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A vector 0 = x ∈ X is said to be orthogonal to the vector 0 = y ∈ X if (x, y) = 0, x and y are called orhogonal vectors, denoted by x ⊥ y. If A, B ⊂ X are non-empty subsets of X then a. b. x ⊥ A, if (x, y) = 0 for each y ∈ A, A ⊥ B, if (x, y) = 0 if x ⊥ y for each x ∈ A and y ∈ B. 3, called the orthogonal sum of A and B. All the elements of X, which stay orthogonal to some non-empty subset A ⊂ X is called the orthoplement of A. )) be an Inner Product Space and let A be an nonempty subset of X, then A⊥ = {x ∈ X | (x, y) = 0 for every y ∈ A} is called the orthoplement of A.

S ∈{S } is an orthonormal set and an upper bound of {S }. Thus by Zorn’s Lemma, there exists a maximal element S0 of {S}. S ⊆ S0 and because of it’s maximality, S0 is an orthogonal base of H. There exists an orthonormal base S0 of a Hilbert Space H. This orthogonal base S0 can be used to represent elements f ∈ H, the so-called Fourier expansion of f . With 63 the help of the Fourier expansion the norm of an element f ∈ H can be calculated by Parseval’s relation . 8 Let S0 = {eα | α ∈ Λ} be an orthonormal base of a Hilbert Space H.

Proof a. Let x, y ∈ A⊥ and α ∈ K, then (x + α y, z) = (x, z) + α (y, z) = 0 for every z ∈ A. Hence A⊥ is a linear subspace of X. Remains to prove: A⊥ = A⊥ . (⇒) The set A⊥ is equal to A⊥ unified with all its accumulation points, so A⊥ ⊆ A⊥ . (⇐) Let x ∈ A⊥ then there exist a sequence {xn } in A⊥ such that limn→∞ xn = x. Hence (x, z) = lim (xn , z) = 0, n→∞ b. c. d. for every z ∈ A. ) So x ∈ A⊥ and A⊥ ⊆ A⊥ . If x ∈ A ∩ A⊥ = ∅ then x ⊥ x, so x = 0. If x ∈ A, and x ⊥ A⊥ means that x ∈ (A⊥ )⊥ , so A ⊂ A⊥⊥ .