The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 12

MATH3961: Metric Spaces Semester 1, 2012

Lecturer: Laurentiu Paunescu

1. Let V be a vector space over C, and let b be an hermitean inner product on V .Let β(u, v) = Re(b(u, v)) for all u, v ∈ V . Show that

(i) β is a real inner product (i.e., that β is real-valued and R-linear in each of itsvariables, β(u, u) > 0 for all u 6= 0 in V , and β(u, v) = β(v, u) for all u, v ∈ V );

(ii) β(u, iu) = 0 for all u ∈ V ; and

(iii) b(u, v) = β(u, v) + iβ(u, iv) for all u, v ∈ V .

Conversely, show that if β is a real inner product on a complex vector space V

such that (ii) holds then (iii) defines an hermitean inner product on V .

Solution: This is straightforward calculation.

2. Let H and H ′ be two Hilbert spaces, with inner products b and b′ and norms ||− ||and || − ||′, respectively. Let A : H → H ′ be a linear function.

Show that A is continuous if and only if A is bounded, i.e.,

A is continuous ⇔ ||A|| = sup{||A(v)||′ | ||v|| = 1} is finite.

Solution: If A is continuous there is a δ > 0 such that ||u−0|| < δ ⇒ ||Au−A0|| <1. Hence ||u|| < 1 ⇒ ||Au|| < 1

δand so ||A|| ≤ 1

δ. Thus A is bounded.

Conversely, if A is bounded and ε > 0 then ||u − v|| < ε

||A||⇒ ||Au − Av|| ≤

||A||.||u− v|| < ε, and so A is continuous.

3. Let A : H → H be a bounded linear operator (from H to itself).

Show that ||A|| = sup{|b(Au, v)| | ||u|| = ||v|| = 1}.

Solution: We may clearly assume A 6= 0, and so ||A|| > 0. The CBS inequalitygives |b(Au, v)| ≤ ||Au||.||v||, and so |b(Au, v)| ≤ ||A||.||u||.||v||.

Hence |b(Au, v)| ≤ ||A|| if ||u|| = ||v|| = 1.

On the other hand, there is a unit vector un such that ||Aun|| ≥ ||A|| − 1

n, for all

n > 1

||A||. Let vn = un

||Aun||be the unit vector in the same direction as un.

Then |b(Aun, vn)| = ||Aun|| > ||A|| − 1

n.

Hence sup{|b(Au, v)| | ||u|| = ||v|| = 1} = ||A||.

4. If v ∈ H define ℓv : H → C by ℓv(u) = b(u, v) for all u ∈ H.

Show that ℓv+w = ℓv + ℓw and ℓzv = zℓv, for all v, w ∈ H and z ∈ C.

Show that ||ℓv|| = ||v||.

Copyright c© 2012 The University of Sydney 1

(Thus ℓ : H → H∗ is an R-linear isometry, if H∗ has the operator norm, as recalledin exercise 2).

Solution: The first two assertions follow immediately from the linearity of b.

The final assertion is clearly true if v = 0, so we may assume v 6= 0.

The CBS inequality gives |ℓv(u)| = |b(u, v) ≤ ||u||.||v||, and so ||ℓv|| ≤ ||v||.

On the other hand, if w = v

||v||then ||w|| = 1 and ℓv(w) = b(w, v) = ||v||.

Hence ||ℓv|| = ||v||.

5. Show that the vector space dimension of ℓ∞2 is uncountable.

(The dimension of ℓ∞2 is clearly infinite. If ℓ∞2 had a countable vector space basisthe Gram-Schmidt process would give a countable orthonormal vector-space basis{vn}n≥1, say. It is easy to see that Σn≥1

1

nvn is in ℓ∞2 but is not a finite linear

combination of the vns.)

Solution: The solution is outlined in the question.

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