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MATH 675 – HOMEWORK 7
SOLUTIONS
Exercise 1. Prove the Polarization Identity: Let x, y ∈ V , an inner product space. If V is a vector space over R, then,
4 〈x, y〉 = ‖x + y‖^2 − ‖x − y‖^2.
If V is a vector space over C, then
4 〈x, y〉 = ‖x + y‖^2 − ‖x − y‖^2 − i(‖x + iy‖^2 − ‖x − iy‖^2 ).
(Hint: The proof is a computation involving the Parallelogram Law.)
For a real vector space,
‖x + y‖^2 − ‖x − y‖^2 = 〈x + y, x + y〉 − 〈x − y, x − y〉 = 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉 − (〈x, x〉 − 〈x, y〉 − 〈y, x〉 + 〈y, y〉) = 2 〈x, y〉 + 2〈y, x〉 = 4 〈x, y〉.
For a complex vector space, note that by the same calculation as above
‖x + y‖^2 − ‖x − y‖^2 = 2〈x, y〉 + 2〈y, x〉
and ‖x + iy‖^2 − ‖x − iy‖^2 = 2〈x, iy〉 + 2〈iy, x〉. Denoting by LHS the left hand side of the identity in question,
LHS = 2 〈x, y〉 + 2〈y, x〉 + 2i〈x, iy〉 + 2i〈iy, x〉 = 2 〈x, y〉 + 2〈y, x〉 + 2(−i^2 )〈x, y〉 + 2i^2 〈y, x〉 = 2 〈x, y〉 + 2〈y, x〉 + 2〈x, y〉 − 2 〈y, x〉 = 4 〈x, y〉
Exercise 2. If M is any subset of a Hilbert space H, define M ⊥^ by
M ⊥^ = {x ∈ H: 〈x, y〉 = 0 for all y ∈ M }.
Then the following hold.
(a) M ⊥^ is a closed subspace of H.
(b) If M is a subspace then (M ⊥)⊥^ = M , the closure of M.
(c) If M ⊆ N then N ⊥^ ⊆ M ⊥.
To see (a) note first that if x, y ∈ M ⊥^ and α, β scalars then given z ∈ M ,
〈z, αx + βy〉 = α 〈z, x〉 + β 〈z, y〉 = 0.
Hence αx+βy ∈ M ⊥^ and M ⊥^ is a subspace of H. To see that it is also closed, let {xn} ⊆ M ⊥^ and suppose that xn → x in H. For any z ∈ M , using the continuity of the inner product,
〈z, x〉 = 〈z, limn xn〉 = limn 〈z, xn〉 = limn 0 = 0.
Hence x ∈ M ⊥^ and M ⊥^ is closed. To see (b), let x ∈ M then for any z ∈ M ⊥, 〈x, z〉 = 0 which means that x ∈ (M ⊥)⊥^ and so M ⊆ (M ⊥)⊥. But by part (a) (M ⊥)⊥^ is closed and hence contains M since M is the smallest closed set containing M. Now suppose that x /∈ M. We must show that it is also outside (M ⊥)⊥^ which means we must find y ∈ M ⊥^ such that 〈x, y〉 6 = 0. To this end note that since M is closed, infz∈M ‖x − z‖ = d > 0 and by the Hahn-Banach theorem there is a continuous linear functional λ on H such that λ(x) = d > 0 and λ(z) = 0 for all x ∈ M. By the Riesz Representation Theorem every bounded linear functional μ on H has the form μ(x) = 〈y, x〉 for some y ∈ H. Hence there is a yλ such that λ(x) = 〈yλ, x〉 = d > 0 and 〈z, yλ〉 = 0 for all z ∈ M. To see (c) suppose that y ∈ N ⊥. Then for all x ∈ N , 〈x, y〉 = 0. But since M ⊆ N the same holds for every x ∈ M. Hence y ∈ M ⊥^ and N ⊥^ ⊆ M ⊥.
Exercise 3. Kolmogorov, Problem 5, p. 238.
Suppose that M is invariant under A, let x ∈ M ⊥^ and fix y ∈ M. Then
〈A∗x, y〉 = 〈x, Ay〉 = 0
since Ay ∈ M by the invariance of M under A. Hence A∗x ∈ M ⊥^ and M ⊥^ is invariant under A∗.
Exercise 4. Kolmogorov, Problem 6, p. 238.
Each of these is a straightforward verification.
