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Let $E, F$ be real Banach spaces. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. Let $I:E \to E$ be the identity map. For $T \in \mathcal L(E)$,

  • we denote by $N(T)$ its kernel and by $R(T)$ its range.
  • we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
  • for $\lambda \in EV(T)$, the set $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.

I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $T \in \mathcal{L}(E)$.

  1. Prove that $\sigma(T) = \sigma(T^*)$.
  2. Give examples showing that there is no general inclusion relation between $E V(T)$ and $E V(T^*)$.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it? I'm also happy to see other approaches.

We need two results from the same book, i.e.,

Let $A: D(A) \subset E \to F$ be an unbounded linear operator that is densely defined and has a closed graph.

Theorem 2.20. The following properties are equivalent:

  • (a) $A$ is surjective, i.e., $R(A)=F$,
  • (b) there is a constant $C$ such that $$ |v| \leq C |A^* v| \quad \forall v \in D (A^*), $$
  • (c) $N (A^*) = \{0\}$ and $R(A^*)$ is closed.

Theorem 2.21. The following properties are equivalent:

  • (a) $A^*$ is surjective, i.e., $R(A^*)=E^*$,
  • (b) there is a constant $C$ such that $$ |u| \leq C |A u| \quad \forall u \in D (A), $$
  • (c) $N (A) = \{0\}$ and $R(A)$ is closed.

  1. Let $\lambda \in \mathbb R$. By Theorem 2.20 and Theorem 2.21, $T+\lambda I$ is bijective IFF $(T+\lambda I)^* = T^*+\lambda I$ is bijective. The claim then follows.

  2. Let $E := \ell^2$ with its canonical norm $|\cdot|_2$. An element $x\in E$ is denoted by $x=(x_1, x_2, \ldots, x_n,\ldots)$. Consider the operators $$ \begin{align*} S_r x &= (0, x_1, x_2, \ldots, x_{n-1}, \ldots), \\ S_{\ell} x &= (x_2, x_3, x_4, \ldots, x_{n+1}, \ldots), \end{align*} $$ respectively called the right shift and left shift. From exercise 6.18 in the same book, we have

  • $S_r$ and $S_\ell$ are adjoints of each other.
  • $EV(S_r) = \emptyset$ and $E V (S_{\ell}) = (-1, 1)$.
Akira
  • 17,367
  • To finish (2) you should compute the $EV$ of $S_r$ and $S_\ell$. The argument for (1) works, but as a matter of style using a theorem that hides all the substance of the proof into a black box, it less elegant than a proof that exposes what is important for the result to hold. – NDB Jul 25 '23 at 20:57
  • You say these are real Banach spaces. Does that mean $\rho(T)$ and $\sigma(T)$ are subsets of $\mathbb R$? Normally, I would expect $\sigma(T)$ to include complex values, and then $\sigma(T^) = \overline{\sigma(T)}$. I suppose for real* operators, the set $\sigma(T)$ is invariant under conjugation, and that is what Brezis intends? – GEdgar Jul 28 '23 at 16:16
  • @GEdgar Yess, $\rho (T)$ and $\sigma (T)$ are subsets of $\mathbb R$. – Akira Jul 28 '23 at 17:25

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