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?

  • $\begingroup$ This math.stackexchange.com/questions/115800/… help you. $\endgroup$ – Davide Giraudo Apr 22 '12 at 18:44
  • $\begingroup$ It works with this definition. Why the unit ball of $X$ is denoted $B_x$ instead of $B_X$? $\endgroup$ – Davide Giraudo Apr 22 '12 at 18:58
  • $\begingroup$ Sorry about that Davide. I changed it to $B_X$. $\endgroup$ – josh Apr 22 '12 at 19:11

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.

  • $\begingroup$ 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). $\endgroup$ – Jochen Wengenroth Apr 26 '12 at 14:33
  • $\begingroup$ That's the way I would have done it. But the question required a proof by definition. $\endgroup$ – Martin Argerami Apr 26 '12 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.