Can someone help me with this question?

Let $\ell^2$ be the space of complex sequences $\{ x_1, x_2, \ldots \}$ that $\sum_{n=1}^{\infty} \lvert x_n \rvert ^2 < \infty$. If $\mu$ be Counting Measure on $\mathbb{N}$, then $\ell^2$ is $L^2(\mathbb{N}, \mu)$, and thus a Hilbert space.

Now suppose that $\{ a_{ij} \}$ is a complex multi index sequence such that $\sum_{i, j} \lvert a_{ij} \rvert ^2 < \infty$. Thus, we can define $T : \ell^2 \rightarrow \ell^2$ \begin{equation} \{ x_n \}_{n=1}^{\infty} \overset{T}{\mapsto} \{ x'_n \}_{n=1}^{\infty}, \qquad x'_n = \sum_{k=1}^{\infty} a_{nk} x_k \ . \end{equation} Show that

  1. $T$ is well defined, means that we have $\{ x'_n \}_{n=1}^{\infty}$,
  2. $T$ is Bounded, and
  3. $T$ is Compact.

Thanks in advance.

Edit: I already proved number 1 and 2.

For number 2, I proved that if $\sum_{n=1}^{\infty} \lvert x_n \rvert ^2 < 1$, then we have $\sum_{n=1}^{\infty} \lvert \sum_{k=1}^{\infty} a_{nk} x_k \rvert ^2 < \infty$.

I couldn't prove number 3 with definition of compact operators, or other equivalent definitions, such as being limit of finite-rank operators.

  • $\begingroup$ What are your thoughts on the problem? What have you tried? $\endgroup$ – msteve Jun 17 '15 at 18:19
  • $\begingroup$ $\sum_{i,j}|x_n|^2<\infty$ doesn't make sense, check your indexes of summation !! $\endgroup$ – Alonso Delfín Jun 17 '15 at 18:28
  • $\begingroup$ @msteve: I edited my post and added what you want. $\endgroup$ – Hossein Moradi Jun 17 '15 at 18:40
  • $\begingroup$ @Alonso Delfin: That was typing error, I correct that. Thanks. $\endgroup$ – Hossein Moradi Jun 17 '15 at 18:41
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    $\begingroup$ For 3 using that T is a limit of finite rank bounded operators works! $\endgroup$ – Alonso Delfín Jun 17 '15 at 19:04

Doing only 3).

For each $n \in \mathbb{N}$ we define the operators $T_n: \ell^2 \to \ell^2$ as follows, for each $x=\{ x_j\}_j \in \ell^2$ put $$ T_n(x) = \left( \sum_{k=1}^\infty a_{1k} x_k, \cdots, \sum_{k=1}^\infty a_{nk} x_k, 0 ,\cdots \right) $$ By Hölder, for any $j\in \mathbb{N}$ $$ \left|\sum_{k=1}^\infty a_{jk} x_k\right| \leq \left( \sum_{k=1}^\infty \left|a_{jk}\right|^2 \right)^{1/2}\left( \sum_{k=1}^\infty|x_k|^2 \right)^{1/2} = \|x\|_2\left( \sum_{k=1}^\infty \left|a_{jk}\right|^2 \right)^{1/2} $$ hence, $$ \|T_n(x)\|^2_2 = \sum_{j=1}^{n} \left|\sum_{k=1}^\infty a_{jk} x_k\right|^2 \leq \|x\|^2_2 \sum_{j=1}^{n} \sum_{k=1}^\infty \left|a_{jk}\right|^2 $$ Thus each $T_n$ is bounded with norm $\| T_n \|^2\leq \sum_{j=1}^\infty\sum_{k=1}^\infty \left| a_{jk}\right|^2 < \infty$. Since $dim(T_n(\ell^2))=n$, all the $T_n$ are bounded operators of finite rank, and therefore they are compact operators.

Lets now consider $x \in \ell^2$ with $\| x\|_2=1$, then $$ \|(T-T_n)(x)\|^2_2 = \sum_{j=n+1}^{\infty} \left|\sum_{k=1}^\infty a_{jk} x_k\right|^2 \leq \|x\|_2^2\sum_{j=n+1}^{\infty} \sum_{k=1}^\infty \left|a_{jk}\right|^2 =\sum_{j=n+1}^{\infty} \sum_{k=1}^\infty \left|a_{jk}\right|^2 $$ so, by the definition of norm of an operator, we get $$ \|T-T_n\| = \sup_{\|x\|_2=1} \left\{\|(T-T_n)(x)\|_2 \right\} \leq \left( \sum_{j=n+1}^{\infty} \sum_{k=1}^\infty \left|a_{jk}\right|^2 \right)^{1/2} \underset{n \to \infty}{\longrightarrow} 0 $$ Therefore $\| T - T_n\| \to 0$ as $n \to \infty$. Hence being $T$ limit of compact operators , $T$ is indeed compact, since the space of compact operators is closed in the space of bounded ones.

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    $\begingroup$ Very nice. Thanks a lot for your help. $\endgroup$ – Hossein Moradi Jun 17 '15 at 19:37
  • $\begingroup$ @HosseinMoradiDavijani No problem I am glad to help !! And welcome to MSE!! $\endgroup$ – Alonso Delfín Jun 17 '15 at 19:43

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