + αz := α. It follows from that y + αz = α z y/α α δ, so that l 0 y + αz δ 1 y + αz for all y S and α R. By the Hahn-Banach theorem, then l 0 can be extended to a functional on all of, which we denote by l. By construction, we have ly j = 0 for all y j, but lz = δ 1 = 0.

2 Problem. A subspace Y in a normed real vector space is called finite-dimensional if there exists a number n N the dimension of Y and vectors {y 1,..., y n } in Y with the property that every element x Y has a unique representation of the form x = α 1 y α n y n for suitable α i R with i = 1,..., n. 3 Prove that every finite-dimensional subspace of a normed real vector space is closed. Hint: Reduce the problem to the fact that R n is complete. Proof. If Y = {0}, then there is nothing to prove. Assume therefore that n 1. Consider the map c: Y R n defined by cα 1 y α n y n := α 1,..., α n. This map is an isomorphism because the representation 3 is unique. Then there exist constants C 1, C > 0 such that n n n C 1 α i y i α i C α i y i. 4 In fact, on the one hand we can use the triangle inequality to estimate n n α i y i α i y i, so C 1 := max i y i 1 will work for the first inequality in 4. Note that we have y i = 0 for all i since otherwise the representation 3 would not be unique. On the other hand, the map α 1,..., α n α 1 y α n y n is continuous because n n n n α i y i β i y i = α 1 β i y i C1 1 α i β i. Therefore we obtain δ := inf { n α i y i : } n α i = 1 > 0. In fact, the inf of a continuous function over a compact set is attained, and the inf cannot be zero because that would contradict that y i = 0 for all i. This implies 4 with C := δ 1. We have therefore shown that a sequence {x k } in Y converges in the norm if and only if the sequence of coefficients {cx k } converges in the l 1 -norm in R n. Therefore closedness of Y is equivalent to completeness of R n with respect to the l 1 -norm. The latter fact is well known.

3 Problem 3. Let, M, μ be a finite measure space. 1 Prove that the map da, B := 1 A 1 B dμ defined for all A, B M with 1 A the characteristic function of A is a pseudo-distance on the σ-algebra M. That is, the map d satisfies all properties of a distance, except that da, B = 0 does not imply A = B. Prove that the pseudo-metric space M, d is complete. Proof. 1 We have that 0 da, B 1 A + 1 B dμ μa + μb < for all A, B M. The symmetry da, B = db, A is clear from the definition. To prove the triangle inequality, note that da, B = μa B + μb A. One can then check that A C = A C c = A B c C c A B C c = A B C C A = C A c = C A c B c B C A c = C A B A B C, B C A, and the union is disjoint. Similar formulas hold for A and B interchanged. We find that μa C + μc B = μ A B C + μ A B C + μ C A B + μ A C B = μa B + μ A B C + μ C A B because A B C A C B = A B c C c A C B c = A B c = A B, and the union is disjoint. A similar formula holds with A and B interchanged. We conclude that da, C + db, C = μa C + μc A + μb C + μc B = μa B + μb A + μ A B C + μ C A B da, B. Consider a sequence of sets {A n } in M that is Cauchy with respect to d. Then the sequence {1 A n} is Cauchy with respect to the L 1, μ-norm, and thus 1 A n f for some f L 1, μ. Recall that strong convergence implies convergence in measure. We define a map φs := s 1 s for all s R, which vanishes exactly for s = 0 or s = 1. Then we have that φ1 A n φf in measure. By dominated convergence note that 0 1 A nx 1 for all x and n N, we get φf dμ = lim φ1 A n dμ = 0, and thus fx {0, 1} for μ-a.e. x. This implies that f = 1 A for some A M. Then da n, A = 1 A n 1 A dμ 0 as n, and so A, d is complete.

4 Problem 4. Let, M, μ be a measure space and consider a sequence of measurable functions f n : R. Assume that there exists a function g L 1, μ with the property that f n g for all n N. Prove that then lim inf f n dμ lim inf f n dμ lim sup f n dμ lim sup f n dμ. 5 Give an example showing that this chain of inequalities may be no longer true if there exists no dominating function g as above. Proof. Consider the sequence of functions h n := f n g. By assumption, we have that h n 0 for all n N, and so by linearity of the integral and Fatou s lemma we obtain lim inf f n dμ g dμ = lim inf hn dμ lim inf h n dμ = lim inf f n dμ g dμ. Since g L 1, μ, the last integral is finite, so we can add it on both sides to get lim inf f n dμ lim inf f n dμ. The lim inf is bounded above by the lim sup, which proves the second inequality in 5. To prove the last inequality, we proceed in a similar fashion. Consider the functions k n := f n +g. By assumption, we have k n 0 for all n N, and so by Fatou s lemma we obtain lim inf f n dμ + g dμ = lim inf kn dμ lim inf k n dμ = lim inf f n dμ + g dμ. But now lim inf an = lim sup a n for any sequence {a n }. Therefore lim sup f n dμ lim sup f n dμ. which proves the result. We used again that the integral over g is finite. To prove that the dominated integrability is needed, consider the case = R, with μ equal to the Lebesgue measure. Consider the functions f n := n1 0,1/n for all n N. Then we have and so On the other hand, we have that lim inf f n = lim sup f n = 0 lim inf f n dμ = μ-a.e., lim sup f n dμ = 0. f n dμ = 1 for all n N. There exists no dominating function g since such a function would have to behave like 1/x as x 0, which is not integrable.

5 Problem 5. Let e denote the exterior outer Lebesgue measure on R n and let Br, x denote the open ball of radius r about x R n. For E R n we define outer density D E x at x by D E x = lim r 0 E Br, x e Br, x e, whenever the limit exists. 1 Show that D E x = 1 for a.e. x E. Show that E is Lebesgue measurable if and only if D E x = 0 for a.e. x E c. Proof. Notice first that if E is Lebesgue measurable then the function χ E x is locally integrable and therefore, by the Lebesgue differentiation theorem, D E x = χ E x for a.e. x R n. Thus { 1 for a.e. x E D E x = 0 for a.e. x R n E. To prove 1 for arbitrary set E we use the fact that there exists a measurable set U such that E U and for every measurable set M we have E M e = U M. Although the construction of U is more or less standard we sketch it below. Suppose first E e <. For every n N there exists an open set G n E with G n E e < 1/n. If U = n G n E, then U = E e < and therefore we can assume that M <. Note that E M e M M E M e M M U M = U M, which shows that holds since E M e U M. For arbitrary E we can write E = k E k where E k = E B0, k has a finite exterior measure. Hence, for every k there exists a measurable set U k E k such that E k M e = U k M. Then U = lim inf U n = n=1 k n U k E. If H k = j k U k U k we have E k H k U k which shows that E k M e = H k M. Letting k in the last equality we obtain. From and the definition of D E x it follows that D E x = D U x for every x R n and since D U x = 1 for a.e. x U we see that D E x = 1 for a.e. x E. To prove it remains to show that if D E x = 0 for a.e. x R n E then E is measurable. If we assume that E is not measurable and take U as above, then U E e > 0 and we get a contradiction since D E x = 1 for a.e. x U.

6 Problem 6. 1 Let, Y, Z be Banach spaces and let B : Y Z be a separately continuous bilinear map, that is, Bx, LY, Z for each fixed x and B, y L, Z for each fixed y Y. Prove that B is jointly continuous, that is, continuous from Y to Z. Is there a nonlinear function f : R R R, which is separately continuous, but not jointly continuous? Proof. 1 Denote B x = Bx, : Y Z and B y = B, y : Z. Then for every x we have B y x = Bx, y = B x y B x y, which shows that sup B y x B x. y =1 By the uniform boundedness principle we conclude that C = sup B y <. y =1 For a nonzero y Y we put y = 1 y y and we see that Bx, y = y Bx, y = y B y x C x y, i.e. Bx, y C x y. Clearly this inequality is also true when y = 0. The continuity of B now follows immediately since Bx, y Bx 0, y 0 Bx, y y 0 + Bx x 0, y 0 C x y y 0 + x x 0 y 0. Yes. The function fx, y = { xy x +y if x, y = 0, 0 0 if x, y = 0, 0 is continuous in each variable separately, but is not continuous at the origin.

7 Problem 7. Let be a real normed space. Prove that the norm is induced by an inner product if and only if the norm satisfies the parallelogram law, i.e. x + y + x y = x + y for every x, y. Proof. If is an inner product space then x + y = x + y, x + y = x + y + x, y, and likewise x y = x y, x y = x + y x, y. Adding these equalities we obtain. Conversely, suppose that holds. We want to show that the map x, y x, y from R defined by x, y = 1 x + y x y 6 is an inner product on. Clearly x, y = y, x and x, x = x 0 with equality if and if x = 0. Thus, it remains to prove that x, y is linear in x. Using the definition 6 we see that x+y, z x, z y, z = x+y+z + z x+z + y+z + x + y x+y. 7 On the other hand, from we see that x + y + z + z = x + y + z + x + y x + z + y + z = x + y + z + x y. Substituting these in the right-hand side of 7 we find proving that x + y, z x, z x, y = x + y It remains to show that x + y x y = 0, x + y, z = x, z + x, y for every x, y. 8 αx, y = α x, y for every x, y and α R. 9 From 8 it follows by induction that 9 holds when α is a positive integer. Replacing x by 1 α x we see that if 9 holds for some α = 0, then it is true also for 1/α. Hence 9 holds for all positive rational numbers α. The case α = 1 follows easily from thus proving 9 for all α Q. We can prove now 9 by a limiting procedure if we can show that x, y is a continuous function of x for every y fixed. From the triangle inequality we see that x, y 1 x + y x y = x y, and therefore x, y = ±x, y x y. For arbitrary α R we consider a sequence of rational numbers {r n } such that r n α. Note that lim r n x, y = αx, y, because r n x, y αx, y = r n αx, y r n α x y 0 as n. The proof of 9 now follows by letting n in r n x, y = r n x, y.

8 Problem 8. Let, M, μ be a finite measure space and let {f n } be a sequence in L p where 1 < p < such that sup n f n p <. Show that if f n 0 a.e., then f n 0 weakly in L p. Proof. Let q be the conjugate exponent to p. Since L q is the dual of L p, we must show that fn g dμ 0 for every g L q. Let M = sup n f n p <. Fix ε > 0. Since dν = g q dμ is a finite measure on, M that is absolutely continuous with respect to μ, there exists δ > 0 such that 1/q if E M and μe < δ, then g dμ q < ε. On the other hand, μ <, f n 0 a.e. and therefore by Egoroff s theorem there exists E M such that μe < δ and f n 0 uniformly on E. Thus, there is some N such that for n N we have f n x < ε E for every x E. Using the above and Hölder s inequality, we see that for n N we have f n g dμ f n g dμ + f n g dμ completing the proof. E f n p E E g q dμ 1/q + M + μ 1/p g q ε, E f n p dμ 1/p g q