Parseval's Identity holds for all $x\in H$ implies $H$ is a Schauder basis

Question

Prove that any set $\{v_j\}_{j \in \mathbb{Z}}$ for which the Parseval identity $\|x\|^2=\sum_{j=1}^\infty |\langle v_j,x\rangle|^2$ holds for every $x \in H$ is a Schauder basis.

I know that a Schauder basis for $X$ if to each vector $x$ in the space there corresponds a unique sequence of scalars $\{c_1,c_2,\dots\}$ such that $x=\sum_{n=1}^\infty c_nx_n$.

The conclusion is obvious but I can't wrap my hea around how to use the Parseval Identity in the proof.

Answer 1

You have probably seen how Parseval's equality for a given $x$ is equivalent to convergence of $\sum_{j}\langle x,v_j\rangle v_j$, but have forgotten: $$ x = \left(x-\sum_{j=1}^{N}\langle x,v_j\rangle v_j\right)+\sum_{j=1}^{N}\langle x,v_j\rangle v_j $$ This is an orthogonal decomposition, meaning that the expression in parentheses on the right is orthogonal to the sum that is written after that. Therefore, by the Pythagorean Theorem, $$ \begin{align} \|x\|^{2} & = \left\|x-\sum_{j=1}^{N}\langle x,v_j\rangle v_j\right\|^{2} + \left\|\sum_{j=1}^{N}\langle x,v_j\rangle v_j\right\|^{2}\\ & = \left\|x-\sum_{j=1}^{N}\langle x,v_j\rangle v_j\right\|^{2} +\sum_{j=1}^{N}|\langle x,v_j\rangle|^{2} \end{align} $$ Now you can see that the sum on the far right converges to $\|x\|^{2}$ iff the first term on the right converges to $0$. In order words, Parseval's equality is equivalent to norm convergence of the vector sum to $x$. I've written this as an ordered sum, but the order does not matter. The uniqueness of the coefficients for any such sum converging to $x$ follows from orthogonality of the $v_j$.

Answer 2

So it is enough to show that if Parseval's Equality holds for all $x\in H$, then any $x\in H$ has a unique set of coefficients $c_n$.

Suppose Parseval's holds for $f,g\in H$ and $f\not=g$ but that they have the same coefficients: $$ f=\sum_{j=1}^\infty \langle f,v_j\rangle v_j \qquad g=\sum_{j=1}^\infty \langle g,v_j\rangle v_j, \qquad \langle f,v_j\rangle=\langle g,v_j\rangle. $$

Then $0\not=f-g\in H$ and \begin{align} f-g&=\sum_{j=1}^\infty \langle f-g,v_j\rangle v_j\\ 0\not=\|f-g\|^2&=\sum_{j=1}^\infty |\langle f-g,v_j\rangle|^2\\ &=\sum_{j=1}^\infty |\langle f,v_j\rangle-\langle g,v_j\rangle|^2\\ &=0, \end{align} which is a contradiction. Thus, the coefficients (relative to the orthonormal set $\{v_j\}$) are uniquely determined by $x\in H$. (Technically, this is true after we have identified functions that only differ on a set of measure zero.)