# Compactness and boundedness of multiplication operator .

I need help with the following problem,

If $1\le p < \infty$ , $(m_k)_{k\in \mathbb N}$and $K \subset \ell^p(\mathbb N)$ and

$T: \ell^p(\mathbb N) \to \ell^p(\mathbb N)$ , define multiplication operator : $(Tx_k)_{k\in \mathbb N} = (m_kx_k)_{k\in N}$ How do i show If $K$ is relative compact then $K$ is bounded and $\lim_{n\to +\infty} \sum_{i=n}^\infty |x_i|^p = 0$ uniformly for all $(x_i)_{i\in \mathbb N} \in K$ , and

$T$ is compact if and only if $(m_k)_{k\in \mathbb N} \in c_0(\mathbb N)$.

What can i say about the spectrum of the operator ?

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Sorry but is there any relation between $K$ and $T$? – Tomás Jan 16 '13 at 21:23
@Tomás : i guess now it makes sense . does it ?? – Theorem Jan 16 '13 at 21:29

If $K$ is relative compact, then $\overline{K}$ is compact which implies that $\overline{K}$ is bounded. Because $K\subset\overline{K}$ we conclude that $K$ is bounded.

Suppose on the contrary that the set $K$ does not have the property of uniformity, i.e. for given $\epsilon>0$ and for all $n$, we can find $x_n$ such that $$\sum_{i=n}^\infty |x_{i,n}|^p>\epsilon$$

By using the fact that $K$ is relative compact, you can extract a subsequence $x_n$ (not relabeled) such that $x_n\rightarrow x$. Can you arrive to a contradiction from here?

Now suppose that $T$ is compact, hence you have that if $u_n\rightarrow u$ weakly then $Tu_n\rightarrow Tu$ strongly. If you take the sequence $e_n$, what happens? On the other hand suppose that $m\in c_0$. Try to aproximate $T$ by a sequence of finite rank operators.

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The first reasoning is based on the boundedness of compact set right ? I don't know if its true in any arbitrary space . – Theorem Jan 16 '13 at 22:38
Any compact space in a metric space is bounded and closed. – Tomás Jan 16 '13 at 22:46
then i can extract a subsequence that converges outside the set $K$. which contradicts the relative compactness of $K$ right ? And i don't get the last two lines , if i take unit vectors $e_n$ i cannot say that it converges strongly . I am not able to understand the last two lines . – Theorem Jan 17 '13 at 16:04
For your first question the answer is: $x_n\rightarrow x$ where $x\in \overline{K}$. This implies that the tail of the sequence $x=(x_1,...,x_n,..)$ converges to zero which is a contradictions with the inequality. For you second question: Note that $e_n\rightarrow 0$ weakly. Then $Te_n\rightarrow T0$ strongly. What is $Te_n$? – Tomás Jan 17 '13 at 16:09
Yes , got my first doubt cleared . $Te_n$ converges strongly to $0$ . But how does that help me to see that $m_k \in c_0$ . I think i am not understanding something here . – Theorem Jan 17 '13 at 16:22