1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $a_i \rightarrow 0$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.