2) Let X be a compact space. Prove that the space C(X) of continuous real-valued functions is a complete metric space.

Transcription

1 University of Bergen General Functional Analysis Problems with solutions 6 ) Prove that is unique in any normed space. Solution of ) Let us suppose that there are 2 zeros and 2. Then = + 2 = 2 + = 2. 2) Let X be a compact space. Prove that the space C(X) of continuous real-valued functions is a complete metric space. Solution of 2) Let {x n } be a Cauchy sequence in C(X); that is for any ε > there exists N N such that () n, m N = x n (t) x m (t) x n x m < ε t X. It implies that the sequence {x n } converges uniformly on X. Since X is compact the limit function x(t) will be continuous. Letting m in () we get x n (t) x(t) < ε t X and n N. This means that {x n } converges uniformly, or in other words converges in the metric of C(X). 3) Let A be a subset of vector space V. Construct a smallest linear subspace (linear manifold) containing A. 4) Show that the set P of all polynomials on [, ] is a linear manifold in C[, ]. Is it closed? Give an example of a closed linear manifold in C[, ]. 5) Let X be a vector space. Let B = {x X : x < }. Prove that B is open and that B = {x X : x }. 6) Suppose that A is a linear operator between two normed vector spaces X and Y. Show that the following are equivalent: () A is continuous at one point, (2) A is continuous at all points, (3) A is uniformly continuous, (4) A is bounded. Solutions of 6) (4) = (3): If A is bounded, then Ax M x for all x X. Then Ax Ay M x y, so A is uniformly continuous. (3) = (2): This is true by the definitions of continuity and uniform continuity. (2) = (): This is true by definition. () = (4): Suppose that A is continuous at x. Then for all ε > there is some δ > such that x x = Ax Ax < ε. Therefore if y < δ then so we can assume x =. Ay = A(y + x ) Ax < ε,

2 2 If we let ε = we have that y < δ = Ay <. Thus, if we tae an arbitrary x (A is linear so A = ) and we consider y = We conclude that A is bounded. Ax = A( 2 x δ y) = 2 x Ay 2 δ δ x. δ 2 x x then 7) Let X be a normed vector space and Z its subspace (linear manifold). Prove that if y X has distance d from Z, then there exists a linear functional Λ : X R such that (The distance referred is d = inf z Z z y ) Λ, Λ(y) = d and Λ(z) =, z Z. Solutions of 7) If y Z, then d(y, Z) = and the zero functional wors, so we may assume that y X \ Z. Consider the subspace Y = {λy + Z : λ R} X. Since y / Z for every x Y, there is a unique α R and z Z such that x = αy + z. Define Λ(x) = Λ(αy + z) = ad. Observe that Λ(z) = for all z Z and that Λ(y) = d. The functional Λ is linear and Λ(x) = αd α y + z = x α by the definition of d. Thus Λ is linear bounded functional with Λ. Applying the Hahn - Banach theorem we extend Λ on all X with the same estimate for the norm. ) 8) Let (C[, ], be a space of continuous functions on [, ] with x = sup x(t). Show t [,] that the functional f(x) = is linear, continuous and bounded. Find its norm. x(t) dt x(t) dt Solutions of 8) Let us show that it is linear. Let α, β R, x, y C[, ]. f(αx + βy) = = α (αx + βy)(t) dt x(t) dt + β = αf(x) + βf(y). (αx + βy)(t) dt y(t) dt α x(t) dt β Now we show that the functional is bounded and thus is continuous. We have f(x) x(t) dt + x(t) dt 2 x y(t) dt for any x C[, ]. If x ε/2 then f ε ε >. We conclude that f 2. Let us show that f = 2. Consider the following sequence of functions if x [, n ), x n (t) = if x ( n, ], nx if x [ n. n ],

3 3 We have f(x n ) = 2 n and x n =. Thus that shows f = 2 9) Verify, that the function x = f(x n ) sup = 2 n N x n m max t [a,b] x() (t) = define a norm in the space C m [a, b] of the continuous functions that have up to the m derivatives x () and the derivatives are also continuous on [a, b]. What does it mean that a sequence converge in this space? ) Let (X, ) be a normed space. Let {x n } and {y n } be two Cauchy sequences in X. Show that the sequence λ n = x n y n converges. Solution of ) Let ε > and N N is such that Since and then if n, m N = x n x m < ε 2 and y n y m < ε 2. x n y n x n x m + x m y m + y m y n x m y m x m x n + x n y n + y n y m, λ n λ m = x n y n x m y m x n x m + y m y n < ε. Since (R, ) is complete the Cauchy sequence {λ n } has a limit Λ R. ) Let c be a space of real sequences x = {x n } n= converging to. Let l be a set of real sequences w = {w } = furnished with the norm w = w n. Prove that c is closed in l. sup n x () n sup n< Solution of ) Let x () c be a sequence converging to ω l. Tae ε > and N N such that ρ sup (x (), ω) = ω n < ε/2 for all > N. For each choose N N such that x () n ε/2 for all n > N. Thus, ω n ω n x () n + x () n ε for n > N and > N, that means that the sequence ω n converges to and ω c. 2) Show the inequality f + g p 2 p ( f p + g p ). 3) Let X = (X, A, µ) be a measure space with σ-algebra A and positive measure µ. Let f L p. Show that ( ) p f p = f p dµ is a norm. (Extend the Minowsii and Hölder inequality from a sum to an integral.) X 4) Let us denote by L (X) a set of measurable functions f on X = (X, A, µ) endowed with the norm Show that the space L (X) is complete. f = ess sup f(t) = inf { M : µ{t : f(t) > M} = } <.

4 4 Solution of 4) Let {f n } be a Cauchy sequence in L (X). Let also and A = {x X : f (x) > f } B m,n = {x X : f n (x) f m (x) > f n f m }. By E we denote the union of A and B m,n over all, m, n N. Measure of E is zero because of the zero measure of all A and B m,n. On the complement of E the sequence {f n } converges uniformly to a bounded function f. (Why?) Define f(x) = for x E. Then f(x) L (X) and we get f n f as n. 5) If f L (X) and g L (X) then show that X fg dµ f g. Solution of 5) Notice that f(x)g(x) g f(x) for almost all x X. Then X f(x)g(x) dµ g f(x) dµ = f g. 6) Let E be a normed space and f E. Define ernel of f by Ker(f) = {x E : f(x) = }. Show that dist(x, Ker(f)) = f(x) f, x E. Solution for 6) Remember that ρ(x, L) = inf x y. If y L, then y L Thus f(x) = f(x y) f x y. f(x) ρ(x, L) = inf x y y L f. Let us show the reciprocal inequality. We tae a sequence {x n } E such that x n = for all n N and n n + f < f(x n). Consider the element y = x f(x) f(x x n) n It is easy to see that y L. Moreover, x y f(x) x n n + f(x) f(x n ) n f. Taing firstly the limit as n, then taing the infimum over all y L, we get the result ρ(x, L) f(x) f. 7) Show that all norms on R n are equivalent. Solution of 7) Let be the Euclidean norm (with respect to a basis e,..., e n ) and let be any other norm on R n. We claim that x x is continuous with respect to the Euclidean norm. Write x = n = c e. Then by the triangle inequality, n x c e ( n max e ) max c M x, =

5 where M = n max e. Thus if x y ε M then x y ε and we conclude that the map is continuous. Now we consider the region K = {x : x = }. This is just the unit sphere in R n with respect to the Euclidean norm, which is compact. The map x x is continuous, so it attains a minimum m and a maximum M on K. Thus, for x K, m x M. Now for x \{}, K and then We see that the norm are equivalent. m x x M = m x x M x. 8) Consider the linear functionals f(x) = x( t) dt and f(x) = x(t2 ) dt over C[, ]. Are they bounded? 9) Let V be a normed space and {f n } V. Prove that f n converges if and only if {f n } converges uniformly on the unit ball B(, ) = {x V : x }. 2) Let V be a normed space and x V. Show that x = sup f(x). f V, f = 2) Given f L p (), p < and ε > there is a bounded measurable function b with b < M and f b p ε. 2) We define the cut-function n for n f(x), f n (x) = f(x) for n f(x) n, n for f(x) n. Then f n n, and the sequence {f n } converges to f almost everywhere. Thus f f n also almost everywhere. Since f n (x) f(x) we have f f n p 2 p f p. Moreover f L p and we get that f f n p p = f f n p as n by the Lebesgue Convergence Theorem. We conclude that given ε > there is b = f n such that b M(= n) (function is bounded) and f b p ε. 22) Let g be an integrable function on [, ] ( g L ([, ]) ). Show that there is bounded measurable function f (f L ([, ])) such that f and fg = g f. [,] Solution of 22) Let us suppose that we defined the linear functional F as in the case of p [, ) for the characteristic function, for the step function ψ, and for the bounded measurable function f. Let g L ([, ]) and ε >. Then there is a bounded measurable function f such that f g < ε. Since f is bounded then F (f) = fg by the previous considerations. Then F (g) fg = F (g) F (f) F f g F ε. thus F (g) = fg since ε is arbitrary. 23) Let c be a space of sequences x = (x,..., x n,...) converging to with the norm x = sup n x n. Show that the dual space is isomorphic to the l of all absolutely summable sequences f = (f,..., f n,...) with the norm f = f n. x x 5

6 6 Solution of 23) Any element f l defines the bounded linear functional F over c by F (x) = f n x n. Since F (x) x f n we have F f n = f. Now we prove the reciprocal inequality. Let us consider the unit elements in c : e = (,,,...,,,...), e 2 = (,,,...,,,...), , e n = (,,,...,,,...), Set Then x (N) c, x (N) and { N f n x (N) = f e n n for f n, for f n =. F (x (N) ) = N f n f n F (e n) = N f n. We conclude that lim N F (x (N) ) = f n = f. Thus F F (x(n) ) x (N) f n = f. Finally, we obtain F = f. We constructed the linear isometry f F of the space l into the space c. Let us show that it is the map onto; that is any functional F c has the representation F (x) = f n x n, where f = {f n } l. For any x = {x n } c we have x = x ne n. The series on the right hand side converges to x in c because of N x x n e n = sup x n as N. n>n The functional F c is continuous, thus F (x) = x n F (e n ). So we have to verify that F (e n) <. Setting x (N) = N F (e n ) F (e e n) n we deduce N N F (e n ) F (e n ) = F (e n ) F (e n) = F (x (N) ) F because of x (N) c, x (N). Since N was arbitrary we conclude that F (e n) <. 24) Given f L p, < p and ε >, there is a step function ψ such that f ψ p < ε. Solution of 24) By the problem we may choose a bounded function b such that f b p < ε/2 and b < M. Any bounded measurable function can be approximated by the step functions, so we can find a

7 7 step function ψ such that b ψ ε/4 except on a set E of the measure µ(e) < δ (we presice δ later). We set ψ(x) = on E. Then b ψ p p = [,] b ψ p = [,]\E b ψ p + b ψ p E if we choose δ = < εp 4 p + M p δ εp 2 p εp M p 4. Consequently b ψ p p < ε/2. Applying the Minowsii inequality we get f ψ p f b p + b ψ p < ε. 25) Study the analogue of Hahn-Banach theorem for complex vector space. Let X be a complex vector space, S a linear subspace, p a real-valued function on X such that p(x + y) p(x) + p(y), and p(αx) = α p(x). Let f be a complex linear functional on S such that f(s) p(s) for all s S. Then there is a linear functional F defined on X such that F (s) = f(s) for s S and F (x) p(x) for all x X. 26) Let l be a space of bounded real-valued sequences x = (x,..., x n,...) with the norm x = x n. Prove that l = l, but l l by using the Hahn - Banach theorem. For the last part, note sup,... that l can be identified with a subset of l by a(b) = a n b n for a = {a n } l, b = {b n } l. Show that not every element of l is of this form. Solution of 26) First note that given any b = {b n } l we obtain a bounded linear functional on l by a(b) = a n b n for a = {a n } l. The sum is well-defined because {b n } is bounded and a n is absolutely convergent. It is clearly linear in {a n } and a(b) b a l. So a b. We showed that any element from l defines a linear functional on l We need to show that any bounded linear functional F l is defined by unique elements from l. Suppose we have F l. Let b i = F (e i ), where e i has in the i-th place, and zeros everywhere else. F is a bounded linear functional, so we have that b i = F (e i ) F, so if we let b = {b i }, then b l. We now claim that F ({a n }) = a n b n. Indeed, consider the following sequence a m l { a m an if n m, n = if n > m. The functional F is linear, so F (a m ) = F ( m ) m a n e n = a n F (e n ) = m a n b n. The series a n is absolutely convergent and {b n } is bounded, so the righthand side above converges to an b n as m. F is continuous on l, and a m a n in l, so the lefthand side converges to F ({a n }), giving us that F comes from the element b l. The last step is to show that the constructed map is an isometry. We already now that b l b. The norm b = sup b n implies that given ε >, there is an N N such that b N > b ε. then b l b(e N ) = b N > b ε = b l b. Now we show that l l. First note that the first paragraph above also shows that we may identify l with a subspace of l, so it is enough to show that subspace in not all of l. Consider the closed subspace c of l consisting of convergent sequences. If {a n } c, define f({a n }) = lim n a n. This defines a bounded linear functional on c, which is closed. The Hahn - Banach theorem says that we can extend it to a bounded linear functional F on l. Observe, that f and that F c =. Suppose that F l considered as a subspace of l, then F ({a n }) = a n b n for some {b n } l. The functional f vanishes on c, so b n = f(e n ) = because e n c. This tells us that F should be the zero functional, but it is not so in the subspace of l corresponding to l.

8 8 27) Let D ([, ]) denote the topological dual of C ([, ]) with the corresponding topology. One calls elements of D ([, ]) distributions or generalised functions. Let ι : L ([, ]) D ([, ]) be the inclusion map defined by ι(φ)[ψ], φ L ([, ]), ψ C ([, ]). () Show that ι indeed maps into D ([, ]), is injective and continuous. One regards L ([, ]) as a subset of D ([, ]). (2) Let τ = ι C ([,]). Show that τ : C ([, ]) D ([, ]) is continuous. (3) Show that d dx : C ([, ]) C ([, ]) has a continuous extension to a map d dx : D ([, ]) D ([, ]), given by d u[ψ] = udψ dx dx, u D ([, ]), ψ C ([, ]). Thus every distribution, in particular every L function, can be differentiated arbitrary many times, in the sense of distributions. Solution of 27) () Suppose f L ([, ]), ψ C ([, ]), then ι(f)[ψ] = fψ f ψ C ψ. [,] So ι(f) D ([, ]). The functional f fψ is continuous that follows from the above inequality. Now if f C([, ]) and ι(f) =, then we claim that f =. Indeed, if f, then {f > } and {f < } are both open and at least one of them is non-empty. Without loss of generality, we may assume there is a nonempty open set on which {f > }. We may find ψ C ([, ]) such that the support supp ψ {f > }, ψ, and ψ(x) > for some x {f > }. (Construct such function!). Then = ι(f)[ψ] = fψ = fψ > a contradiction, so that f =. Thus ι C([,]) is injective. For f L ([, ]), ι(f) =, we consider the functions f (x) = 2 x+ x f(y) dy = 2 {f>} The functions f are continuous because if x n x, we have f (x n ) f (x) ( x f(y) dy + 2 x n f(x + y) dy. x+ x n + ) f(y) dy because f L ([, ]), with a similar statement holding for x n x. Lebesgue s theorem tells us that f f almost everywhere. Now, for ψ C ([, ]) we have that = 2 ι(f )[ψ] = 2 f(x)ψ(x y) dxdy = 2 f(x + y)ψ(x) dydx = (ι(f))[ψ( y)]dy = by Fubini s theorem (since fψ is integrable on the product space). Then ι(f ) =, so f =. The sequence f f almost everywhere, so f = almost everywhere and thus f =, so ι is injective. (2) We just need to show that the inclusion j : C ([, ]) L ([, ]) is continuous, but we have for ψ C ([, ]) ψ = ψ 2π ψ so it is continuous, so then τ = ι j is continuous. (3) In order to show that the map d dx : D ([, ]) D ([, ]) is continuous then it is enough to show that for each ψ, the map u ( d dψ dxu)[ψ] is continuous. But this functional is just u u( dx ), and

9 9 so it is continuous. We now wish to show that d dx indeed extends the map from C ([, ]). Indeed, if ψφ C ([, ]), we have ( d (τφ))[ψ] = (τφ)[dψ dx dx ] = φ dψ dφ dx = dx ψ = τ(dφ dx )[ψ] by integration by part. 28) Suppose V is an inner product space (positive definite scalar product) either over R or over C. () Prove that the inner product can be recovered from the norm by the polarisation identity: (x, y) = 4 ( x + y 2 x y 2 ) if the field if R, and (x, y) = 4 ( x + y 2 x y 2 ) i 4 ( x + iy 2 x iy 2 ) if the field is C. (2) Prove that a normed vector space (V, ) is a n inner product space (with the induced norm begin ) if and only if the norm satisfies the parallelogram law: x + y 2 + x y 2 = 2 x y 2 x, y V. (3) Conclude that the standard norms on l p and L p (R), p 2, do not arise from inner products. Solution of 28) () For the R case, we just expand the right hand side and use the symmetry of the inner product: 4 ( x + y 2 x y 2 ( ) ) = 4 (x, x) + (y, y) + (x, y) + (y, x) (x, x) (y, y) + (x, y) + (y, x) = 2( (x, y) + (y, x) ) = (x, y). For the other case, we again expand the right side, using the relation we just proved: 4 ( x + y 2 x y 2 ) i 4 ( x + iy 2 x iy 2 ) ) ( ) = 2( (x, y) + (y, x) i 2 (x, iy) + (iy, x)) = (x, y) i2 2 (x, y) + i2 2 (y, x) = (x, y). (2) If the norm comes from the inner product, then we have x + y 2 + x y 2 = 2(x, x) + 2(y, y) + (x, y) + (y, x) (x, y) (y, x) = 2( x 2 + y 2 ). Now we suppose that the norm satisfies the parallelogram law. Assume the field is C (the argument is similar to R),and define the inner product via the polarisation identity from part (). If x, y, z V, we

10 have that (writing x + y = x + y+z 2 + y z 2 and similar for x + z): (x, y) + (x, z) = ( x + y 2 + x + z 2 x y 2 x z 2) 4 i ( x + iy 2 + x + iz 2 x iy 2 x iz 2) 4 = ( x + y + z y z 2 2 x y + z 2 2 y z 2 2) i ( x + i y + z i y z 2 2 x i y + z 2 2 i y z 2 2) = ( x + y + z y + z 2 2 x y + z 2 2 y + z 2 2) i ( x + i y + z i y + z 2 2 x i y + z 2 2 i y + z 2 2) = ( x + y + z 2 + x 2 x (y + z) 2 x 2) 4 i 4( x + i(y + z) 2 + x 2 x i(y + z) 2 x 2) = (x, y + z). This holds for all x, y, z V, so, in particular, (x, ny) = n(x, y), and so if r = p q Moreover, by the polarisation identity, Q, then (x, ry) = r(x, y). (x, iy) = 4 ( x + iy 2 x iy 2) i 4 ( x y 2 x + y 2) = i(x, y). Combining these two results gives that if α Q + iq, then (x, αy) = α(x, y). If α C, then there are α n Q + iq, α n α, so (x, α n y) (x, αy) because all the norms above must converge. This gives us that α(x, y) = lim n α n(x, y) = lim n (x, α ny) = (x, αy), so (x, y) is linear. Observe that because i(x iy) = x iy, we have that (y, x) = (x, y), and that (x, x) = 4 (2 x 2 ) = i 4 ( + i 2 x 2 i 2 x 2 ) = x 2, so this shows that the norm is induced by (, ) and that it is also positive definite, so it is an inner product. (3) Let f = χ [,/2], g = χ [/2,] L p (R), so then f + g 2 = f g 2 =, and ( f 2 = g 2 = 2 ) 2/p, so the parallelogram law does not hold unless p = 2. (For p =, the same examples wor, but the norms are slightly different.) ( ) 2/p, Let a = (/2, /2,,,...), b = (/2 /2,,,...) l p, then a 2 = b 2 = ( 2 )p + ( 2 )p while a + b 2 = a b 2 =. So the parallelogram law does not hold unless p = 2. 29) Let T t be the operator T t (φ)(x) = φ(x + t) on L 2 (R). What is the norm of T t? What does it converge to as t and in what topology?

11 Solution of 29) First we claim that for each phi C (R) L 2 (R), T t φ wealy as t. Indeed, let ψ L 2(R) = L 2 (R). Fix ε >. Since ψ L 2 (R) there is some K such that ( /2 ψ(x) dx) 2 < ε. x >K We let M be such that supp φ [ M, M]. Then if t > M +K we have that if φ(x), then x t < K, and so we have that (T t φ, ψ) = φ(x + t)ψ(x) dx = φ(x)ψ(x t) dx M ( /2 φ 2 = φ(x)ψ(x t) dx ψ(x t) dx) C φ ε, M so (T t φ, ψ) as t. Now for f L 2 (R), we choose φ C (R) with compact support such that φ f 2 ε, so then we have (T t f, ψ) (T t (f φ), ψ) + (T t φ, ψ) 2ε for large enough t. Thus T t in the wea operator topology, assuming that it is bounded, which we show now. Foe any φ L 2 (R), we have that φ(x + t) 2 dx = φ(x) 2 dx = T t φ = φ 2 = T t =. R R This also shows that T t does not converge to anything in the strong operator topology (and hence not in norm, either) because it must converge to, but T t φ = φ. 3) Let f L ([, ]). Let g be a bounded measurable function on [, ]. Then gf is in L ([, ]). Solution of 3) Let ϕ n be a sequence of step functions converging both L and almost everywhere to f. Let ψ n be a sequence of a step functions converging pointwise to g. Then ϕ n ψ n is a sequence of step functions and as n this sequence converges almost everywhere to fg. Changing f and g on a set of measure (e.g. giving them the value ), we can assume that this convergence is pointwise everywhere. If C is a bound for g, i.e. g(x) C for all x, then fg C f. We conclude that fg L ([, ]) by the Dominated Convergence Theorem.