Question #163044

Prove that every (non-zero) Hilbert space H has an orthonormal basis.


1
Expert's answer
2021-02-24T07:31:07-0500

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,...}B = \{v_1, v_2, ...\} and x be an element of H. We'll show that the series xn=k=1nx,vkvkx_n=\sum_{k=1}^n\langle x, v_k\rangle v_k converges to x.

Let ε>0\varepsilon>0 and xk=1nλkvk<ε|x-\sum_{k=1}^n\lambda_kv_k|<\varepsilon (it is possible to find such linear combination, since span B is dense in H). Then xxn,vk=x,vkxn,vk=x,vkx,vk=0\langle x-x_n,v_k\rangle=\langle x,v_k\rangle-\langle x_n,v_k\rangle=\langle x,v_k\rangle-\langle x,v_k\rangle=0, hence xxn,xnk=1nλkvk=0\langle x-x_n,x_n-\sum_{k=1}^n\lambda_kv_k\rangle=0 and, by Pythagorean theorem, xk=1nλkvk2=(xxn)+(xnk=1nλkvk)2=xxn2+xnk=1nλkvk2|x-\sum_{k=1}^n\lambda_kv_k|^2=|(x-x_n)+(x_n-\sum_{k=1}^n\lambda_kv_k)|^2 = |x-x_n|^2+|x_n-\sum_{k=1}^n\lambda_kv_k|^2 , which implies xxn2=xk=1nλkvk2xnk=1nλkvk2<ε2|x-x_n|^2=|x-\sum_{k=1}^n\lambda_kv_k|^2-|x_n-\sum_{k=1}^n\lambda_kv_k|^2<\varepsilon^2 and xxn<ε|x-x_n|<\varepsilon.

For instance, if we take k=1nλkvk=xn1\sum_{k=1}^n\lambda_kv_k=x_{n-1} , then we have xxn2=xxn12xnxn12xxn12|x-x_n|^2=|x-x_{n-1}|^2-|x_n-x_{n-1}|^2\leq |x-x_{n-1}|^2 which implies xxnxxn1|x-x_n|\leq|x-x_{n-1}|.

Therefore, the sequense xxn|x-x_n| is monotonous, non-negative and starting from some n is lesser than ε\varepsilon. This means xnx_n converges to x and, therefore, B is an orthonormal basis in H.


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