# Is the limit of compact operators again compact?

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?

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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).

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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