Let $\ell^2$ be the set of real number sequences $\{a_n\}$ such that $\sum a_n^2 <\infty$. Let $\langle a_n,y\rangle \rightarrow \langle a,y\rangle$ for some $a\in \ell^2$ and for all $y\in \ell^2$.

Would you help me to prove that $\{\| a_n\|\}$ bounded and for a fixed $i$, $a_i^{(n)} \to a^{(n)}$ where $x^{(n)}$ denote the $n$-component of $x$.


Answer to the first part of your question easily follows from uniform boundedness principle.

For each $n\in\mathbb{N}$ we define linear operator (in fact linear functional) $$ T_n:\ell_2\to\mathbb{C}:y\mapsto\langle y, a_n\rangle $$ Let's compute their norm. From Cauchy Schwarz inequality it follows that for all $y\in\ell_2$ we have $$ |T_n(y)|=|\langle y,a_n\rangle|\leq\Vert a_n\Vert\Vert y\Vert $$ hence $$ \Vert T_n\Vert\leq\Vert a_n\Vert\tag{1} $$ On the other hand $\Vert T_n(a_n)\Vert=\langle a_n,a_n\rangle=\Vert a_n\Vert^2$, so $$ \Vert T_n\Vert\geq \Vert T(a_n)\Vert/\Vert a_n\Vert=\Vert a_n\Vert\tag{2} $$ From $(1)$ and $(2)$ it follows that $$ \Vert T_n\Vert=\Vert a_n\Vert\tag{3} $$ Now fix $y\in \ell_2$, then by assumption $\lim\limits_{n\to\infty} T_n(y)=\lim\limits_{n\to\infty}\langle y,a_n\rangle=\langle a,y\rangle$. This means that the sequnce $\{T_n(y):n\in\mathbb{N}\}$ is convergent and, as the consequence is bounded. Thus for each $y\in\ell_2$ we have $$ \sup\{| T_n(y)|:n\in\mathbb{N}\}<+\infty $$ Then from uniform boundedness principle it follows that $$ \sup\{\Vert T_n\Vert:n\in\mathbb{N}\}<+\infty $$ Using $(3)$ we conclude that $\{\Vert a_n\Vert:n\in\mathbb{N}\}$ is bounded. In fact there is more general and much more short proof, but it involves additional usage of standard embedding into the double dual which is isometric for normed spaces.

As for the second part of the question. It is much more easier. Fix $i\in\mathbb{N}$ and consider vector $y_i\in\ell_2$ such that $$ y_i(k)=\begin{cases}1&\quad\text{ if }\quad k=i\\0&\quad\text{ if }\quad k\neq i\end{cases} $$ Obviously $$ \langle y_i,a_n\rangle=\sum\limits_{k=1}^\infty a_n(k)\overline{y_i(k)}=a_n(i)\\ \langle y_i,a\rangle=\sum\limits_{k=1}^\infty a(k)\overline{y_i(k)}=a(i) $$ Since for all $y\in\ell_2$ we have $\lim\limits_{n\to\infty}\langle y,a_n\rangle=\langle y,a\rangle$ then $$ \lim\limits_{n\to\infty} a_n(i)=\lim\limits_{n\to\infty}\langle y_i,a_n\rangle=\langle y_i,a\rangle=a(i) $$ This s exactly what we wanted to prove.


There is NO absolutely elementary proof for your question without appealing to any theorem. Want you or not, you MUST use completeness of $\ell_2$. For incomplete inner product spaces the first result doesn't hold. For counterexample see answer to the question What facts about the weak topology fail in spaces that aren't Banach?. Well, one can modify my proof so it won't use uniform boundedness principle, but this modification will significantly enlarge the proof and in fact one WILL prove uniform boundedness principle for this particular case, but without explicitly giving things their names.

  • $\begingroup$ Forgive me: it seems to me to understand that $T_n$ is conjugate-linear because, if I correctly understand, $T_n(\alpha y)=\bar{\alpha}T_n(y)$. So I think that the uniform boundedness principle also applies to conjugate-linear functionals: have I understood? $\infty$ thanks!!! $\endgroup$ – Self-teaching worker Sep 12 '14 at 18:03
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    $\begingroup$ @DavideZena, nice catch! I've fixed the proof $\endgroup$ – Norbert Sep 12 '14 at 19:00

To show that a weakly convergent sequence is norm-bounded, usually the uniform boundedness principle is applied, see also this link. The convergence in the $n$-the component follows immediately, if you choose $y$ as the $n$-th unit vector in $l^2$.

  • $\begingroup$ Is there elementary proof Vobo? $\endgroup$ – beginner Sep 17 '12 at 23:14
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    $\begingroup$ @beginner, I think you can't find elementary proof without applying to some deep theorem. If you don't want to use any theorems and want to prove boundedness by hand then you need take into account completeness of $\ell^2$. In fact you will repeat the proof of uniform boundedness principle, but for your particular case. $\endgroup$ – Norbert Sep 18 '12 at 4:20
  • $\begingroup$ @Norbert: I would agree, I haven't seen this proven without the main ideas of the uniform boundedness principle. $\endgroup$ – Vobo Sep 18 '12 at 16:25

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