Functional Analysis (H) Midterm USTC-2015F haha Your name: Solutions

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1 Functional Analysis (H) Midterm USTC-215F haha Your name: Solutions 1.(2 ) You only need to anser 4 out of the 5 parts for this problem. Check the four problems you ant to be graded. Write don the definitions of We call a function p : X R on a topological vector space X a seminorm if... for any x, y X and all scalars α, one has p(x + y) p(x) + p(y) and p(αx) = α p(x). We say a subset E in a topological vector space X is bounded if... for any neighborhood U of in X, there exists s > such that E tu for all t > s. A family Λ = {L α } of continuous linear operators from a topological vector space X to a topological vector space Y is said to be equicontinuous if... for any neighborhood V of in Y, one can find a neighborhood U of in X so that L α (U) V for any L α Λ. The eak topology on the dual space X of a topological vector space X is defined to be... the eakest topology on X so that for any x X, the map ev x : X F, L L(x) is continuous. The annihilator of a vector subspace M in a Banach space X is defined to be the set... M = {x X x, x =, x M}. 1

2 2.(2 ) You only need to anser 4 out of the 5 parts for this problem. Check the four problems you ant to be graded. Write don the statement of the folloing theorems The Hahn-Banach theorem: Let X be a real vector space, p : X R a quasi-seminorm. Suppose Y X is a subspace, and l : Y R is a linear functional on Y that is dominated by p. Then there exists a linear functional L : X R on X that extends l and is dominated by p. The closed graph theorem: Let X, Y be F -spaces, and L : X Y be a linear operator hose graph Γ L is closed in X Y. Then L is continuous. The uniform boundedness principle: Let X be a Banach space, Y a normed vector space, and Λ a family of continuous linear operators from X to Y. Suppose the family Λ is pointise bounded, i.e. for any x X, sup L Λ Lx <. Then there exists a constant C so that L < C for any L Λ. The Banach-Alaoglu theorem: Let X be a topological vector space, and V X a neighborhood of in X. Then the polar set K = {L X Lx 1, x V } is compact in the eak-* topology. The Riesz representation theorem (for Hilbert space): Let H be a Hilbert space. Then for any L H, there exists a unique y H so that Lx = (x, y) for any x H. 2

3 3.(2 ) You only need to anser 4 out of the 5 parts for this problem. Check the four problems you ant to be graded. For each of the folloing statements, rite don an example (No detail is needed!) A normed vector space hich is not a Banach space. The set of all polynomials ith the supreme norm or C([, 1]) ith the L 1 -norm etc A linear functional that is not continuous. Let X = C([, 1]) equipped ith the L 1 -norm, and let L(f) = f( 1 2 ). A topological vector space that is not locally convex. L p ([, 1]) ( < p < 1), ith metric d(f, g) = 1 f(x) g(x) p dx. A Banach space hose closed unit ball contains no extreme points. c = {x = (a 1, a 2, a 3,... ) l lim n a n = } l, equipped ith the l norm. A separable Banach space hose dual space is not separable. l 1 = {x = (a 1, a 2, a 3,... ) n a n < }, equipped ith the standard l 1 norm. 3

4 [ You only need to anser 3 out of the folloing 4 problems!] [ 2 points each. Check the three problems you ant to be graded!] Anser the folloing problems 4. For any f L 2 (R) e define Lf be the function (a) Prove: L L(L 2 (R), L 2 (R)). (b) Find the operator norm L. (c) Find the kernel and the range of L. (d) Find the adjoint L. (Lf)(x) = f( x ). (a) L is ell-defined because if f L 2 (R), i.e. f is a measurable function defined on R so that R f(x) 2 dx <, then Lf is again a measurable function on R, and R (Lf)(x) 2 dx = f(x) 2 dx + f( x) 2 dx = 2 L is linear because for any f, g L 2 (R) and any scalars α and β, L(αf + βg)(x) = (αf + βg)( x ) = α(lf)(x) + β(lg)(x). L is continuous because it is bounded: for any f L 2 (R), ( Lf 2 = 1/2 Lf(x) dx) 2 = R ( 2 f(x) 2 dx <. f(x) 2 dx) 1/2 2 f 2. (b) The last line above shos L 2. On the other hand, if e take f to be a square integrable function on R that equals for all x <, then ( Lf 2 = So L 2. So L = 2. 1/2 Lf (x) dx) 2 = R ( 2 f (x) 2 dx) 1/2 = 2 f 2. (c) The kernel of L consists of all the measurable and square integrable functions on R hich equals for almost every x >. The range of L consists of all the even functions on R that are measurable and square integrable. 4

5 (d) For any f, g L 2 (R), (Lf, g) = So the adjoint of L is = R Lf(x)g(x)dx = ( f(x) g(x) + g( x) L g(x) = f(x)g(x)dx + ) dx = (f, L g). { g(x) + g( x), x,, x <. f( x)g(x)dx 5. Prove the folloing statements: (a) Let X be a topological vector space and let L : X R be a linear functional on X. Then L is continuous if and only if ker(l) is closed. (b) Any infinitely dimensional Frechét space contains a subspace that is not closed. (a) If L is continuous, then ker(l) = L 1 () is closed since {} is closed in R. Conversely suppose ker(l) is closed, then X/ker(L) is a topological vector space. More over, the map L : X/ker(L) R, [x] = x + ker(l) L([x]) = L(x) is ell-defined since [x 1 ] = [x 2 ] = x 1 x 2 ker(l) = L(x 1 ) = L(x 2 ) = L([x 1 ]) = L([x 2 ]). Obviously L is linear and surjective. It is also injective since L([x]) = = L(x) = = x ker(l) = [x] = X/ker(L). So L is a linear isomorphism from X/ker(L) to R. So L is a homeomorphism, and in particular L is continuous. So L is continuous since it is the combination of continuous maps L = L π, here π is the canonical projection: L : X π X/R L R. (b) Let {x 1, x 2, x 3, } be a countable sequence of linearly independent vectors in the given Frechét space. Let X n = span{x 1,, x n } and let X = n=1 X n. Then X is a vector subspace of the given Frechét space. We claim theat X is not closed. Suppose X is closed, 5

6 then it is complete, and thus a complete metric space. But each X n is a finite dimensional vector subspace of X, hich has to be closed. Claim: Each X n contains no interior points, since if X n contains an interior point x, then is also an interior point of X n = X n x. In other ords, X n contains a neighborhood of in X. As a consequence, X n contains the hole of X since any neighborhood of is absorbing. This conflicts ith the fact that X n is finite dimensional hile X is infinite dimensional. Back to the proof. We see that X = n X n is a countable union of nohere dense subsets. So X is of first category. This conflicts ith Baire s category theorem since (if X is closed, then) X is a complete metric space. 6. Prove the folloing statements: (a) Let X, Y be Banach spaces, and L : X Y is a continuous linear operator. Prove: If x n x, then Lx n Lx. (b) In a Hilbert space H, a sequence x n converges to x (in norm topology) if and only if x and x n x. x n (a) Since L is a continuous linear operator defined on the hole of X, L is ell-defined on the hole of Y. In other ords, for any y Y, e have L y X. So So Lx n Lx. Lx n, y = x n, L y n x, L y = Lx, y. (b) If x n x in the norm topology, then x n x since the norm topology is stronger than the eak topology, and x n x since the norm function is continuous ith respect to the norm topology. Conversely, suppose x n x and x n x. Then (x n, x) (x, x) = x 2 since the function L x : H F, y (y, x) is a linear functional, and (x, x n ) x 2 since (x, x n ) = (x n, x). So e get x n x 2 = (x n x, x n x) = x n 2 (x, x n ) (x n, x) + x 2 n. So x n x in norm topology. 7. Let X = l be the Banach space of all bounded sequences x = (a 1, a 2, a 3, ) of real numbers, ith norm x = sup n a n. Prove: (a) There exists L X ith L = 1 such that if x = (a 1, a 2, a 3, ) X and lim a n = a exists, then L(x) = a. 6

7 (b) There is no sequence y = (b 1, b 2, b 3, ) l 1 such that Lx = n=1 a nb n for all x X. (a) Let Y X be the subset { Y = x = (a 1, a 2, a 3, ) l } lim a n = a exists. n Then Y is obviously a subspace of X. Define a linear functional l on Y by l : Y F, l((a 1, a 2, )) = lim n a n. The linearity of l is also trivial to check. It is continuous since l(x) = lim a n sup a n = x. n Note that if e choose x = (a, a, ) be the constant sequence, then x Y, x = a and l(x) = a. So as a continuous linear functional on Y, e have l = 1. By the Hahn-Banach theroem, there exists L X ith L = l = 1 that extends l, i.e. if x = (a 1, a 2, a 3, ) l and lim a n = a exists, then L(x) = a. (b) Suppose such a y = (b 1, b 2, ) l 1 exists. For each k e let x k = (,,, 1,, ), here 1 is at the kth entry and is everyhere else. Then x k Y and l(x k ) = for each k. So Lx k =. But Lx k = n a nb n = b k. So b k = for each k. So y = (,, ) and thus Lx = for each x. This contradicts ith the fact that L = 1. n 7

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