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Let A be a bounded linear operator on $l^2$ defined by A($a_n$)=($\frac{1}{n} a_n$). Would you help me to prove that A is compact operator. I guess the answer using an approximation by a sequences of finite range operator.

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Which definition of compact operator are you starting from? There are several equivalent definitions. – Christopher A. Wong Nov 5 '12 at 11:01
Try projections onto the space generated by $\{e_k\}_{k=1}^n$? – Hui Yu Nov 5 '12 at 11:03
up vote 3 down vote accepted

Let $$T_k((a_n)_n))(j)=\begin{cases} \frac 1ja_j&\mbox{ if }j\leq k;\\ 0&\mbox{ otherwise.} \end{cases}$$ This gives a linear operator, and the range of $T_k$ is generated by $e_1,\dots,e_k$, a finite dimensional space. For $a\in\ell^2$, $$\lVert T(a)-T_k(a)\rVert=\sum_{j\geq k+1}\frac 1j|a_j|\leq\frac 1{k+1}\sum_{j\geq 1}|a_j|,$$ so $\lVert T-T_k\rVert\leq\frac 1{k+1}$ and $T$ is compact as the limit in operator norm of compact operators.

Note that more generally, we can define $T(a)(k):=d_ka_k$, where $d_k\to 0$, and $T$ from $\ell^p$ to $\ell^p$, where $1\leq p<\infty$, and this will give a compact operator by the same argument.

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