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in the Higson and Roe's book "analytic K-homology" just after the definition of the Fredholm operator there is a remark (2.1.3 you can see at it onlin at Google books (click here)) which claims that in the definition of Fredholm operators the condition of being closed for image of operator is superfluous. It seems that in the proof $cokernel(T)$ is supposed to be $\frac{H_2}{Image(T)}$ instead of $\frac{H_2}{\overline{Image(T)}}$ which cause the newly defined operator $\tilde{T}$ to be immediately surjective. (which is not obviouse for me when I take $coker(T)= \frac{H_2}{\overline{Image(T)}}$ (as it is expected when we consider coker at the category of Hilbert spaces with bounded operators)).

I was wondering if there is anybody help me to find out what is goning on in this proof or prove or disprove the statement in a clear way?

Edit:

So You agree with me that this proof is true just if we take $cokernel_1(T)=\frac{H_2}{Image(T)}$.

Now here is the question: what if we put $cokernel_2(T)=\frac{H_2}{\overline{Image(T)}}$? I mean if $cokernel_2(T)$ is finite dimension can we say that $cokernel_1(T)$ is finite dimensional also?

In other word: Is there any bounded operator $T:H_1\to H_2$ which has finite dimensional $coker_2(T)$ (please note the index 2) and $ker(T)$ but it is not a Fredholm operator?

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You might also look at Theorem 1.4.7 of Murphy's "C*-algebras and operator theory", which says more or less what Chad Groft's answer does –  Yemon Choi Feb 10 '11 at 5:38
3  
Yes, the cokernel has to be taken in the "algebraic" sense 1, not in the "functional analytic" sense 2. To see this, take e.g. the inclusion of the Sobolev space $H^1$ in $L^2$, which is dense, so it has zero cokernel in the second sense. –  Theo Buehler Feb 11 '11 at 1:19
    
thank you Theo. this is exactly what I was looking for. –  madmath Feb 11 '11 at 3:03

1 Answer 1

No, the proof is correct, and it's taking $\mathrm{coker}\ T = H_2/\mathrm{im}\ T$. The essence of the argument is that $V\oplus\mathbb{C}^n$ contains $V$ as a closed subspace, regardless of the norm on $V$, and that $H_2$ is "close enough" to $(\mathrm{im}\ T)\oplus\mathbb{C}^n$.

The full argument goes like this: Let $T: H_1\to H_2$ be Fredholm (the $H_i$ are Hilbert spaces). WLOG $T$ is injective, otherwise replace $H_1$ with $(\mathrm{ker}\ T)^\perp$. Choose $v_1,\dots, v_n\in H_2$ which represent a basis of $\mathrm{coker}\ T$, and use them to define a linear function $\mathbb{C}^n\to H_2$. Combine this with $T$ to get a linear map $\overline{T}: H_1\oplus\mathbb{C}^n\to H_2$.

Now $\overline{T}$ is a linear isomorphism, so its inverse $\widetilde{T}$ is also a linear isomorphism. $\overline{T}$ is also continuous, so the graph of $\overline T$ is a closed subset of $H_1\oplus\mathbb{C}^n\oplus H_2$, so the graph of $\widetilde T$ is a closed subset of $H_2\oplus H_1\oplus\mathbb{C}^n$, which makes $\widetilde T$ continuous by the Closed Graph Theorem. (This last is not an obvious statement; it's a consequence of the Baire Category Theorem). Finally, $\mathrm{im}\ T$ is precisely the pullback of $H_1$ through $\widetilde T$, making it closed.

As to why most sources don't present it this way… I think in most cases where somebody wants to prove that a given operator $T$ is Fredholm, it's just easier to establish that $\mathrm{coker}\ T$ is finite-dimensional after proving that $\mathrm{im}\ T$ is closed.

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That "im T + C^n" is supposed to be at the end of the first paragraph. Not sure why it didn't show up there. –  Chad Groft Feb 10 '11 at 3:58
    
@Chad Groft: I wanted to add to your explanation, that this proof also works perfectly in any Banach space. All you need is the theorem of Hanh-Banach, from which you get, that every finate dimensional subspace (here it is coker T) splits. It is remarkable, that one doesn't need, that the kernel is finate dimensional, too! –  kostja Aug 28 '11 at 16:54
    
To summarize: if a bounded, linear operator between Banach spaces has finite-dimensional co-kernel, then it has closed image. –  Alain Valette Aug 28 '11 at 21:15

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