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Prove that the operator $T:\ell^1\rightarrow\ell^1$ which maps $x=(x_1,x_2,\dots)$ to $\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\dots\right)$ is compact.

For an arbitrary sequence $x^{(N)}\in\ell^1$ one would have extract a convergent subsequence of $T x^{(N)}$. Maybe via the diagonal argument?

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    The sequence $x^{(N)}$ you choose is not arbitrary (its norm is uniformly bounded by a constant, for example $1$. In this case, we can find a limit for $(Tx^{(N})_j$? Now we have to show that the convergence is in $\ell^1$.2012-07-15

2 Answers 2

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Define $T_j(x):=\left(x_1,\frac{x_2}2,\dots,\frac{x_j}j,0,\dots,0\right).$ It's a compact operator (because it's finite ranked) and $T(x)-T_j(x)=\left(0,\dots,0,\frac{x_{j+1}}{j+1},\dots\right),$ hence $\lVert T(x)-T_j(x)\rVert_{\ell^1}=\sum_{k=j+1}^{+\infty}\frac{|x_k|}{k}\leq \frac 1{j+1}\sum_{k=j+1}^{+\infty}|x_k|\leq \frac 1{j+1}\lVert x\rVert_{\ell^1},$ which proves that $\lVert T-T_j\rVert\leq \frac 1{j+1}$.

To conclude, notice that a norm limit of compact operators is compact.

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    Can someone share a link where there is a proof of "norm limit of compact operators is compact".?2013-12-10
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Check that it's the limit of the following finite rank operators and invoke the theorem that a uniform limit of finite rank operators on a Banach Space is a compact operator.

$T_n(x_1,x_2....)=(x_1,\frac{x_2}2...\frac{x_n}n,0,0,0,0)$

You can verify this by checking that $T_n$ is a Cauchy sequence in $L(\mathcal l_1)$