If $H$ is an $\infty$-dimensional Hilbert space and $T:H\to{H}$ is a compact operator with closed range, how do I show that $T$ has finite rank, without using the open-mapping theorem? (The open-mapping theorem is not in my lecture notes).

The definitions I have in my lecture notes are:

(Let $B(H)$ denote the space of all bounded operators mapping $H\to{H}$, $K(H)$ denote the space of all compact operators mapping $H\to{H}$, $R(H)$ denote the space of all finite rank operators mapping $H\to{H}$).

  • $T\in{B(H)}$ is compact if the closure of $T(B(0,1))$ is a compact set.
  • $T\in{B(H)}$ has finite rank if $Range(T)=T(H)$ is finite-dimensional.

I'm not sure how to do the proof, but I think that the following propositions in my lecture notes could be useful:

  • $T\in{R(H)}$ iff $T\in{B(H)}$ is the norm limit of a sequence of finite rank operators, i.e. $K(H)$ is the closure of $R(H)$.
  • Let $T\in{R(H)}$. Then there is an orthonormal set $\{e_1,...,e_L\}$ s.t. $$Tu=\sum\limits_{i,j=1}^{L}{c_{ij}(u,e_j)e_i}$$ where $c_{ij}$ are complex numbers.

Thank you in advance.


Sketch: Find a way to write the range of $T$ as a countable union of compact sets $K_i$. Then the Baire category theorem will guarantee that one of the $K_i$ has nonempty interior (relative to the range of $T$). This means the range of $T$ is locally compact, hence finite dimensional.

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  • $\begingroup$ I don't have the Baire Category Theorem in my lecture notes. Any other suggestions would be appreciated. $\endgroup$ – Maths Student May 26 '12 at 18:22
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    $\begingroup$ @MathsStudent: What do you have? $\endgroup$ – Nate Eldredge May 26 '12 at 20:39
  • $\begingroup$ Other than the results I mentioned in my original post, I have that: T restricted to the orthogonal complement of kerT is a bijective and continuous map (although this is in the Fredholm operators section so I'm not sure if it only holds for Fredholm operators), and I have that finite rank operators are compact, and also that K(H) is a 2-sided ideal in B(H) and is closed under taking adjoints. $\endgroup$ – Maths Student May 27 '12 at 11:29
  • $\begingroup$ My question is an assignment question, and any results I use in my answer which are not in the lecture notes, I will have to prove. Although perhaps it will be easier for me to prove either the OMT or the BCT in my answer than find another way to do the proof. $\endgroup$ – Maths Student May 27 '12 at 11:41

The key thing to prove (or know) is that if $V$ is a Hilbert space whose closed unit ball is compact, then $V$ is finite-dimensional.

(The same is true with the word Hilbert replaced by the word Banach, but the Hilbertian case is simpler to prove, because every Hilbert space has an orthogonal basis.)

I assume that you have seen this result in your notes; if not, it is not too hard to prove once one recalls that in a metric space the notions of compactness and sequential compactness coincide.

Now you say you want to avoid the open mapping theorem for Banach spaces. I think this is OK because soft algebraic arguments tell us that $T:H\to H$ factors as $H \to H/\ker(T) \to \overline{{\rm Im} T} \to H$ where each arrow is continuous and linear, and the second arrow is a bijection. So now apply the result I originally mentioned with $V= \overline{{\rm Im} T}$.

Filling in the remaining details seems to me like an instructive exercise, so I leave it to you.

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    $\begingroup$ I think this still requires the open mapping theorem or something similar: even if $T$ were bijective, you still need to show $T^{-1}$ is continuous. $\endgroup$ – Nate Eldredge May 26 '12 at 18:02
  • $\begingroup$ @NateEldredge Hmm, I thought I had a way round this but it turns out there is a gap (for essentially the reason you describe). The issue is of course how to deduce that the unit ball of the Hilbert space T(H) is compact from the given assumptions without invoking OMT. Perhaps one can do something by considering the restriction of T to $(\ker T)^\perp$ but I admit I can't see a way through right now $\endgroup$ – user16299 May 26 '12 at 19:29
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    $\begingroup$ It seems it will be essential to use completeness, since this is not true for operators $H \to K$ where $K$ is incomplete. $\endgroup$ – Nate Eldredge May 26 '12 at 20:38

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