Consider $E$ the space of all the functions defined on $\Bbb R$ which admit a representation of the form $x(t) = \sum_{r \in \Bbb R}^* c_r e^{irt}$, where $\sum^*$ indicates that only a finite number of the $c_r$ is non-zero. Then, we have this inner product: $$\langle x,y \rangle = \lim_{T\to \infty} \frac{1}{2T} \int_{-T}^T x(t)\overline{y(t)}\,{\rm d}t.$$ I managed to check that this really is an inner product on $E$. Also, I know that $\{ e^{irt} \mid r \in \Bbb R\}$ is an orthonormal family. I am supposed to check that $(E, \langle \cdot,\cdot \rangle)$ is not a Hilbert space. I have something here, but I am not $100\%$ sure. For each $n$, take $$x_n(t) = \sum_{k=1}^n \frac{1}{k}e^{ikt}$$ Clearly $x_n$ is in $E$ for all $n$. To check that the sequence is Cauchy, I computed $$\|x_{n+p}-x_n\| = \sqrt{ \sum_{k=n+1}^{n+p}\frac{1}{k^2} },$$ which seems to go to zero as $n\to +\infty $, whatever may $p$ be, since $$(n\to +\infty) \iff (n+p \to +\infty)\iff (n+1 \to +\infty).$$ Also, if $x_n$ is to converge, it will be to $$ t \mapsto \sum_{n \geq 1} \frac{1}{n}e^{int},$$ although I'm not completely sure on this. My intuition with this inner product is near zero.

My main problem is:

The family of the $e^{irt}$ being orthonormal really ensures the uniqueness of the $c_r$, so that the above function is not in $E$?

  • $\begingroup$ I think you made a mistake here : $\|x_{n+p}-x_n\| = \sqrt{ \sum_{k=n+1}^{n+p}\frac{1}{k^2} }$. Then $(x_n)$ is a Cauchy sequence, but converge to an element wich is not in $E$. $\endgroup$ – Patissot May 3 '15 at 16:31
  • $\begingroup$ Yes, I got that result, it was a typo, sorry. Thanks for pointing. $\endgroup$ – Ivo Terek May 3 '15 at 16:33
  • $\begingroup$ If $\sum c_re_r(t)=\sum c_r'e_r(t)$ (where $e_r(t)=e^{irt}$) then $\langle \sum c_re_r, e_i \rangle =c_i=\langle \sum c_re_r, e_i \rangle =c_i'$ so the coefficients $(c_r)$ are uniquely determined. $\endgroup$ – Patissot May 3 '15 at 16:38
  • $\begingroup$ Ok, guess this settles it, then. Thanks. You can upgrade your comments to an answer if you want to :) $\endgroup$ – Ivo Terek May 3 '15 at 16:40

Assume that for all $t$ we have $\sum _{r\geq 1} c_re_r(t)=\sum _{r\geq 1} c_r'e_r(t)$. Then we have $\langle \sum _{r\geq 1}c_re_r, e_i \rangle =c_i=\langle \sum _{r\geq 1}c_re_r, e_i \rangle =c_i'$ for all $i\geq 1$ (by linearity and orthogonality). This show uniqueness of coefficients.

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