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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 $\ell^1$. – Davide Giraudo Jul 15 '12 at 14:27
up vote 7 down vote accepted

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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You can diagonalize directly in this example (it's instructive to do this in detail), and maybe it's also worth mentioning that the statement "a norm limit of compact operators is compact" using the definition given in the OP is usually also proved via diagonalization. – t.b. Jul 15 '12 at 14:31
Can someone share a link where there is a proof of "norm limit of compact operators is compact".? – Maths Student Dec 10 '13 at 22:15

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.


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

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