Answer to Question #163044 in Functional Analysis for Ebenezer

Question

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

Answer 1

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

Let $\varepsilon>0$ and $|x-\sum_{k=1}^n\lambda_kv_k|<\varepsilon$ (it is possible to find such linear combination, since span B is dense in H). Then $\langle x-x_n,v_k\rangle=\langle x,v_k\rangle-\langle x_n,v_k\rangle=\langle x,v_k\rangle-\langle x,v_k\rangle=0$, hence $\langle x-x_n,x_n-\sum_{k=1}^n\lambda_kv_k\rangle=0$ and, by Pythagorean theorem, $|x-\sum_{k=1}^n\lambda_kv_k|^2=|(x-x_n)+(x_n-\sum_{k=1}^n\lambda_kv_k)|^2 = |x-x_n|^2+|x_n-\sum_{k=1}^n\lambda_kv_k|^2$ , which implies $|x-x_n|^2=|x-\sum_{k=1}^n\lambda_kv_k|^2-|x_n-\sum_{k=1}^n\lambda_kv_k|^2<\varepsilon^2$ and $|x-x_n|<\varepsilon$.

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

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