Let $H$ be an inner product space, $e_n\;(n \in \Bbb N )$ be the orthonormal system of $H$. Here I want to prove that for any $f \in H$ , $\langle f, e_n\rangle_H \to 0$ as $n \to \infty$.
The bracket means the inner product.
Let $H$ be an inner product space, $e_n\;(n \in \Bbb N )$ be the orthonormal system of $H$. Here I want to prove that for any $f \in H$ , $\langle f, e_n\rangle_H \to 0$ as $n \to \infty$.
The bracket means the inner product.
Let $a_n = \langle f, e_n \rangle$. One has that $||f||^2 = \sum_{i = 1}^\infty |a_i|^2$. Since $||f|| < \infty$, one must have that $|a_i| \rightarrow 0$. So $a_i \rightarrow 0$.
In response to Ann's doubt under William's answer: if $\,\{e_i\}\,$ is an orthonormal basis of $\,H\,$, then we have that $\forall\,\,f\in H\,\,,\,f=\sum_{i\in \Bbb N} \langle\,f,ei\,\rangle e_i\Longrightarrow ||f||^2=\sum_{i,j\in\Bbb N} \langle\,f,ei\,\rangle\langle\,f,e_j\,\rangle\langle\,e_i,e_j\,\rangle=$ $=\sum_{i\in\Bbb N}\langle\,f,e_i\,\rangle^2 $ Now, as $\,||f||^2<\infty\,$, the last series above converges and thus its general term's sequence converges to zero.