Mathematisches Institutder Universitat MunchenProf. L. ErdosDr. T. Ø. Sørensen

Functional AnalysisExercises from class

Winterterm 2003/04January 22 & 26, 2004

Exercise 64: Let X and Y be normed spaces. Let T : D(T ) → Y be a linearoperator (here, D(T ) ⊂ X is the domain of T - it may be that D(T ) 6= X).The operator T is called closed if the graph of T defined by

Γ(T ) ={〈x, y〉 ∈ X × Y

∣∣ x ∈ D(T ), y = Tx}⊂ X × Y

is a closed subset of (X × Y, ‖ · ‖X×Y ) with ‖〈x, y〉‖X×Y := ‖x‖X + ‖y‖Y .

a) Let T : D(T ) → Y be closed. Show that the null-space (that is, the set{x ∈ D(T ) |Tx = 0Y }) is a closed subspace of X.

b) Let T : D(T ) → Y be defined by Tx = 0Y . Is T closed?

Exercise 65: Extend the Hellinger-Toeplitz theorem to include pairs ofoperators A, B satisfying (Ax, y) = (x, By) - that is, prove the following:Let A, B be two everywhere-defined linear operators on a Hilbert space Hwith (x, Ay) = (Bx, y) for all x, y ∈ H. Then both A and B are bounded.

Exercise 66: Let {xn}n∈N be a weakly convergent sequence in a finite-dimensional normed space. Show that {xn}n∈N is norm-convergent.

Exercise 67: Let the functions {xn}n∈N, {yn}n∈N ⊂ C([0, 1]) be given by

xn(t) =

2nt, t ∈ [0, 1

2n),

2− 2nt, t ∈ [ 12n

, 1n)

0, t ∈ [ 1n, 1].

, yn(t) =

{1− nt, t ∈ [0, 1

n),

0, t ∈ [ 1n, 1].

Draw the graphs of xn and yn, and show that xn ⇀ 0, but yn 6⇀ 0. (Hints:For xn, assume not, and construct a subsequence {xnk

}k∈N such that 0 ≤∑K1 xnk

≤ 4 for all K. For yn, find a suitable linear functional).

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Mathematisches Institutder Universitat MunchenProf. L. ErdosDr. T. Ø. Sørensen

Functional AnalysisExercises from class

Winterterm 2003/04January 22 & 26, 2004

Exercise 68: Proof the following:Mazur’s theorem (Hilbert space case): Let H be a Hilbert space. As-sume {xn}n∈N ⊂ H satisfies xn ⇀ x for some x ∈ H. Then there exists asequence of convex linear combinations,

yn =

N(n)∑j=1

λ(n)j xj (λ

(n)j ≥ 0 ,

N(n)∑j=1

λ(n)j = 1)

such that ‖yn − x‖ → 0.(Hint: Take the mean of a suitable subsequence).

Exercise 69: Definition: A sequence {kn}n∈N ⊂ C([−1, 1]) is said to tendto the δ-function if for all f ∈ C([−1, 1]),

limn→∞

∫ 1

−1

f(t)kn(t) dt = f(0). (1)

Prove the following:Theorem (Toeplitz): The sequence {kn}n∈N ⊂ C([−1, 1]) tends to the δ-function in the sense of (1) if and only if it satisfies the following conditions:

(i) limn→∞∫ 1

−1kn(t) dt = 1.

(ii) For every C∞-function g whose support does not contain 0,

limn→∞

∫ 1

−1

g(t)kn(t) dt = 0. (2)

(iii) There is a constant c such that for all n ∈ N,∫ 1

−1

|kn(t)| dt ≤ c. (3)

(Hint: assume first that f(0) = 0, and approximate f by a C∞-functionuniformly on [−1, 1].

Thomas Østergaard Sørensen

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