Let $(e_n)$ be an orthonormal sequence in an inner product space $X$. Then, for every $x \in X$, we have $$ \sum_{n=1}^\infty \vert \langle x, e_n \rangle \vert^2 \ \leq \ \Vert x \Vert^2.$$

Now $\ell^2$ is the inner product space of all sequences $x\colon=(\xi_n)$ of complex numbers such that $$\sum_{n=1}^\infty \vert \xi_n \vert < +\infty,$$ with the inner product defined by $$ \langle x, y \rangle \colon= \sum_{n=1}^\infty \xi_n \overline{\eta_n} \ \ \ \mbox{ for all } \ x \colon= (\xi_n), \ y \colon= (\eta_n) \in \ell^2. $$ So the norm on $\ell^2$ is given by $$\Vert x \Vert \colon= \sqrt{ \langle x, x \rangle} = \sqrt{ \sum_{n=1}^\infty \vert \xi_n \vert^2 }. $$

Now can we give an example of an orthonormal sequence $(e_n)$ in $\ell^2$ and an element $x \in \ell^2$ such that $$\sum_{n=1}^\infty \vert \langle x, e_n \rangle \vert^2 \ < \ \Vert x \Vert^2? $$

I've been trying different $x$ with the following orthonormal sequence: $$e_n \colon= (\delta_{nj}) \ \ \ \mbox{ for each } \ n= 1, 2, 3, \ldots, $$ where $$ \delta_{nj} \colon= \begin{cases} 1 \ & \mbox{ if } \ j =n ; \\ 0 \ & \mbox{ if } \ j \neq n. \end{cases} $$ But I've had little success so far.

However, if we takt a proper subsequence of this sequence, say the sequence $(e_{2n})$, then $x = (1/n)$ would do the job. Am I righrt?

Is there any $x$ that would work (i.e. give the strict inequality) even with the full sequence $(e_n)$?

  • $\begingroup$ Right. Also with $e_{2n}$ you could take $x = e_1.$ $\endgroup$ – zhw. Apr 28 '15 at 6:56

You're right : take an orthonormal basis, like the one you took, and remove elements. This will do the trick.

Here's the detailed thing. Let $(e_n)$ be the basis given by $e_n = \delta_{i,n}$. It can be easily shown it's a basis and it is clear that it's orthonormal. Now take the orthonormal family $(e_1,e_2)$. And then, take $x$ to be an element in $l^2$ whose third coordinate is nonzero, like $x = (1,1,1,0,0...)$. It is clear that $x$ is in $l^2$ and you can verify that $$ |\langle e_1,x \rangle|^2 + |\langle e_1,x \rangle|^2 = 1 + 1 = 2 $$ while $$||x||^2 = (1 + 1 + 1) = 3 $$

These kind of things will become clear when you'll go through Hilbert space theory. In fact, there are two well-known theorems : Bessel's theorem states that, for every orthonormal family $(e_i)_{i \in I}$ in a Hilbert space $H$, and for every $x \in H$, we have $$\sum_{i \in I}|\langle e_i, x \rangle |^2 \leq ||x||^2 $$ while in Fact, Parseval's formula states that if $(e_i)_{i \in I}$ is also a Hilbert basis, then we have equality. So the answer to your second question is no, as the family $(e_n)$ you gave is also a Hilbert basis of $l^2$, so for every $x$ we have the equality $$||x||^2 = \sum_{n=1}^{\infty} |\langle x, e_n \rangle |^2 $$ This can be easily checked, because $|\langle x, e_n \rangle |^2 = |x_i|^2$ as a consequence of the definition of $(e_n)$.


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