4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ For 3 using that T is a limit of finite rank bounded operators works! $\endgroup$ – Alonso Delfín Jun 17 '15 at 19:04
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

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.