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 $a_i \rightarrow 0$.

  • $\begingroup$ Thank you William, but would you give me some more details about the equality $\|f \|^2 = \sum | a_i |^2 ?$ $\endgroup$ – Ann Jul 30 '12 at 4:20
  • $\begingroup$ @Ann It is called Parseval's Identity. You can find its proof in any Functional Analysis Textbook, for example Real Analysis by Stein page 165. $\endgroup$ – William Jul 30 '12 at 4:26
  • $\begingroup$ She can even google it. There are thousands of pages dealing with it. $\endgroup$ – DonAntonio Jul 30 '12 at 5:25
  • $\begingroup$ Since $(e_n)_{n\in\mathbb N}$ is only given to be an orthonormal system, not a basis, one can in fact only assert $||f||^2 \geq \sum_{i = 1}^\infty |a_i|^2$, but this suffices for the argument. And so it is Bessel's inequality rather than Parseval's identity. $\endgroup$ – Marc van Leeuwen Jul 30 '12 at 5:42
  • $\begingroup$ @MarcvanLeeuwen Thanks. I assumed that $\{e_n\}$ were an orthonormal basis. $\endgroup$ – William Jul 30 '12 at 6:13

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.


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