We recall that a pre-Hilbert space $H$ is a normed vector space whose norm is derived from an inner product denoted $\langle .,. \rangle$. We call Hilbert basis of $H$ any family $B = (b_i)_{i \in \mathbb{N}}$ such that: (i) the family is orthonormal: for all $i$ and $j$ in $\mathbb{N}$, $\langle b_i, b_j \rangle = 1$ if $i = j$ and $0$ otherwise. (ii) every element $x$ of $H$ can be written: $x = \sum_{i=0}^{+\infty} \langle x, b_i \rangle b_i$, that is $$\lim_{N \rightarrow +\infty} \left\| x - \sum_{i=0}^{N} \langle x, b_i \rangle b_i \right\| = 0$$ Show that if $B = (b_i)_{i \in \mathbb{N}}$ is a Hilbert basis of $H$, then $$\forall x \in H, \quad \|x\|^2 = \sum_{i=0}^{+\infty} |\langle x, b_i \rangle|^2$$
We recall that a pre-Hilbert space $H$ is a normed vector space whose norm is derived from an inner product denoted $\langle .,. \rangle$. We call Hilbert basis of $H$ any family $B = (b_i)_{i \in \mathbb{N}}$ such that:
(i) the family is orthonormal: for all $i$ and $j$ in $\mathbb{N}$, $\langle b_i, b_j \rangle = 1$ if $i = j$ and $0$ otherwise.
(ii) every element $x$ of $H$ can be written: $x = \sum_{i=0}^{+\infty} \langle x, b_i \rangle b_i$, that is
$$\lim_{N \rightarrow +\infty} \left\| x - \sum_{i=0}^{N} \langle x, b_i \rangle b_i \right\| = 0$$
Show that if $B = (b_i)_{i \in \mathbb{N}}$ is a Hilbert basis of $H$, then
$$\forall x \in H, \quad \|x\|^2 = \sum_{i=0}^{+\infty} |\langle x, b_i \rangle|^2$$