Question 1: Is a compact operator $T:X\to Y$ 's restriction on a subspace $Z\subset X$ still compact? (I think I've got the answer)

I think the compact operator's restriction on any subspace must still be compact since for any bounded sequence $\{ x_n\}\subset Z$, it's also bounded in $X$ since the subspace shares the norm.

However a subspace can be equipped with a different norm. For example, it's not immediately clear whether the injection $i: (C^\infty,\| \cdot\|_\infty )\to (L^2,\| \cdot\|_{L^2})$ is compact. But we know that the injection: $i_2: (C^\infty,\| \cdot\|_{L^2} )\to (L^2,\| \cdot\|_{L^2})$ is compact by the above argument and $i_1: (C^\infty,\| \cdot\|_\infty )\to (C^\infty,\| \cdot\|_{L^2} )$ is continuous. Therefore $i=i_2\circ i_1$ is compact. (Such argument won't work for $H^1\to L^2$, you can try by yourself)

The lesson is we can't discuss the operator's compactness ignoring the norm of the space. The definition of compactness heavily depends on them(bounded, convergent). Maybe someone can furnish us with an example that the operator is compact in one norm but not in the other.

Question 2: From Fredholm alternative, we know that the range of $\lambda I -T$ is always closed if $\lambda \neq 0$. So how about the range of a compact operator itself? Is it always closed or open? Are there any examples?


Your intuition about the restriction of a compact operator to a subspace looks correct to me. A linear operator $T : X \to Y$ is compact if, for every bounded sequence $\{x_n\}$ in $X$, the sequence $\{Tx_n\}$ has a subsequence which is Cauchy in $Y$. So, if you have a subspace $Z$ of $X$, any bounded subsequence in $Z$ is also a bounded subsequence in $X$, so its image under $T$ will have a Cauchy subsequence in $Y$ too.

For your second question, I don't think the range of a compact operator is always closed. Consider the operator $T : \ell^2 \to \ell^2$,

$ (Tx)_n = e^{-|n|}x_n$.

(Pretty sure that's a compact operator.) For all $y$ in the range of $T$, $y_n$ decays exponentially fast as $n \to \pm\infty$. Now consider the sequence $z_n = (1+n^2)^{-1}$, which is in $\ell^2$; since it doesn't decay exponentially fast, it can't be in the range of $T$. On the other hand, there is a sequence of points in the range of $T$ which converge to $z$. If you're stumped, I can write down what the sequence is.

Someone who's less rusty on functional analysis can probably give you a deeper insight here. You should note that an operator have a closed range and being a closed operator are two different things. For example, it's shown here that an operator is closed if its domain is complete with respect to the graph norm

$\|x\|_T = \sqrt{\|x\|^2+\|Tx\|^2}$.

You can verify that, by this criterion, the counterexample I gave you before is a closed operator, even though its range isn't closed.

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    $\begingroup$ Your $T$ is indeed compact. It is the norm limit of its finite rank truncations. $\endgroup$ – Julien May 14 '13 at 22:53
  • $\begingroup$ The sequence is $x_n^N=\begin{cases} e^{|n|}(1+n^2)^{-1},&|n|<N \\ 0,&|n|>N \end{cases}$? $\endgroup$ – user33869 May 14 '13 at 23:09
  • $\begingroup$ Bingo! Also: that sequence isn't bounded in the graph norm, so you can't use it to say that the operator isn't closed. @julien: Thanks for pointing that out -- the upshot is that the set of compact operators is closed in the strong operator topology, right? $\endgroup$ – korrok May 15 '13 at 4:45
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    $\begingroup$ The compact operators are closed in the operator norm topology. They are not closed in the sot in general. For instance, on $\ell^2(\mathbb{N})$, the identity is the sot limit of the finite rank projections onto the span of $\{e_0,\ldots,e_n\}$. Yet the identity is not compact. $\endgroup$ – Julien May 15 '13 at 13:41

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