# prove that a complete orthonormal sequence is an orthonormal basis

From book "Introduction to Hilbert spaces with applications" by Lokenath Debnath, and Piotr Mikusinski

(Complete orthonormal sequence) An orthonormal sequence $$(x_n)$$ in an inner product space $$E$$ is said to be complete if for every $$x \in E$$ we have $$x = \sum_{n=1}^\infty \langle x, x_n \rangle x_n.$$

(Orthonormal basis) An orthonormal system $$B$$ in an inner product space $$E$$ is called an orthonormal basis if every $$x \in E$$ has a unique representation $$x = \sum_{n=1}^\infty \alpha_n x_n,$$ where $$\alpha_n \in \mathbb{C}$$ and $$x_n$$'s are distinct elements of $$B.$$

theorem Let $$(x_n)$$ be an orthonormal sequence in a Hilbert space $$H,$$ and let $$(\alpha_n)$$ be a sequence of complex numbers. Then the series $$\sum_{n=1}^\infty \alpha_n x_n$$ converges if and only if $$\sum_{n=1}^\infty |\alpha_n|^2 < \infty$$ and in that case $$\begin{eqnarray} \label{orthonormal_sequence_H_series_converges_iff_coefficients_in_l2_equality}\left \lVert \sum_{n=1}^\infty \alpha_n x_n \right \rVert^2 = \sum_{n=1}^\infty |\alpha_n|^2. \end{eqnarray}$$

To show that a complete orthonormal sequence $$(x_n)$$ is an orthonormal basis. It suffices to show the uniqueness. Indeed, if $$x = \sum_{n=1}^\infty \alpha_n x_n \ \text{and} \ x = \sum_{n=1}^\infty \beta_n x_n,$$ then $$0 = \|x-x\|^2 = \|\sum_{n=1}^\infty \alpha_n x_n - \sum_{n=1}^\infty \beta_n x_n\|^2 = \|\sum_{n=1}^\infty (\alpha_n - \beta_n) x_n\|^2 = \sum_{n=1}^\infty |\alpha_n - \beta_n|^2$$ by the last theorem. This means that $$\alpha_n = \beta_n$$ for all $$n \in \mathbb{N},$$ proving the uniqueness.

My question is how he use that the sequence is a complete orthonormal sequence in his proof and how he can make use of the last theorem.

My proof is if $$(x_n)$$ is a complete orthonormal sequence then for every $$x \in H,$$ $$x = \sum_{n=1}^\infty \langle x, x_n \rangle x_n$$. To prove that $$(x_n)$$ is an orthonormal basis, it sufficies to show that $$\langle x, x_n \rangle$$ is the only possible coefficients for the expansion of $$x.$$ Suppose that $$x = \sum_{n=1}^\infty \alpha_n x_n$$ then we want to prove that $$\alpha_n = \langle x, x_n \rangle$$. does this work?? $$\langle x, x_n \rangle = \langle \sum_{k=1}^\infty \alpha_k x_k, x_n\rangle = \sum_{k=1}^\infty \alpha_k\langle x_k, x_n\rangle = \alpha_n$$