A few days ago, I found myself needing to use the Fredholm alternative in functional analysis:
Theorem 1 (Fredholm alternative) Let be a Banach space, let be a compact operator, and let be non-zero. Then exactly one of the following statements hold:
- (Eigenvalue) There is a non-trivial solution to the equation .
- (Bounded resolvent) The operator has a bounded inverse on .
Among other things, the Fredholm alternative can be used to establish the spectral theorem for compact operators. A hypothesis such as compactness is necessary; the shift operator on , for instance, has no eigenfunctions, but is not invertible for any unit complex number . The claim is also false when ; consider for instance the multiplication operator on , which is compact and has no eigenvalue at zero, but is not invertible.
It had been a while since I had studied the spectral theory of compact operators, and I found that I could not immediately reconstruct a proof of the Fredholm alternative from first principles. So I set myself the exercise of doing so. I thought that I had managed to establish the alternative in all cases, but as pointed out in comments, my argument is restricted to the case where the compact operator is approximable, which means that it is the limit of finite rank operators in the uniform topology. Many Banach spaces (and in particular, all Hilbert spaces) have the approximation property that implies (by a result of Grothendieck) that all compact operators on that space are almost finite rank. For instance, if is a Hilbert space, then any compact operator is approximable, because any compact set can be approximated by a finite-dimensional subspace, and in a Hilbert space, the orthogonal projection operator to a subspace is always a contraction. (In more general Banach spaces, finite-dimensional subspaces are still complemented, but the operator norm of the projection can be large.) Unfortunately, there are examples of Banach spaces for which the approximation property fails; the first such examples were discovered by Enflo, and a subsequent paper of by Alexander demonstrated the existence of compact operators in certain Banach spaces that are not approximable.
I also found out that this argument was essentially also discovered independently by by MacCluer-Hull and by Uuye. Nevertheless, I am recording this argument here, together with two more traditional proofs of the Fredholm alternative (based on the Riesz lemma and a continuity argument respectively).
— 1. First proof (approximable case only) —
In the finite-dimensional case, the Fredholm alternative is an immediate consequence of the rank-nullity theorem, and the finite rank case can be easily deduced from the finite dimensional case by some routine algebraic manipulation. The main difficulty in proving the alternative is to be able to take limits and deduce the approximable case from the finite rank case. The key idea of the proof is to use compactness to establish a lower bound on that is stable enough to allow one to take such limits. There is an additional subtlety (pointed out in comments) that when is not a Hilbert space, it is not necessarily the case that can be approximated by finite rank operators; but a modification of the argument still suffices in this case.
Fix a non-zero . It is clear that cannot have both an eigenvalue and bounded resolvent at , so now suppose that has no eigenvalue at , thus is injective. We claim that this implies a lower bound:
Proof: By homogeneity, it suffices to establish the claim for unit vectors . Suppose this is not the case; then we can find a sequence of unit vectors such that converges strongly to zero. Since has norm bounded away from zero (here we use the non-zero nature of ), we conclude in particular that has norm bounded away from zero for sufficiently large . By compactness of , we may (after passing to a subsequence) assume that the converge strongly to a limit , which is thus also non-zero.
On the other hand, applying the bounded operator to the strong convergence (and using the fact that commutes with ) we see that converges strongly to . Since converges strongly to , we conclude that , and thus we have an eigenvalue of at , contradiction.
Remark 1 Note that this argument is ineffective in that it provides no explicit value of (and thus no explicit upper bound for the operator norm of the resolvent ). This is not surprising, given that the fact that has no eigenvalue at is an open condition rather than a closed one, and so one does not expect bounds that utilise this condition to be uniform. (Indeed, the resolvent needs to blow up as one approaches the spectrum of .)
From the lower bound, we see that to prove the bounded invertibility of , it will suffice to establish surjectivity. (Of course, we could have also obtained this reduction by using the open mapping theorem.) In other words, we need to establish that the range of is all of .
Let us first deal with the easy case when has finite rank, so that is some finite-dimension . This implies that the kernel has codimension , and we may thus split for some -dimensional space . The operator is a non-zero multiple of the identity on , and so already contains . On the other hand, the operator maps the -dimensional space to the -dimensional space injectively (since avoids and is injective), and thus also surjectively (by the rank-nullity theorem). Thus contains , and thus (by the short exact sequence ) is in fact all of , as desired.
Finally, we deal with the case when is approximable. The lower bound in Lemma 2 is stable, and will extend to the finite rank operators for large enough (after reducing slightly). By the preceding discussion for the finite rank case, we see that is all of . Using Lemma 2 for , and the convergence of to in the operator norm topology, we conclude that is dense in . On the other hand, we observe that the space is necessarily closed, for if converges to a limit , then (by Lemma 2 and the assumption that is Banach) will also converge to some limit , and so . As is now both dense and closed, it must be all of , and the claim follows.
— 2. Second proof —
We now give the standard proof of the Fredholm alternative based on the Riesz lemma:
Proof: By the Hahn-Banach theorem, one can find a non-trivial linear functional on which vanishes on . By definition of the operator norm of , one can find a unit vector such that . The claim follows.
The strategy here is not to use finite rank approximations (as they are no longer available), but instead to try to contradict the compactness of by exhibiting a bounded set whose image under is not totally bounded.
Let be a compact operator on a Banach space, and let be a non-zero complex number such that has no eigenvalue at . As in the first proof, we have the lower bound from Lemma 2, and we know that is a closed subspace of ; in particular, the map is a Banach space isomorphism from to . Our objective is again to show that is all of .
Suppose for contradiction that is a proper closed subspace of . Applying the Banach space isomorphism repeatedly, we conclude that for every natural number , the space is a proper closed subspace of . From the Riesz lemma, we may thus find unit vectors in for whose distance to is at least (say).
Now suppose that . By construction, all lie in , and thus . Since lies at a distance at least from , we conclude the separation proeprty
But this implies that the sequence is not totally bounded, contradicting the compactness of .
— 3. Third proof —
Now we give another textbook proof of the Fredholm alternative, based on Fredholm index theory. The basic idea is to observe that the Fredholm alternative is easy when is large enough (and specifically, when ), as one can then invert using Neumann series. One can then attempt to continously pertrb from large values to small values, using stability results (such as Lemma 2) to ensure that invertibility does not suddenly get destroyed during this process. Unfortunately, there is an obstruction to this strategy, which is that during the perturbation process, may pass through an eigenvalue of . To get around this, we will need to abandon the hypothesis that has no eigenvalue at , and work in the more general setting in which is allowed to be non-trivial. This leads to a lengthier proof, but one which lays the foundation for much of Fredholm theory (which is more powerful than the Fredholm alternative alone).
Fortunately, we still have analogues of much of the above theory in this setting:
- (Finite multiplicity) is finite-dimensional.
- (Lower bound) There exists such that for all .
- (Closure) is a closed subspace of .
- (Finite comultiplicity) has finite codimension in .
Proof: We begin with finite multiplicity. Suppose for contradiction that was infinite dimensional, then it must contain an infinite nested sequence of finite-dimensional (and thus closed) subspaces. Applying the Riesz lemma, we may find for each , a unit vector of distance at least from . Since , we see that the sequence is then -separated and thus not totally bounded, contradicting the compactness of .
The lower bound follows from the argument used to prove Lemma 2 after quotienting out the finite-dimensional space , and the closure assertion follows from the lower bound (again after quotienting out the kernel) as before.
Finally, we establish finite comultiplicity. Suppose for contradiction that the closed subspace had infinite codimension, then by properties of already established, we see that is closed and has infinite codimension in for each . One can then argue as in the second proof to contradict total boundedness as before.
Remark 2 The above arguments also work if is replaced by an invertible linear operator on , or more generally by a Fredholm operator.
We can now define the index to be the dimension of the kernel of , minus the codimension of the range. To establish the Fredholm alternative, it suffices to show that for all , as this implies surjectivity of whenever there is no eigenvalue. Note that Note that when is sufficiently large, and in particular when , then is invertible by Neumann series and so one already has index zero in this case. To finish the proof, it suffices by the discrete nature of the index function (which takes values in the integers) to establish continuity of the index:
Lemma 5 (Continuity of index) Let be a compact operator on a Banach space. Then the function is continuous from to .
Proof: Let be non-zero. Our task is to show that
for all sufficiently close to .
In the model case when is invertible (and thus has index zero), the claim is easy, because can be inverted by Neumann series for close enough to , giving rise to the invertibility of .
Now we handle the general case. As every finite dimensional space is complemented, we can split for some closed subspace of , and similarly split for some finite-dimensional subspace of with dimension .
From the lower bound we see that is a Banach space isomorphism from to . For close to , we thus see that is close to , in the sense that one can map the latter space to the former by a small perturbation of the identity (in the operator norm). Since complements , it also complements for sufficiently close to . (To see this, observe that the composition of the obvious maps
is a small perturbation of the identity map and is thus invertible for close to .)
Let be the projection onto with kernel . Then maps the finite-dimensional space to the finite-dimensional space . By the rank-nullity theorem, this map has index equal to . Gluing this with the Banach space isomorphism , we see that also has index , as desired.
Remark 3 Again, this result extends to more general Fredholm operators, with the result being that the index of a Fredholm operator is stable with respect to continuous deformations in the operator norm topology.