## Agreg 2012 Analysis Part 2

**Part 2. Some elements of Spectral Analysis**

In this part we prove that the spectrum of a bounded linear operator is non-empty, and we look at the characteristics of the spectrum of a compact operator.

Let be a complex Banach space which is not reduced to . (it is known that ) For we define as the set of those such that is bijective, and denote .

We define the spectrum by . In particular, if is an eigenvalue for we have and therefore . (but note that may contain elements which are not eigenvalues)

1. Suppose that . Prove that and .

2. Prove that if then and

3. Prove that is an open set in and for every the application is analytic in a neighborhood of any point .

4. Deduce that for every , is a non-void and compact.

From here on, until the end of this part, we assume that is of infinite dimension and is a compact operator.

5. Let be a closed subspace of . Prove that there exists such that and . Deduce from here that there exists a sequence of elements in such that for every we have and when .

6. Prove that and for every eigenvalue of is of finite dimension.

7. Consider such that is injective.

i) Arguing by contradiction prove that there exists such that for we have . Deduce that for every closed subset we have that is also closed.

ii) Suppose that is not surjective. Prove that for every we have , with strict inclusion.

iii) Deduce that there exists a sequence of elements such that and .

iv) Conclude that is surjective and therefore the spectrum of is composed of and eventually the eigenvalues of .

**Hints:** 1. First note that if then the operator is well defined (any absolutely convergent series is convergent in a Banach space). Moreover, if then . Thus we have

and therefore is the inverse of .

2. If then satisfies the hypothesis of 1. so there exists the inverse of . But then so . Moreover, the resolvent satisfies

which converges to as .

3. Here we need to be able to see what happens to if we perturb a bit.

Note that if is small enough, by 1., we obtain that the right hand side of the above inequality is in fact an invertible operator, and in consequence, for small enough is still in . This proves that is open and

which leads to

Evaluating both operators in the previous relation at and composing with a linear functional we get exactly the fact that is a complex function differentiable at . Since holomorphic function are analytic, we are done here.

4. Suppose that the spectrum of is void. Then the resolvent is the entire , and the functions defined above are entire and bounded (use 2. and 3.), and by Liouville’s theorem constant. Using 2. we see that the constant is in fact zero. But this is a contradiction, because there are non-trivial functionals , i.e there exist such that and we can pick to find that .

5. Since we can pick . Denote . Pick now such that . The element will be the needed example: indeed, for we have

We construct the needed sequence by induction. At each step we pick .

6. Pick the sequence constructed in 5. and consider . By compacity of it follows that has a convergent subsequence . Suppose now that is bijective. Then we know that is continuous, and therefore , which means that contains a convergent subsequence. But this is impossible since cannot have a Cauchy sub sequence by construction ()

For the second part, if has infinite dimension, then we can find a sequence such that and . Therefore , and since is compact has a convergent subsequence (here we use ). This contradicts the existence of , since cannot have Cauchy subsequences.

7. i) Suppose that for every we can find such that and . We know that has a convergent subsequence . Then we have

and as we obtain that converges to . The initial inequality implies that must converge to , but this is impossible as for every .

Consider now closed. Pick such that . Then and we obtain that

From here we deduce that is Cauchy, it has a limit and so is also in the image of .

ii) The hypothesis implies that is a bijective linear operator (note that is closed so is also a Banach space). Denote . We obviously have the inclusion . Suppose now that we could have equality for some . Then we have

which means that is surjective. This means that the inclusion is always strict.

iii) We use 5. to construct the desired sequence by induction. Since is a closed proper subspace of at each step we can choose with and .

iv) We want to reach a contradiction to the fact that is compact, and one way to do this is to prove that does not have any Cauchy subsequences. We have for

but since

we have

This contradicts the fact that is compact. The initial assumption in ii) was that is not surjective, and this led to a contradiction. Therefore any non-zero component of the spectrum is necessarily an eigenvalue.