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My question is: does this hold for any normed space $X$ or only for Banach spaces:

If $X$ is a Banach space then $K(X)$ (space of compact operators) equals $B(X)$ (space of bounded operators) if and only if $X$ is finite dimensional.

$\color{\grey}{\text{I know that the closed unit ball in a *normed space* $X$ is compact if and only if $X$ is finite}}$

$\color{\grey}{\text{ dimensional and I also know that the identity map is not compact if the unit ball is not.}}$

Student
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2 Answers2

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That the closed unit ball is compact is indeed a characterization of 'finite dimensional'. Take for instance the space $X$ of all bounded functions $f:\mathbb R\to\mathbb R$ equipped with the maximum norm $||f||_\infty:=\sup_{x\in\mathbb R}|f(x)|$. Then one can show that $(X, ||\cdot||_\infty)$ is a (infinite dimensional) Banach space. To show that the unit sphere is not compact, consider the sequence $f_n$ of functions of $X$, given by \begin{align*} f_n(x)=\begin{cases}1,&\mbox{if } x=n \\ 0,&\mbox{if } x\neq n\end{cases}. \end{align*} Then each $f_n$ is bounded such that $||f_n||_\infty=1$. But $||f_m-f_n||_\infty=1$ whenever $m\neq n$, so $(f_n)$ cannot have a convergent subsequence.

Now, take for example the operator $T:X\to X,\ f\mapsto f$ (so $T$ is the identity). Then $T$ is obviously bounded but cannot be compact, since the closed unit ball is mapped to itself, but it is not compact as shown above.

sranthrop
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  • Why did I get a downvote? At least a comment would be nice... – sranthrop Jun 02 '14 at 07:47
  • I agree that your answer is basically correct and I don't know why you got a downvote. I don't think you defined the functionals as you intended though. Perhaps you meant to have a infinite basis and functionals which took value $0$ at all but one basis element. But then you seem to be proving a statement about the dual space $X^{\ast},$ which is usually much bigger than $X$ anyway. – Geoff Robinson Jun 02 '14 at 07:58
  • Sorry, I am having trouble understanding whether you are saying that the statement holds or does not hold in any normed space. – Student Jun 02 '14 at 08:06
  • Thx! I am not sure if I got your comment correctly, but all I wanted to do was to give an example of an infinite dimensional Banach space and a sequence of elements of this space, where you cannot extract a convergent subsequence from, so my space is not (sequentially) compact. – sranthrop Jun 02 '14 at 08:08
  • Okay. Could you please elaborate how this relates to my question? I don't yet understand. – Student Jun 02 '14 at 08:15
  • As @GeoffRobinson mentioned, even the identity is not compact in a infinite dimensional space. I'll add this to my answer. – sranthrop Jun 02 '14 at 08:21
  • Does the theorem in my question hold for Banach spaces only or for normed spaces in general? – Student Jun 02 '14 at 10:25
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The assumption that the unit ball is compact is indeed the issue. In general, a metric space is compact if and only if it is complete and totally bounded. The condition of being totally bounded is stronger than being bounded (basically, it means that for any $\varepsilon >0,$ the space can be covered by finitely many open balls of that radius (if you recall, that takes some effort to prove for a closed interval in $\mathbb{R} )).$ It is indeed the case that the closed unit ball of a complete real normed linear space is compact if and only if it is finite dimensional, and this can be found in many functional analysis texts. Note that the unit ball of an infinite dimensional normed linear space $X$ contains a sequence with no convergent subsequence (whether or not $X$ is complete- if $X$ is not complete it contains a Cauchy sequence which does not converge, and the elements can be taken from the closed unit ball of $X$ if required), so even the identity operator is not compact.

Geoff Robinson
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  • Sorry, I am having trouble understanding whether you are saying that the statement holds or does not hold in any normed space. – Student Jun 02 '14 at 08:05
  • @Student: The answer I gave means that for an infinite dimensional normed linear space, even the identity operator is not compact, since the unit ball contains a sequence with no convergent subsequence. – Geoff Robinson Jun 02 '14 at 08:19
  • Yes I know. But what does it imply for the statement in my question? – Student Jun 02 '14 at 08:24
  • Well, the identity operator is certainly always a bounded operator ( it has norm $1$), but is not compact when $X$ is infinite dimensional. – Geoff Robinson Jun 02 '14 at 09:11
  • Yes, I added to my question that I know this. But does it imply that the statement in my question is true in normed spaces that aren't complete? – Student Jun 02 '14 at 10:24
  • You keep changing the question: I don't think it matters whether $X$ is complete or not. It should be even easier to find a bounded sequence with no convergent subsequence in the unit ball of $X$ if $X$ is not complete- by definition, there is a Cauchy sequence of elements of $X$ which does not converge: it is bounded, and has no convergent subsequence. – Geoff Robinson Jun 02 '14 at 10:58
  • No I have not changed the question once. I merely deleted additional information, you can confirm this by checking the edit history. – Student Jun 02 '14 at 11:00
  • I am still not clear: does your last comment imply that the theorem in my question does hold or that it does not hold? Note that my question does not contain the theorem about the compactness of the unit ball. – Student Jun 02 '14 at 11:02
  • Given your last comment I now believe the answer is affirmative. – Student Jun 02 '14 at 11:05
  • If an operator is compact, the image under it of any bounded sequence has to have a convergent subsequence. Hence if the identity operator is compact, every bounded sequence should have a convergent subsequence. If the space is not complete, this is never the case. – Geoff Robinson Jun 02 '14 at 11:08