Continuous and residual spectrum of a multiplication operator on $\ell^p$ in the non-Hilbert case $p\ne2$

Question

Let $a\in\ell^\infty$, $p\in[1,\infty)$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ It's easy to show that $\sigma_p(T)=\{a_n:n\in\mathbb N\}$.

How can we determine $\sigma_c(T)$ and $\sigma_r(T)$? And is there a nice characteriation of $\sigma_p(T')$?

It really easy to show that $\mathbb C\setminus\overline{\sigma_p(T)}\subseteq\rho(T)$.

But beyond that, I'm only able to answer the question in the case $p=2$. In that case $T$ is self-adjoint and hence $\sigma_r(T)=\emptyset$ from which we can conclude that $\sigma_c(T)=\overline{\sigma_p(T)}\setminus\sigma_p(T)$.

Answer 1

First I will compute the approximate point spectrum, $\sigma_{ap}(T)$ which is the set of $\lambda\in\mathbb{C}$ so that $T-\lambda I$ is not bounded below.

Obviously $\sigma_{p}(T)\subset\sigma_{ap}(T)$. Using the fact that in any Banach algebra the set of invertible elements is open, together with the fact that $B(X)$ is a Banach algebra for any Banach space $X$ one can conclude that $\sigma_{ap}(T)$ is a closed subset. Since $\sigma_p(T)=\{a_n:n\geq1\}$ as you have already observed, we have that $\overline{\{a_n:n\geq1\}}\subset\sigma_{ap}(T)$. I will show the other inclusion, i.e. that $\sigma_{ap}(T)\subset\overline{\{a_n:n\geq1\}}$. Equivalently, I will show that $\mathbb{C}\setminus\overline{\{a_n:n\geq1\}}\subset\mathbb{C}\setminus\sigma_{ap}(T)$.

Let $\lambda\in\mathbb{C}\setminus\overline{\{a_n:n\geq1\}}$. Then there exists a disk in the complex plane centered at $\lambda$ that does not get intersected by $\{a_n:n\geq1\}$, so there exists $\varepsilon>0$ so that $|a_n-\lambda|\geq\varepsilon$ for all $n\geq1$. Note that for $x=(x_n)\in\ell^p$ we have that $$\|(T-\lambda I)x\|_p^p=\sum_{n=1}^\infty|a_nx_n-\lambda x_n|^p=\sum_{n=1}^\infty|a_n-\lambda|^p\cdot|x_n|^p\geq\varepsilon^p\cdot\|x\|_p^p$$ so $T-\lambda I$ is bounded below, i.e. $\lambda\not\in\sigma_{ap}(T)$.

Corollary: $\sigma_{ap}(T)=\overline{\{a_n:n\geq1\}}$.

Now we compute the compression spectrum of $T$, $\sigma_{cp}(T)$ which is the set of $\lambda\in\mathbb{C}$ so that $T-\lambda I$ does not have dense range.

Lemma: Let $A:X\to X$ be a bounded operator from a Banach space into itself. Then $\sigma_{cp}(A)=\sigma_{p}(A^*)$, where $A^*$ is the adjoint operator $A^*:X^*\to X^*$.

Proof of the lemma: Let $\lambda\in\sigma_{cp}(A)$, so $A-\lambda I_X$ is an operator that does not have dense range. Set $Y=\overline{\text{range}(A-\lambda I_X)}$. Then $Y$ is a proper, closed subspace of $X$. By Hahn-Banach there exists a non-trivial functional $\phi\in X^*$ so that $\phi\vert_Y=0$. Recall that taking the adjoint of operators is a (contra-variant) functor, so $I_{X^*}=(I_X)^*$ and note that$(\mu B)^*=\mu B^*$ for all bounded operators $B$ and scalars $\mu$. But then $(T^*-\lambda I_X)=(T-\lambda I_X)^*$ and $$(T^*-\lambda I_{X^*})(\phi)=(T-\lambda I_X)^*(\phi)=\phi\circ(T-\lambda I_X)=0.$$ Therefore $\phi\in\ker(T^*-\lambda I_{X^*})$, so $\lambda$ is an eigenvalue of $T^*$.

Conversely, if $\lambda\in\sigma_p(A^*)$, then there exists a non-zero functional $\phi\in X^*$ so that $A^*(\phi)=\lambda\cdot\phi$, so $\phi(Ax)=\phi(\lambda x)$ for all $x\in X$. Therefore $\phi$ is zero on the range of $A-\lambda I_X$ and since $\phi$ is continuous it is also zero on the closure of the range of $A-\lambda I_X$. If $A-\lambda I_X$ has dense range, then $\phi=0$, a contradiction. $\blacksquare$

Let us now apply our lemma to our operator.

The dual of $\ell^p$ is $\ell^q$, where $q$ is the conjugate exponent of $p$ (i.e. $1/p+1/q=1$). We identify $$\ell^q\equiv\{\omega_x: x\in\ell^q\}$$ where $\omega_x(y)=\sum_{n=1}^\infty y_nx_n$ for all $y=(y_n)\in\ell^p$, where $x=(x_n)$. Note that if $x\in\ell^q$ then $T^*(\omega_x)=\omega_x\circ T$, so $$T^*(\omega_x)(y)=\omega_x(Ty)=\omega_x\big((a_ny_n))=\sum_{n=1}^\infty a_ny_nx_n=\sum_{n=1}^\infty y_n\cdot(a_nx_n)=\omega_{Sx}(y)$$ where $S:\ell^q\to\ell^q$ is the multiplication operator with the sequence $\{a_n\}_{n=1}^\infty\in\ell^\infty$. Therefore $T^*(\omega_x)=\omega_{Sx}$, so $T^*$ is identified with the operator $S:\ell^q\to\ell^q$. Since $S$ is a multiplication operator by a sequence of $\ell^\infty$ we already know its point spectrum, so we have that $\sigma_{p}(S)=\{a_n:n\geq1\}$, so by our lemma we conclude that $\sigma_{cp}(T)=\{a_n:n\geq1\}$.

Now the relations $$\sigma_r(T)=\sigma_{cp}(T)\setminus\sigma_p(T)$$ and $$\sigma_c(T)=\sigma_{ap}(T)\setminus(\sigma_r(T)\cup\sigma_p(T))$$ yield the result: $\sigma_r(T)=\emptyset$, $\sigma_c(T)=\{\text{ the accumulation points of }(a_n)\}\setminus\{a_n:n\geq1\}$.

A final comment: Note that I am only using the Banach space adjoint. This is slightly different than the Hilbert space adjoint, you can see more details about their (non-essential) difference in this post. For example, when $p=2$ and we have a Hilbert space in our hands you can use the Hilbert-space version of the lemma which says that if $A\in B(H)$ is a bounded operator on a Hilbert space, then the operator $T-\lambda I_H$ does not have dense range if-f $\bar{\lambda}$ is an eigenvalue of $T^*$, where $T^*$ is the Hilbert space adjoint.

I hope this answers your question!

Answer 2

The crucial fact that makes the Hilbert space case easy to deal with is that $$ \text{Ker(T)}^\perp = \overline{\text{Ran}(T^*)}. $$ So, when $T$ is self-adjoint and $\lambda $ is real, it immediately follows that $T-\lambda $ is injective iff $\text{Ran}(T-\lambda )$ is dense.

Even though there is no notion of self-adjointness on a Banach space such as $\ell ^p$, when $p\neq 2$, the same principle can be easily adapted, as long as we are dealing with diagonal operators, that is, operators of the form $T(x)=ax$, for $a\in \ell ^\infty $, as in the original post.

In fact, for such an operator, it is very easy to prove that the following are equivalent:

1. $T$ is injective,
2. $a_n\neq 0$, for every $n$,
3. $T$ has dense range.

Noticing that $T-\lambda $ is a diagonal operator, whenever $T$ itself is, this shows that the residual spectrum of a diagonal operator is always empty, so the conclusion follows just as easily as in the case of self-adjoint operators on Hilbert spaces.

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EDIT. Let me break down my answer in order to try to make it a bit clearer. Before I start, let me say that the arguments below are completely elementary and, in particular, make no use of transposed operators.

Lemma. Let $1\leq p<\infty $, and let $T$ be a diagonal operator on $\ell ^p$, namely an operator of the form $$ T(x_1, x_2, x_3, \ldots ) = (a_1x_1, a_2x_2, a_3x_3, \ldots ), $$ where $a = (a_1, a_2, a_3, \ldots ) \in \ell ^\infty $. Then the following are equivalent:

1. $T$ is injective,

2. $a_n\neq 0$, for every $n$,

3. $T$ has dense range.

Proof. (1) $\Rightarrow$ (2). For each $n$, let $e_n$ the $n^{th}$ canonical basis vector of $\ell ^p$. Then, since $T$ is injective, we have that $$ 0\neq T(e_n) = a_ne_n, $$ so clearly $a_n\neq 0$.

(2) $\Rightarrow$ (1). Suppose that $T(x)=0$. Then $a_nx_n=0$, for all $n$, and, since $a_n\neq 0$, we deduce that $x_n=0$, whence $x=0$, as well.

(2) $\Rightarrow$ (3). Observing that $$ T\left(\frac{e_n}{a_n}\right) = e_n, $$ we see that $e_n$ lies in the range of $T$. Consequently also $$ \text{span}\{e_n:n\geq 1\}\subseteq \text{Ran}(T), $$ so it follows that $\text{Ran}(T)$ is dense.

(3) $\Rightarrow$ (2). For each $n$, let $E_n$ be the subspace of $\ell ^p$ given by $$ E_n=\{x\in \ell ^p: x_n=0\}. $$ Notice that $E_n$ is the kernel of the continuous linear functional $$ x\in \ell ^p\mapsto x_n\in \mathbb R, $$ so $E_n$ is a proper closed subspace.

If $a_n=0$, then clearly $$ \text{Ran}(T)\subseteq E_n, $$ so $\text{Ran}(T)$ cannot be dense, contradicting (3). QED

Back to the question, observe that if $T$ is a diagonal operator, as in the above Lemma, then so is $T-\lambda I$, for every scalar $\lambda $. Therefore $T-\lambda I$ is injective iff it has a dense range. We then conclude, as in the case of self-adjoint operators on Hilbert's space, that the residual spectrum of $T$ is empty.