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In class, we started talking about operators on Banach spaces after covering the Arzela-Ascoli Theorem. We defined a continuous operator $T\colon X \to Y$ to be compact if $\overline{T(B_X)}^{Y}$ is compact, where $B_X$ is the unit ball in $X$.

Consider the Hilbert space $\ell_2 = \{ |x_n|^2 : \sum |x_n|^2 < \infty \}$. Given a bounded sequence $(a_n)^{\infty}_{n=1}$, define a linear operator $A\colon \ell_2 \to\ell_2$ by $A((x_n)) = (a_nx_n)$. With the above definition, how can I show that the operator $A$ is compact if and only if $\lim a_n = 0$? Do I need an equivalent definition of compactness, or can it be done using the above definition?

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  • $\begingroup$ This math.stackexchange.com/questions/115800/… help you. $\endgroup$ Commented Apr 22, 2012 at 18:44
  • $\begingroup$ It works with this definition. Why the unit ball of $X$ is denoted $B_x$ instead of $B_X$? $\endgroup$ Commented Apr 22, 2012 at 18:58
  • $\begingroup$ Sorry about that Davide. I changed it to $B_X$. $\endgroup$
    – josh
    Commented Apr 22, 2012 at 19:11

1 Answer 1

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This is how I would do it:

First assume that $\{a_n\}$ does not converge to zero. This means that there exists $\varepsilon>0$ and a subsequence $\{a_{n_k}\}_k$ with $|a_{n_k}|\geqslant\varepsilon$. Now consider the sequence of vectors $\{e_k\}$, where $e_k$ has a 1 in the $n_k$ position, and zero elsewhere. Then $Te_k$ is the sequence with $a_{n_k}$ in the $n_k$-entry and zeroes elswhere. So $\|Te_k-Te_j\|_2\geqslant\sqrt2\varepsilon$; considering the balls of radius $\varepsilon/2$ centered on the $Te_k$, we produce an infinite number of disjoint balls in $\overline{T(B_X)}$, which shows that $\overline{T(B_X)}$ is not compact, i.e. $T$ is not compact. This proves that if $T$ is compact, then the sequence goes to zero.

Now assume that $\lim a_n=0$. Let $y_1,y_2,\ldots$ be a sequence in $\overline{T(B_X)}$. Fix $\varepsilon>0$. Then we can get a sequence $x_1,x_2,\dots$ in $B_X$ with $\|y_j-Tx_j\|_2<2^{-j}\varepsilon$ for all $j$. Fix $n_0$ such that $|a_n|<\sqrt{\varepsilon/8}$ when $n\geqslant n_0$. Now, for each $k=1,\ldots,n_0$, consider the sequence of $k^{\rm th}$ entries of the sequence $\{x_j\}_j$. As this is a finite number of sequences in the unit ball of $\mathbb{C}$, there is a subsequence $\{x_{j_h}\}_h$ such that its first $n_0$ entries converge. So we can find $h$ such that, for $\ell=1,\ldots,n_0$, $$ |x_{j_{h+m}}(\ell)-x_{j_h}(\ell)|<\frac{\sqrt\varepsilon}{2^{(\ell+1)/2}K^{1/2}}\ \ \ \text{ for all }m $$ (i.e. $\{x_{j_h}\}$ is Cauchy in its first $n_0$ coordinates). Then $$ \|Tx_{j_{h+m}}-Tx_{j_h}\|_2^2=\sum_{\ell=1}^{n_0}|a_\ell(x_{j_{h+m}}(\ell)-x_{j_h}(\ell))|^2 +\sum_{\ell=n_0+1}^\infty|a_\ell(x_{j_{h+m}}(\ell)-x_{j_h}(\ell))|^2 \\ \leqslant\frac{\varepsilon}2+\frac{\varepsilon}8\,\|x_{j_{h+1}}-x_{j_h}\|_2^2 \leqslant\frac{\varepsilon}2+\frac{\varepsilon}8\,2^2=\varepsilon. $$ We have shown that $\{Tx_{j_h}\}_h$ is Cauchy, and so it is convergent in $\overline{T(B_X)}$. The sequence $\{y_{j_h}\}_h$ gets arbitrarily close to this sequence, so it is also convergent. So $T$ is compact.

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  • $\begingroup$ A quick alternative to prove compactness if $a_n\to 0$ is to show that $A$ is the limit (in the the space of all operators with the operator norm) of finite rank operators (which are always compact). $\endgroup$ Commented Apr 26, 2012 at 14:33
  • $\begingroup$ That's the way I would have done it. But the question required a proof by definition. $\endgroup$ Commented Apr 26, 2012 at 15:50

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