ANALYSIS HW 5 CLAY SHONKWILER

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1 ANALYSIS HW 5 CLAY SHONKWILER Let X be a normed linear space and Y a linear subspace. The set of all continuous linear functionals on X that are zero on Y is called the annihilator of Y and denoted by Y. Show that Y is a closed linear subspace of the dual space, X, of X. Show that Y = Y. Proof. To see that Y is a linear subspace, suppose l, l 2 Y. Then, if y Y (l + l 2 )(y) = l (y) + l 2 (y) = so l + l 2 Y. Also, if c R, (cl )(x) = cl (x) =, so cl Y. To see that Y is closed, suppose l k Y and l k L. Let ɛ >. Then, for any y Y, there exists N N such that, for n > N, Hence, l n (y) L(y) < ɛ/2. L(y) = L(y) = l n (y) L(y) < ɛ/2. Since L(y) is smaller than any positive number, it must be the case that L(y) =, meaning L Y, so Y is closed. Now, if Y signifies the completion of Y, then it is clear that Y Y, since Y Y. On the other hand, suppose L Y. Let y Y. Then there exists a sequence y j Y such that y j y. Since L is continuous, y j y L(y j ) L(y). Since y j Y, L(y j ) = for all j N, so their limit L(y) =. Since our choice of y was arbitrary, we see that L Y, so Y Y. Since containment goes in both directions, we conclude that Y = Y. 2 Consider C[, ] with the uniform norm and let K C[, ] be the set of functions with the property that /2 f(t)dt /2 f(t)dt =. Show that K is a closed convex set not containing the origin but K has no point closest to the origin.

2 2 CLAY SHONKWILER Proof. Suppose there exists a sequence f n K such that the f n converge to f in C[, ]. Now, since this convergence is uniform, we are justified in switching limits and integrals: /2 f(t)dt /2 f(t)dt = /2 lim n f n (t)dt /2 lim n f n (t)dt /2 = lim n f n (t)dt lim ( n /2 f n(t)dt /2 = lim n f n (t)dt ) /2 f n(t)dt =. Hence, f K, so K is closed. Also, for α and f, g K, /2 (αf(t) + ( α)g(t))dt /2 (αf(t) + ( α)g(t))dt ( /2 = α f(t)dt ) ( /2 f(t)dt /2 + g(t)dt ) /2 g(t)dt ( /2 g(t)dt /2 g(t)dt ) = α + α =. Hence, K is convex. Finally, consider the sequence f n as drawn below: Then, certainly, f n K. Now, for each n, f n unif = max f n (x) = n n 2. Hence, inf f n unif = lim n K such that f unif. Then /2 n n 2 /2 f(t)dt /2 f(t)dt dt =. However, suppose there exists f /2 dt = /2 + /2 =. Since f K, this inequality must actually be an equality, meaning that { t /2 f(t) = t > /2 However, this is not a continuous function, so f / K. Hence inf f unif = f K but this infimum is never achieved, so there is no point in K closest to the origin. 3 Let K L (, ) be the set of all f with f(t)dt =. Show that K is a closed convex set (of L (, )) that does not contain the origin but there are infinitely many points in K that minimize the distance to the origin.

3 ANALYSIS HW 5 3 Proof. To see that K is closed, suppose that f is the limit of a sequence f k K. Let ɛ >. Then there exists N N such that, if n > N, f f n < ɛ. Now, f(t)dt = f(t)dt f n(t)dt = (f(t) f n(t))dt f(t) f n(t) dt = f f n < ɛ. Since f(t)dt is arbitrarily small, we conclude that f(t)dt =, which is to say that f K. Hence, K is closed. To see that K is convex, suppose f, g K. Then, for α (αf(t) + ( α)g(t))dt = (αf(t) + g(t) αg(t))dt = α f(t)dt + g(t)dt α g(t)dt = α + α =. Hence, K is convex. Certainly the origin is not in K, since dt =. However, for any f K, f = f = f(t) dt f(t)dt = Furthermore, if g K such that f >, g = g = g(t) dt = g(t)dt = There are infinitely many such g, so there are infinitely many points in K that minimize the distance to the origin. 4 Consider points x = (x, x 2 ) in R 2 with the norm x = x + x 2. Let the subspace V be the x axis and define the linear functional l by l((x, )) = x. a) Show that l is a bounded linear functional on V. Proof. Let (x, ), (y, ) V. Then l(c(x, )+d(y, )) = l((cx +dy, )) = cx +dy = cl((x, ))+dl((y, )) for any c, d R, so l is certainly linear on V. Now, to see that l bounded, we need only note that, if x = (x, ) V so l(x) x. l(x) = l((x, )) = x = x + = x

4 4 CLAY SHONKWILER b) Find all extensions of l to a continuous linear map on R 2. Answer: Consider functionals of the form F c ((x, x 2 )) = x + cx 2 for c. Then, for x V, F c (x) = F c ((x, )) = x + c() = x = l(x). Furthermore, for any y = (y, y 2 ) R 2, F c (y) = F c (y, y 2 ) = y + cy 2 y + c y 2 y + y 2 = y. Since, F c (, ) = = (, ), we see that, in fact, F c = l. c) Construct a related example for L (, ). Construction: Let V = {f L : f(t) = for t [/2.)} and define l by l (f) = /2 f(t)dt. Then l is certainly linear and bounded on V. To see this, suppose f, g V, c, d R. Then l (cf+dg) = /2 cf(t)+dg(t)dt = c For boundedness, if f V, /2 l (f) = f(t)dt Now, any functional of the form L(f) = /2 /2 /2 f(t)dt+d f(t) dt = /2 f(t)dt + c f(t)dt /2 g(t)dt = cl (f)+dl (g). f(t) dt = f. where c will be an extension of l to all of L (, ). Since there are infinitely many such, we see that the Hahn-Banach extension is far from unique. 5 Let l, l denote the Banach spaces of all complex sequences x = {x j } with the usual norms x = x j < and x = sup x j < and let c be the subspace of sequences in l that converge to zero. a) If z l show that l(x) := z j x j defines a bounded linear functional on c with l = z. Proof. Let x l. Then l(x) = zj x j zj x j ( ) z j x = zj x = z x,

5 so l is bounded. Furthermore, if x, y l, c, d R, ANALYSIS HW 5 5 l(cx + dy) = z j (cx j + dy j ) = c z j x j + d z j y j = cl(x) + dl(y), so l is linear. To see that l = z, let ɛ >. Then, since z j converges, we know there exists N N such that, if n > N, z j < ɛ/2. j=n+ Let x = (x j ) j=, where x j = z j for j N, x j = for j > N. Then x c and l(x) z = z j x j z j = z j < ɛ/2. Hence, we see that l = z. N+ b) Every bounded linear functional on c is as above. In other words, l is the dual space of c. Proof. Suppose l is a bounded linear functional on c. Denote by e j the vector with a in the j th position and zeros everywhere else. Let x c. Then l(x) = l ( x je j ) = x jl(e j ) x j l(e j ) x l(e j ) = x l(e j) = c x where l(e j ) = c <, so {l(e j )} l. Hence, we see that the dual space c l. Since we showed the reverse containment in part (a), we conclude that c = l. c) Show that l is the dual space of l. Proof. Let x l. Define l(z) := x j z j. Then, for y, z l, c, d C, l(cy + dz) = x j (cy j + dz j ) = c x j y j + d x j z j = cl(y) + dl(z), so l is linear on l. Furthermore, if z l, l(z) = x j z j xj z j ( ) x z j = x zj = x z, so l is bounded. Furthermore, just as in part (a), l = x. Now, if l is a bounded linear functional on l and x l, then ( ) l(x) = l xj e j xj l(e j ) l x j = l x j. Since we ve shown a correspondance between the linear functionals on l and the elements of l, we conclude that l is the dual space of l.

6 6 CLAY SHONKWILER d) Show that l is contained in the dual space (l ) of l, but is not all of (l ) since there are elements in (l ) that are zero on all of c. 6 a) With the positive linear functional l(f) := R f(t)dt on C c(r) show that the set {x R : x < } is measurable. Proof. Let A = {x R : x < }. Let ɛ >. Define Ω := ( ɛ/2, ). Then Ω A = (Ω A)\(Ω A) = ( ɛ/2, )\[, ) = ( ɛ/2, ). Then M (Ω A) = inf Vol(O) = Vol(( ɛ/2, )) = sup Ω A O,O open (I(g)) g C c(ɛ/2,) where g(x). Suppose f(x) is on ( ɛ/2, ) and zero everywhere else. Then M (Ω A) = sup (l(g)) l(f) = dx = ɛ/2 < ɛ. g C c( ɛ/2,) ɛ/2 However, this is just the definition of measurable, so we see that A is measurable. b) With the positive linear functional l(f) := R 2 f(x, y)dxdy, on C c (R 2 ) show that the set {(x, y) R 2 : x, < y < } is measurable. Proof. Let B = {(x, y) R 2 : x, < y < } and let ɛ >. Define Then Ω := [, ] [, ]. Ω B = ([, ] [, ]) ([, ] (, )) = ( (, )) ( (, )). Now, M (Ω B) = inf Vol(O) Ω B O where O is open. Now, consider the sequence of open sets O n = (( /n, /n) (, )) (( /n, + /n) (, )). Then Vol(O n ) = sup g Cc(On), g(x) (l(g)) /n /n dydx + +/n /n dydx = 2/n + 2/n = 4/n. Certainly, the sequence 4/n, so we see that M (Ω B) inf Vol(O 4 n) = inf n N n N n =. Since M (S)/geq for any set S, we conclude that Therefore, B is measurable, M (Ω B) =.

7 ANALYSIS HW 5 7 c) With the positive linear functional l(f) := f() + 2f() on C c (R) discuss which sets are measurable. Can you compute their measures? Answer: Suppose S is a set containing neither nor. Then we can construct an open set Ω containing S such that Ω contains neither nor. If f C c (Ω), then f() = f() =, meaning l(f) = f() + 2f() =. Hence M (S) = inf S O,O open Vol(O) Vol(Ω) = sup g Cc(Ω), g(x) (l(g)) =. Therefore, S has measure. On the other hand, if S is a set containing but not, then we can construct an open set Ω containing S such that Ω does not contain. If f C c (O) where O is an open set not containing, then f() =, so l(f) = f() + 2f() = f(). Furthermore, if O is an open set containing zero, then there exists a function h O C c (O) such that h O () =. Thus, M (S ) = inf S O,O open Vol(O) ( ) inf S O,O open sup g Cc(O), g(x) (l(g)) inf S O,O open(l(h O )) =. Thereforre, S has measure. A parallel argument demonstrates that if S contains but not, then the measure of S is 2. Finally, if S contains both and, then M (S ) = inf Vol(O). S O,O open If Q is any open set containing S, there exists f Q C c (Q) such that f Q () = f Q () = and f(x). Hence, l(f Q ) = f Q () + 2f Q () = 3, so Vol(Q) = sup (l(g)) 3. g C c(q), g(x) If we let g Q denote the function that is on Q and zero outside, then it is also clear that Vol(Q) = sup (l(g)) (l(g Q )) = 3. g C c(q), g(x) Therefore, for any open Q containing S, Vol(Q) = 3. Therefore, M (S ) = inf Vol(O) = 3, S O,O open so S has measure 3.

8 8 CLAY SHONKWILER 7 In a Hilbert space H, a map U : H H is called unitary if (i). Ux, Uy = x, y for all x, y in H. (ii). U is onto. Let Q := {x H : Ux = x} = ker(i U) of all points left fixed by U. a) Show that in finite dimensions property (ii) is a consequence of property (i). Proof. Let H be finite dimensional and let U be unitary. Suppose x, y H such that Ux = Uy. Then = Ux Uy = U(x y), which is to say that = U(x y), U(x y) = x y, x y x y =. Hence, x = y, so U is injective. Since U : H H is injective, it must be surjective, since H is finite-dimensional. b) In the Hilbert space l 2, show that the right shift S : (x, x 2,...) (, x, x 2,...) has property (i) but not property (ii). Proof. Let x, y l 2. Then x = (x, x 2,..., ), y = (y, y 2,..., ) for x i, y i C. Then x, y = x i y i = + x i y i = Sx, Sy, i= so S fulfills property (i). However, There is no z l 2 such that so S is not unitary. i= Sz = (,,,..., ), c) View functions on the circle S as functions periodic with period 2π. If H = L 2 (S ) and α is real, show that the map L : f(x) f(x + 2πα) is unitary. Proof. Suppose f L 2 (S ). Then L(f(x 2πα)) = f(x 2πα + 2πα) = f(x), so L is surjective. Now, let f, g L 2 (S ). Then Lf, Lg = S (Lf)(Lg) = 2π f(x + 2πα)g(x + 2πα)dx = 2π+2πα 2πα f(u)g(u)du = S (f)(g) = f, g where we let u = x + 2πα. Therefore, L is unitary. d) If α is irrational, find the set Q. Answer: If α is irrational, then, if f Q, f(x) = Lf = L m f = f(x + 2πmα)

9 ANALYSIS HW 5 9 for all m Z. Since Q L 2 (S ), f must also be 2π-periodic. Hence, either 2πmα = 2π for some integer m or 2πα = n2π for some integer n. Dividing by 2π on both sides of each, we see that mα = or α = n. Since α is irrational, neither of these is possible, so we conclude that Q is empty if α is irrational. e) If α is rational, find the set Q. Answer: If α is rational, then, again, if f Q, f(x) = Lf = L m f = f(x + 2πmα) for all m Z. Since α is rational, α = p/q for p Z, q Z. Also, if f Q, f must be 2π-periodic. Hence, the elements of Q will be precisely those functions which are 2π nq -periodic, where n Z. f) If U is unitary, show that U = U and hence that Ux = x if and only if U x = x. Consequently ker(i U) = ker(i U). Proof. Let x, y H. By definition, x, Uy = U x, y = UU x, Uy. Hence, = x, Uy UU x, Uy = x UU x, Uy. Since our choice of y was arbitrary, we see that it must be the case that x UU x =, or x = UU x U x = U x. Since our choice of x was arbitrary, this means that U = U. Hence, Ux = x if and only if x = U x = U x, and so ker(i U) = ker(i U). g) Using (ImageL) = kerl for any bounded linear operator L in a Hilbert space, conclude that H = ker(i U) Image(I U). Proof. From problem above, we know that Image(I U) = (Image(I U)) = ker(i U) = ker(i U), where we use the given fact to get the second equality and part (g) to get the third. Now, since we are in a Hilbert space, (ker(i U)) = (Image(I U) ) = Image(I U). Applying this result to the conclusion proved in problem 5 of last week s homework, we see that H = ker(i U) (ker(i U)) = ker(i U) Image(I U).

10 CLAY SHONKWILER 8. (Continuation) von Neumann s mean ergodic theorem states that for any f H, one has N () lim U n x = P x N N where P is the orthogonal projection onto the set Q := {x H : Ux = x} = ker(i U). Equation () is clarly true for all x in Q = ker(i U). a) Prove that for all z Image(I U), that is z = (I U)x for some x H, then (2) N N U n z = N (x U N x). Proof. We see immediately that N U n z = N U n (I U)x = N U n x U n+ x = (x Ux) + (Ux U 2 x) (U N x U N x) = x U N x. Hence, N N U n z = N (x U N x). b) Conclude that (2) holds for all z Image(I U). Proof. First, note that, for all x H, Ux 2 = Ux, Ux = x, x = x 2, so Ux = x. Now, let z Image(I U). Then there exists a sequence {z j } Image(I U) such that z j z. Let ɛ >. Then there exists M N such that, if k > M, z z k < ɛ/2. Also, from part (a), we know that for all k there exists N N such that, if n > N, n U m z k < ɛ/2. n

11 ANALYSIS HW 5 Let k > M, n > N. Therefore, n n U m z = n n U m (z z k ) + n n U m z k n n U m (z z k ) + n n z k n n U m (z z k ) + n n U m z k = n n z z k + n n z k = z z k + n n z k < ɛ/2 + ɛ/2 = ɛ. Hence, for all z Image(I U), N N U n z. c) Use the last part of the previous problem to show that P z = if and only if z Image(I U). Thus (2) proves () for all x that satisfy P x = and thus for all x in the Hilbert space. Proof. Certainly, as we ve shown in (b) above, if z Image(I U), then N lim N U n z =. Furthermore, as we saw in 7(g), if z Image(I U), then z (ker(i U) ), meaning P z =. Hence, z Image(I U). On the other hand, if P z =, then z (ker(i U) ) and, therefore, by 7(g), z z Image(I U). By (b), then, N P z = = lim U n z. N 9 Let Ω R n be a bounded open set and a(x) > and f(x) be smooth functions periodic with period 2π. Let H (S ) be the completion of smooth 2π periodic functions on the circle S in the norm φ 2 H := ( φ 2 + φ 2 )dx. S

12 2 CLAY SHONKWILER Call u H (S ) a weak solution of u + a(x)u = f on S if [u v + a(x)uv]dx = S fvdx S for all v H (S ). a) (uniqueness) Show that the equation u + a(x)u = f on S has at most one weak solution. Proof. Suppose there exist two weak solutions, u and u 2. Then = S fvdx S fvdx = S [u v + a(x)u v]dx S [u 2 v + a(x)u 2 v]dx = S [(u u 2 )v + a(x)(u u 2 )v]dx for all v H (S ). Particularly, when v = u u 2, v = u u 2, so S [(u u 2) 2 + a(x)(u u 2 )]dx =. Since a(x) >, this means that u u 2 =, or u = u 2, so any weak solution must be unique. b) (existence) Show that the equation u + a(x)u = f on S has a solution. Proof. Define the norm 2 H such that, for v H (S ), v 2 H = [(v ) 2 + a(x)v 2 ]dx. S Define the functional l(v) := S fv. Then l is certainly linear and l(v) = fv f L 2 v L 2 c f L 2 v 2 H S by the Holder inequality and since this norm is equivalent to the L 2 norm, so l is a bounded linear functional on the Hilbert space. Hence, by the Riesz-Frechet Representation Theorem, there exists u H (S ) such that l(v) = v, u. Therefore fv = l(v) = v, u = [v u + a(x)vu]dx, S S so u is a weak solution. c) Show that if b(x) is a smooth function on S with b(x) sufficiently small, then the equation u + b(x)u + a(x)u = f on S has exactly one weak solution. Proof. A weak solution u to this equation must satisfy [u v bv u + a(x)vu]dx = fv S S for all v H (S ).

13 ANALYSIS HW 5 3 To show uniqueness of weak solutions, suppose u and u 2 are weak solutions. Then, if we let w = u u 2, = S fv S fv = S [u v b(x)v u + a(x)vu ]dx S [u 2 v b(x)v u 2 + a(x)vu 2 ]dx = S [w v b(x)v w + a(x)vw]dx for all v. Particularly, when v = w = [(w ) 2 b(x)w w + a(x)w 2 ]dx. S If we integrate the second term by parts, we will get = [(w ) 2 B(x)(w ) 2 + a(x)w ]dx = [( + B(x))(w ) 2 + a(x)w 2 ]dx S S where B(x) = x b(t)dt. Now, if b(x) is suffiently small for all x S, then B(x) >, which in turn implies that ( + B(x)) >. Since a(x) >, all terms in the above integral are nonnegative, so w conclude that w =, meaning u = u 2. Therefore, any weak solution to this equation is unique. To prove existence, we now need to define another new norm, 2 H, where, for v H (S ), v 2 H = S [(v ) 2 b(x)v v + a(x)v 2 ]dx. Let l(v) = S fv. Then, again, l is linear and, using Holder, l(v) = fv f L 2 v L 2 c f L 2 v 2 H S since this norm is equivalent to the L 2 norm. So l s a bounded linear functional on the Hilbert space. Now, in order to construct a weak solution u, we need to introduce a bilinear form B(, ), where, for v, w H (S ), B(v, w) = [v w b(x)v w + a(x)vw]dx. S Now, we need to check the following: (i) B is bounded (ii) There exists b > such that B(y, y) b y 2 for all y H (S ). To see (i), let v, w H (S ). Then B(v, w) = S [v w b(x)v w + a(x)vw]dx S v w dx + S b(x)v wdx + S a(x)vwdx = S a(x)vwdx av L 2 w L 2 c v 2 H w 2 H. To prove (ii), we merely note that B(v, v) = [(v ) 2 b(x)v v + a(x)v 2 ]dx = v, v = v 2, S

14 4 CLAY SHONKWILER so B(v, v) v 2 for all v. Therefore, by the Lax-Milgram Lemma, there exists u such that l(v) = B(v, u). This means that fv = l(v) = B(v, u) = [v u b(x)v u + a(x)vu]dx, S S meaning u is a weak solution of the given equation. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu