Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(T_n)_{n \in \mathbb{N}} \subset \mathcal{L}(\mathcal{X}, \mathcal{Y})$ where $T_n$, $n \in \mathbb{N}$, is compact.

Now, assuming that $(T_n)_{n \in \mathbb{N}}$ has a limit $T \in \mathcal{L}(\mathcal{X}, \mathcal{Y})$ with respect to the operator norm, e.g. $\|T_n-T\| \rightarrow 0$.

Is $T$ also compact in the case where

  • $\mathcal{X}$ and/or $\mathcal{Y}$ are Hilbert spaces,

  • $\mathcal{X}$ and/or $\mathcal{Y}$ are Banach spaces?

I think the statement is true as long as $\mathcal{Y}$ is a Banach space ($\mathcal{X}$ can be any normed space). Is that correct?

If $\mathcal{K}(\mathcal{X}, \mathcal{Y})$ denotes the set of all compact operators from $\mathcal{X}$ to $\mathcal{Y}$, then it is known that $\mathcal{K}(\mathcal{X}, \mathcal{Y})$ is closed in $\mathcal{L}(\mathcal{X}, \mathcal{Y})$ if $\mathcal{Y}$ is a Banach space. So if $(T_n)_{n \in \mathbb{N}} \subset \mathcal{L}(\mathcal{X}, \mathcal{Y})$ with $\|T_n-T\| \rightarrow 0$, we have $T \in \mathcal{L}(\mathcal{X}, \mathcal{Y})$.

Is there a simple counter example for the case that $\mathcal{Y}$ is not a Banach space?

share|improve this question
add comment

1 Answer

I think the statement is true as long as $\mathcal{Y}$ is a Banach space ($\mathcal{X}$ can be any normed space). Is that correct?

Yes, that is correct. In a complete metric space - such as a Banach space - a subset is relatively compact if and only if it is totally bounded. Showing that $T(B_\mathcal{X})$ is totally bounded if $T$ is the norm-limit of compact operators is not hard.

You need to show that for every $\varepsilon > 0$, there is a finite set of $\varepsilon$-balls that cover $T(B_{\mathcal{X}})$. Let $\varepsilon > 0$. There is an $n$ such that $\lVert T_n - T\rVert < \delta := \frac{\varepsilon}{2}$. By the compactness of $T_n$, there are finitely many $y_1,\, \ldots,\, y_k \in \mathcal{Y}$ with

$$T_n(B_\mathcal{X}) \subset \bigcup_{j = 1}^k B_\delta(y_j).$$

Then I claim that

$$T(B_\mathcal{X}) \subset \bigcup_{j=1}^k B_\varepsilon(y_j).$$

Indeed, let $x \in B_\mathcal{X}$. Then there is an $1 \leqslant i \leqslant k$ with $T_n(x) \in B_\delta(y_i)$, and hence

$$\lVert T(x) - y_i\rVert \leqslant \lVert T(x) - T_n(x)\rVert + \lVert T_n(x) - y_i\rVert \leqslant \lVert T - T_n\rVert\cdot\lVert x\rVert + \lVert T_n(x) - y_i\rVert < \delta + \delta = \varepsilon.$$

Is there a simple counter example for the case that $\mathcal{Y}$ is not a Banach space?

Yes, there is. Take $\mathcal{X} = \mathcal{Y} = \{x \in \ell^2 \colon \bigl(\exists n\in \mathbb{N}\bigr)(k > n \Rightarrow x_k = 0)\}$, and let

$$T_n(x) = \sum_{k=1}^n \frac1k x_k.$$

$T_n$ has finite dimensional range, hence is compact, $\lVert T_n - T\rVert \to 0$ for

$$T(x) = \sum_{k=1}^\infty \frac1k x_k,$$

but $T(B_\mathcal{X})$ is not relatively compact because its completion is not contained in $\mathcal{X}$ (the completion contains sequences with infinitely many nonzero terms).

share|improve this answer
    
That's a fine direct proof, thanks also for the example. However, if I have the Lemma that the set of compact operators is a closed subspace of $L(\mathcal{X},\mathcal{Y})$, the statement is also a direct corollary, right? –  Amarus Jul 11 '13 at 2:40
    
Yes, that's exactly what closed means. –  Daniel Fischer Jul 11 '13 at 10:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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