Home > Functional Analysis > Agreg 2012 Analysis Part 2

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 {E} be a complex Banach space which is not reduced to {\{0\}}. (it is known that {E'\neq \{0\}}) For {T \in \mathcal{L}(E)} we define {res(T)} as the set of those {\lambda \in \Bbb{C}} such that {\lambda I-T} is bijective, and denote {R_\lambda(T)=(\lambda I-T)^{-1} \in \mathcal{L}(E)}.

We define the spectrum by {\sigma(T)=\Bbb{C} \setminus res(T)}. In particular, if {\lambda} is an eigenvalue for {T} we have {\ker(\lambda I-T)\neq \{0\}} and therefore {\lambda \in \sigma(T)}. (but note that {\sigma(T)} may contain elements which are not eigenvalues)

1. Suppose that {\|T\|<1}. Prove that {1 \in res(T)} and {(I-T)^{-1}=\sum_{k=0}^\infty T^k}.

2. Prove that if {|\lambda |> \|T\|} then {\lambda \in res(T)} and

\displaystyle \lim_{|\lambda| \rightarrow \infty} \|R_\lambda(T)\|=0.

3. Prove that {res(T)} is an open set in {\Bbb{C}} and for every {x \in E,\ell\in E'} the application {\phi : \lambda \mapsto \ell(R_\lambda(T)x)} is analytic in a neighborhood of any point {\lambda_0 \in res(T)}.

4. Deduce that for every {T \in \mathcal{L}(E)}, {\sigma(T)} is a non-void and compact.

From here on, until the end of this part, we assume that {E} is of infinite dimension and {T \in \mathcal{L}(E)} is a compact operator.

5. Let {M \neq E} be a closed subspace of {E}. Prove that there exists {u \in E} such that {\|u\|=1} and {d(u,M) \geq 1/2}. Deduce from here that there exists a sequence {(x_n)_{n \in \Bbb{N}}} of elements in {E} such that for every {n \in \Bbb{N}} we have {\|x_n\|=1} and {\|x_n-x_m\| \geq 1/2} when {n \neq m}.

6. Prove that {0 \in \sigma(T)} and for every eigenvalue {\lambda \neq 0} of {T} {\ker(\lambda I-T)} is of finite dimension.

7. Consider {\lambda \in \Bbb{C}} such that {\lambda I-T} is injective.

i) Arguing by contradiction prove that there exists {\alpha >0} such that for {x \in E} we have {\|(\lambda I-T)x\| \geq \alpha \|x\|}. Deduce that for every closed subset {\mathcal{F} \subset E} we have that {(\lambda I-T)(\mathcal{F})} is also closed.

ii) Suppose that {\lambda I-T} is not surjective. Prove that for every {n} we have {im((\lambda I-T)^{n+1})\subset im((\lambda I-T)^n)}, with strict inclusion.

iii) Deduce that there exists a sequence of elements {(x_n)} such that {\|x_n\|=1,\ x_n \in im((\lambda I-T)^n)} and {d(x_n ,im((\lambda I-T)^{n+1})\geq 1/2}.

iv) Conclude that {(\lambda I-T)} is surjective and therefore the spectrum of {T} is composed of {0} and eventually the eigenvalues of {T}.

Hints: 1. First note that if {\|T\|<1} then the operator {S=\sum_{k=0}^\infty T^k} is well defined (any absolutely convergent series is convergent in a Banach space). Moreover, if {S_n = \sum_{k=0}^n T^k} then {S_n(I-T)=I-T^{n+1}}. Thus we have

\displaystyle \|S(I-T)-I\|\leq \|(S_n-S)(I-T)\|+\|T^{n+1}\| \rightarrow 0

and therefore {S} is the inverse of {I-T}.

2. If {|\lambda|>\|T\|} then {T/\lambda} satisfies the hypothesis of 1. so there exists the inverse {S_\lambda} of {I-T/\lambda}. But then {(\lambda I-T)(S_\lambda/\lambda)=I} so {\lambda \in res(T)}. Moreover, the resolvent {S_\lambda/\lambda} satisfies

\displaystyle \|S_\lambda/\lambda\|\leq \frac{1}{|\lambda|} \sum_{k=0}^\infty \|T\|^k/|\lambda^k|=\frac{1}{|\lambda|}\frac{1}{1-\|T\|/|\lambda|}

which converges to {0} as {|\lambda| \rightarrow \infty}.

3. Here we need to be able to see what happens to {R_\lambda(T)} if we perturb {\lambda} a bit.

\displaystyle R_\lambda(T) (I-(\lambda+\varepsilon)T)=I-\varepsilon R_\lambda(T)T.

Note that if {\varepsilon} 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 {\varepsilon} small enough {\lambda+\varepsilon} is still in {res(T)}. This proves that {res(T)} is open and

\displaystyle R_{\lambda}(T)=(I-\varepsilon R_\lambda(T)T)R_{\lambda+\varepsilon}(T)

which leads to

\displaystyle \frac{R_{\lambda+\varepsilon}(T)-R_\lambda(T)}{\varepsilon} = R_\lambda(T)TR_{\lambda+\varepsilon}(T).

Evaluating both operators in the previous relation at {x} and composing with a linear functional {\ell} we get exactly the fact that {\lambda \mapsto\ell(R_\lambda(T)x)} is a complex function differentiable at {\lambda}. Since holomorphic function are analytic, we are done here.

4. Suppose that the spectrum of {T} is void. Then the resolvent is the entire {\Bbb{C}}, 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 {\ell}, i.e there exist {\ell,y} such that {\ell(y)\neq 0} and we can pick {x=(\lambda I-T)y} to find that {\ell(y)\neq 0}.

5. Since {M\neq E} we can pick {u \notin M}. Denote {r=d(x,M)>0}. Pick now {x \in M} such that {d(u,x)<2r}. The element {y=(u-x)/\|u-x\|} will be the needed example: indeed, for {z \in M} we have

\displaystyle \|\frac{u-x}{\|u-x\|} -z\| =\frac{1}{\|u-x\|} \|u-(x+z\|u-x\|)\| \geq r/(2r)=\frac{1}{2}.

We construct the needed sequence by induction. At each step we pick {M=\overline {\text{span}\{x_1,..,x_n\}}}.

6. Pick the sequence {x_n} constructed in 5. and consider {y_n=Tx_n}. By compacity of {T} it follows that {y_n} has a convergent subsequence {y_{n_k} \rightarrow y}. Suppose now that {T} is bijective. Then we know that {T^{-1}} is continuous, and therefore {T^{-1} y_{n_k} \rightarrow T^{-1} y}, which means that {x_n} contains a convergent subsequence. But this is impossible since {(x_n)} cannot have a Cauchy sub sequence by construction ({\|x_n-x_m\| \geq 1/2})

For the second part, if {\ker(\lambda I-T)} has infinite dimension, then we can find a sequence {(x_n) \subset \ker(\lambda I-T)} such that {\|x_n-x_m\|\geq 1/2} and {\|x_n\|=1}. Therefore {Tx_n=\lambda x_n}, and since {T} is compact {(Tx_n)=\lambda (x_n)} has a convergent subsequence (here we use {\lambda \neq 0}). This contradicts the existence of {(x_n)}, since {(x_n)} cannot have Cauchy subsequences.

7. i) Suppose that for every {n} we can find {x_n} such that {\|x_n\|=1} and {\|(\lambda I-T)x_n\| \leq \frac{1}{n}}. We know that {(Tx_n)} has a convergent subsequence {y_{n_k} \rightarrow y}. Then we have

\displaystyle \|\lambda x_{n_k}-y\| \leq \|\lambda x_{n_k}-Tx_{n_k}\|+\|Tx_{n_k} -y\|

and as {n \rightarrow \infty} we obtain that {x_{n_k}} converges to {y/\lambda}. The initial inequality implies that {x_{n_k}} must converge to {0}, but this is impossible as {\|x_n\|=1} for every {n}.

Consider now {\mathcal{F} \subset E} closed. Pick {(y_n) \subset (\lambda I-T)(\mathcal{F})} such that {y_n \rightarrow y}. Then {y_n = (\lambda I-T)x_n} and we obtain that

\displaystyle \|y_n-y_m\| \geq \alpha \|x_n -x_m\|.

From here we deduce that {x_n} is Cauchy, it has a limit {x \in E} and {(\lambda I-T)x=y} so {y} is also in the image of {\lambda I-T}.

ii) The hypothesis implies that {\lambda I-T : E \rightarrow F \subset E (F\neq E)} is a bijective linear operator (note that {F} is closed so {F} is also a Banach space). Denote {Y_n=im (\lambda I -T)^n}. We obviously have the inclusion {Y_{n+1} \subset Y_n}. Suppose now that we could have equality {Y_{n+1}=Y_n} for some {n}. Then we have

\displaystyle [(\lambda I-T)^n]^{-1}(\lambda I-T)(\lambda I-T)^n(E)=E

which means that {\lambda I-T} is surjective. This means that the inclusion {Y_{n+1} \subset Y_n} is always strict.

iii) We use 5. to construct the desired sequence by induction. Since {Y_{n+1}} is a closed proper subspace of {Y_n} at each step we can choose {x_n \in Y_n} with {\|x_n\|=1} and {d(x_n,Y_{n+1})\geq 1/2}.

iv) We want to reach a contradiction to the fact that {T} is compact, and one way to do this is to prove that {(Tx_n)} does not have any Cauchy subsequences. We have for {n<m}

\displaystyle \|Tx_n-Tx_m\| = \|(T-\lambda I)x_n -(T-\lambda I)x_m -\lambda x_m+\lambda x_n\|

but since

\displaystyle (T-\lambda I)x_n-(T-\lambda I)x_m -\lambda x_m \in Y_{n+1}

we have

\displaystyle \|Tx_n -Tx_m\| \geq |\lambda|/2.

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

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: