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I have to check that the following operator $T$ is compact:

Define $T:l^{2} \to {l^2}$ by $Tx=y=(\eta_{j})$, where $x=(\zeta_{j})$ and

$(\eta_{j})=\sum_{k=1}^\infty \alpha_{jk} \zeta_{j}$ and $\sum_{j=1}^\infty \sum_{k=1}^\infty |\alpha_{jk}|^{2} \lt \infty$

Show that $T$ is compact.

My attempt:

I constructed a sequence of operators,$(T_{n})$,

$T_{n}:l^{2} \to {l^2}$ defined by,

$T_{n}x=(\sum_{k=1}^\infty \alpha_{1k} \zeta_{1},\sum_{k=1}^\infty \alpha_{2k} \zeta_{2},.......,\sum_{k=1}^\infty \alpha_{nk} \zeta_{n},0,0,....) $ and then i proved that each $T_{n}$ is bounded and the range of $T_{n}$ is finite and therefore each $T_{n}$ is a compact operator.

Then I considered,

\begin{align} ||(T_{n}-T)x||^{2}& = ||(\sum_{k=1}^\infty \alpha_{(n+1)k} \zeta_{n+1},\sum_{k=1}^\infty \alpha_{(n+2)k} \zeta_{n+2},.....)||^{2}\\ & = \sum_{j=n+1}^\infty \sum_{k=1}^\infty |\alpha_{jk} \zeta_{j}|^{2}\\ &= \sum_{j=n+1}^\infty \sum_{k=1}^\infty |\alpha_{jk}|^{2} |\zeta_{j}|^{2}\\ \end{align}

Now how do i use the condition,$\sum_{j=1}^\infty \sum_{k=1}^\infty |\alpha_{jk}|^{2} \lt \infty$ to prove that $T$ is compact? I know that once I get that $(T_{n})$ is uniformly operator convergent then compactness will follow, but how do i get that $(T_{n})$ converges to $T$ from here?

Also in a subpart of this question I have to illustrate using an example that that the condition given in the above problem is sufficient for compactness but not necessary, I don't quite understand what am i required to do here ?

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2 Answers 2

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For the first part, use the fact that for every $\epsilon > 0, \exists N_0 \in \mathbb{N}$ such that $$ \sum_{j=n+1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{k,j}|^2 < \epsilon \quad\forall n\geq N_0 $$ (because the tail of a convergent series goes to zero).

For the second part, take $$ \alpha_{k,j} = \begin{cases} \frac{1}{k^{1/2}} &: k=j \\ 0 &: \text{ otherwise} \end{cases} $$ and check that the corresponding operator is compact even though the square-summability condition is not satisfied.

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  • $\begingroup$ nice example for the second part $\endgroup$ Commented Sep 5, 2015 at 3:37
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For the first Q : You have a typo in your 3rd line, def'n of $(\eta_j)$ should be $$(\eta_j)=\sum_{k=1}^{\infty} \alpha_{j,k} \zeta_k$$ and also in what follows.As for the Q, let $\alpha_n=(\alpha_{n.k})_{k \in N}$ which belongs to the Hilbert space, otherwise $T$ is unbounded. Denote the inner product of vectors $x,y$ as $(x|y)$.We have, for $x \ne 0$, $$ ||(T_n-T) x||^2 /||x||^2= \left(\sum_{j>n} |(\alpha_j | x)|^2\right)/||x||^2$$ $$\le \left( \sum _{j>n} ||\alpha_j||^2 ||x||^2 \right) /||x||^2$$ $$= (\sum_{j>n} \sum_{k \in N} |\alpha_{j,k}|^2) $$ which tends to $0$ as $ n \to \infty$ because $\sum_{j,k \in N} |\alpha_{j,k}|^2 < \infty$ (which follows from some basic properties of absolutely summable real sequences.) Therefore $\lim_{n \to \infty} ||T_n-T||=0$.

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