Math 172 Problem Set 5 Solutions

Transcription

1 Math 172 Problem Set 5 Solutions 2.4 Let E = {(t, x : < x b, x t b}. To prove integrability of g, first observe that b b b f(t b b g(x dx = dt t dx f(t t dtdx. x Next note that f(t/t χ E is a measurable function on 2 since the pull-back of the function t f(t/t to a function F (t, x = f(t/t is measurable by Corollary in the text. Hence Tonelli s theorem (Theorem implies that this last integral is equal to b t f(t t b dxdt = f(t t x b tdt = f(t dt <. This proves that g is integrable, and a nearly identical argument shows that b g(xdx = b f(tdt Consider the set E = {(x, y d+1 : y < f(x }. To show that this set is measurable, consider an increasing sequence of simple functions s i which converge to f pointwise (thus in L 1 by the dominated convergecne theorem. One easily checks that the set of points under the graph of a simple function is a measurable subset of d+1. Letting U i be the set of points under graph of s i, we have that E = i U i so that E is measurable. Moreover, since the U i are nested, we have by monotonicity that m(u i = s i m(e = f. Now the proof of the assertion is a straightforward application of Corollary 2.3.3, for we have E y = {x d : (x, y E} = {x d : f(x > y }. 1

2 Therefore, the corollary tells us that f = m(e = m(e y dy = since E y = for y <. m(e α dα 2.21a The function (x, y f(x y is measurable by Proposition in the text, and the function (x, y g(y is also measurable by Propsition in the text. Hence, as the product of measurable functions, F : (x, y f(x yg(y is also measurable on 2d. 2.21b To show that the integrability of f and g imply that F is integrable, we have ( F = f(x yg(y dxdy = f(x y g(y dx dy 2d d ( ( d ( = g(y f(x y dx dy = g(y dy f(x dx < d d d d where we have used Tonelli s theorem in the second equality and the translational invariance of the Lebesgue integral in the final equality. We thus conclude that F is integrable on 2d. 2.21c Assuming that f and g are integrable, we know from part b above that F is integrable as well. Hence a direct application of Fubini s theorem implies that F x is integrable for a.e. x, which means that the convolution (f g(x = f(x yg(ydy = d F x (ydy d is defined for a.e. x. 2.21d We have f g L 1 = f g = d d f(x yg(ydy dx f(x yg(y dydx d 2d where we have used Tonelli s theorem and the triangle inequality. By part b, we know that this last integral is equal to f L 1 g L 1. Clearly the can be changed to an = if both f and g are non-negative. 2.21e Since f(xe 2πix ξ = f(x, we have that ˆf(ξ = f(xe 2πix ξ dx f(x dx = f L 1, d d which proves that ˆf is bounded. Moreover, as in the proof of Proposition we observe that since e 2πix ξ is continuous in ξ for every x, if ξ n ξ we have that 2

3 f(xe 2πix ξn f(xe 2πix ξ. But since each function f n (x := f(xe 2πix ξn is dominated by the integrable function f, the dominated convergence theorem tells us that ˆf(ξ n = f(xe 2πix ξn dx d f(xe 2πix ξ dx = ˆf(ξ, d which proves that ˆf is continuous. Finally, we prove the identity (f g(ξ = ˆf(ξĝ(ξ. We have ( (f g(ξ = (f g(xe 2πix ξ dx = f(x yg(ydy e 2πix ξ dx d d ( d = f(x ye 2πiy ξ g(ye 2πiy ξ dy e 2πix ξ dx d ( d = f(x ye 2πi(x y ξ dx g(ye 2πiy ξ dy d d = ˆf(ξg(ye 2πiy ξ dy = ˆf(ξĝ(ξ. d where we have used part b and Fubini s theorem in the fourth inequality. Problem 2.4a As the problem suggests, consider the upper triangular matrix ( 1 a L =. 1 Clearly we have L 1 = ( 1 a 1 and so χ L(E (x, y = χ E (L 1 (x, y = χ E (x ay, y as claimed. Now since L(E is measurable (by exercise 1.8, the function χ L(E is measurable as well, and so we have ( m(l(e = χ L(E = χ L(E (x, ydx dy ( = χ E (x ay, ydx dy where we have used Tonelli s theorem to obtain the second equality. Assume for now that m(e <. Then the function χ E is integrable, and hence Fubini s theorem tells us that χ y E is integrable for almost every y. We may thus use the translational invariance of the Lebesgue integral to express the final integral above as ( χ E (x, ydx dy = χ E = m(e,, 3

4 as claimed in the problem. This argument extends easily to an upper (lower unitriangular d d matrix. In particular, suppose L us upper unitriangular. One easily shows that its inverse is also upper unitriangular, so we may write 1 a 12 a 1d L 1 1 a 2d = We thus have χ L(E (x 1,..., x d = χ E (L 1 (x 1,..., x d = χ E (y 1,..., y d, y i = x i + d j=i+1 Hence, by Tonelli s theorem, we have ( m(l(e = χ (y2,...,y d E (x 1 + a 12 x 2 + a 1d x d dx 1 dx 2 dx d. a ij x j. Once again by translational invariance, we have that this integral is equal to ( χ E (x 1, y 2,..., y d dx 1 dx 2 dx d. Another d 1 applications of Tonelli s theorem and translational invariance gives that this integral is equal to d χ E = m(e. An essentially identical argument shows that m(l(e = m(e for lower unitriangular matrices, so we are done with the first part of this problem. Problem 2.4b Suppose we had an LDU decomposition (which is a slight variant of the standard LU decomposition from linear algebra L = L 1 L 2 where L 1 is an upper (lower unitriangular matrix, is a diagonal matrix, and L 2 is a lower (upper unitriangular matrix as in the problem. We write a 11 a 22 =..... a dd It s easy to see that (E = E δ where δ = (a 11, a 22,..., a dd. Hence we know from (an easy generalization of exercise 1.7 that m( (E = a 11 a 22 a dd m(e = det m(e 4

5 and hence m(l(e = m(l 1 L 2 (E = m( L 2 (E = det m(l 2 (E = det m(e. Since det = det L, this proves the claim in this case. In general, this decomposition might fail to exist, but we may always find a permutation matrix P such that P L = L 1 L 2. Since a permutation just amounts to a relabeling of the axes, it is easy to see that m(p L(E = m(l(e, so we are finished as long as m(e <. If not, replace E in the argument above with E n = E B n where B n is the closed ball of radius n. This implies either m(l(e n or m(l(e n = for all n, depending on det L. Either way, the general statement holds by monotonicity. Problem 5 (i. It s easy to see that f per, g per are measurable functions on d. For example, we know that for every a, the set E = f 1 ([, a is a measurable subset of := [ π, π] d. Hence fper([, 1 a is the union of all translates of E by 2πn, where n = (n 1,..., n d has integer coordinates. Thus since f per, g per are measurable, we argue exactly as in exercise 2.21a above to conclude that the function (x, y f per (x yg per (y is measurable on I 2d. Problem 5 (ii. We first note that a straightforward generalization of exercise 2.3 (see problem set 3 implies that f per (x y dx = f(x dx. (.1 Therefore, as exercise 2.21b above, we have ( f per (x yg per (y dxdy = f per (x y dx g(y dy I 2d I ( d ( = f(x dx g(y dy = f L1 ( g L1 ( <, where in the first equality we have used Fubini s theorem and in the second we have used (.1. Problem 5 (iii. Define a function F : 2d by F (x, y = f(x yg(yχ I 2d. We know from (ii that F is integrable, so Fubini s theorem tells us that F x is integrable for a.e. x. This means that F x (ydy = f per (x yg per (ydy = f g(x d is well-defined for a.e. x as claimed. 5

6 Problem 5 (iv. We have f g L1 ( = f g = f per (x yg per (ydy dx f per (x yg per (y dydx where we have used the standard triangle inequaltiy for the Lebesgue integral. Now by applying Tonelli s theorem and (ii above, we see that this last expression is equal to f L1 ( g L1 (, proving the first claim. Now observe that the sign which appears in the triangle equality above can be replaced by an = sign if both f per and g per (and hence both f and g are non-negative a.e. Problem 5 (v. We showed in (iii that f g is defined a.e., and we claim that at such points we have (f g(x = f per (x yg per (ydy = f per (ug per (x udu. To see this, observe that since f g L 1 by assumption, Fubini s theorem implies that we may express this integral as π = π π π xd π x1 π x d +π xd +π = x d π f per (x 1 y 1,..., x d y d g per (y 1,..., y d dy 1 dy d x 1+π x1+π x 1 π f per (u 1,..., u d g per (x 1 u 1,..., x d u d ( 1 d du 1 du d f per (u 1,..., u d g per (x 1 u 1,..., x d u d du 1 du d = f per (ug per (x udu as claimed, where the last equality follows from (an easy generalization of the translational invariance established in exercise 2.3. This last integral, where it is defined, is precisely the definition of (g f(x, and we know that it exists a.e. Hence f g = g f away from a set of measure zero, and therefore we must have f g g f L 1 ( =. Problem 5 (vi. We first show that if f per C 1, then h (f g = ( h f g, where h is the directional derivative in the direction of h d. For any h d, consider a sequence = t n h where {t n } is a decreasing sequence t n. We have (f g(x + (f g(x = [ fper (x + y f per (x y ] g per (ydy. (.2 6

7 The mean value theorem tells us that the absolute value of the integrand is bounded above by the function (sup h f g per which is integrable on. Since the pointwise limit of the integrand as n is ( h f per (x yg per (y, and the limit on the left hand side of (.2 is clearly h (f g, the dominated convergence theorem tells us that h (f g = ( h f g. Since f per is bounded and g is integrable, it is easy to see that ( h f g is bounded. We can show that this function is (uniformly continuous as well, for since h f per is uniformly continuous, for every ɛ > there exists a δ > such that x x < δ h f per (x y h f per (x y < ɛ. Therefore if x x < δ, by the triangle inequality ( h f g(x ( h f g(x f per (x y f per (x y g(y dy < ɛ g L 1. Since h (f g = ( h f g is bounded and continous for any h, we conclude that f g is C 1 on the interior of. Hence (f g per is C 1 on the set d \( +2πZ d For x + 2π ˆm where ˆm Z d, we have (f g per (x + (f g(x h (f g per (x = lim (f g(x + 2π ˆm (f g(x = lim h ( n fper (x + 2π ˆm f per (x = lim g(ydy = lim ( fper (x + f per (x g(ydy by the periodicity of f per. Hence we may apply the same argument at x as for points interior to, and therefore (f g per is C 1 at x, and therefore on all of d. Of course, if f per C k then we apply the argument above and a the standard inductive argument to find that (f g per C k. Moreover, if we instead have g C k, we use part (v and the argument above to ensure that all x-derivatives fall on g, and so the dominated convergence theorem again implies that f g C k. 7