Since H is separable, there exists a countable dense set M, a subset of H. We may select a linear independent countable subset M' of M with the same span (which will be dense in H, since it contains M).
Applying to M' the Gram-Shmidt orthogonalization method, we obtain an orthonormal set B with span B is dense in H. We are to show that B is an orthonormal basis in H.
Let B={v1,v2,...} and x be an element of H. We'll show that the series xn=∑k=1n⟨x,vk⟩vk converges to x.
Let ε>0 and ∣x−∑k=1nλkvk∣<ε (it is possible to find such linear combination, since span B is dense in H). Then ⟨x−xn,vk⟩=⟨x,vk⟩−⟨xn,vk⟩=⟨x,vk⟩−⟨x,vk⟩=0, hence ⟨x−xn,xn−∑k=1nλkvk⟩=0 and, by Pythagorean theorem, ∣x−∑k=1nλkvk∣2=∣(x−xn)+(xn−∑k=1nλkvk)∣2=∣x−xn∣2+∣xn−∑k=1nλkvk∣2 , which implies ∣x−xn∣2=∣x−∑k=1nλkvk∣2−∣xn−∑k=1nλkvk∣2<ε2 and ∣x−xn∣<ε.
For instance, if we take ∑k=1nλkvk=xn−1 , then we have ∣x−xn∣2=∣x−xn−1∣2−∣xn−xn−1∣2≤∣x−xn−1∣2 which implies ∣x−xn∣≤∣x−xn−1∣.
Therefore, the sequense ∣x−xn∣ is monotonous, non-negative and starting from some n is lesser than ε. This means xn converges to x and, therefore, B is an orthonormal basis in H.
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