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

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 $$\sigma_c(T)=\overline{\sigma_p(T)}\setminus\sigma_p(T)$$.

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.

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.

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.

• Thank you for your answer. (a) I've got one question in the Hilbert case: I'm not sure whether your $T$ is supposed to be an arbitrary operator, but please note that the $T$ (which I will now denote by $T_a$) is not self-adjoint, since $T_a^\ast=T_{\overline a}$ (hence it is self-adjoint iff $a$ is a real sequence). – 0xbadf00d Feb 21 at 17:08
• (b) Regarding the second part of your answer: I'm not sure whether I'm missing something, but isn't my $T$ given in the question precisely of the form ($T_ax=ax$ for all $x\in\ell^p$, where $a\in\ell^\infty$) you are asking for? (And note that I didn't edit the question; hence there is no "original" post). In any case, we may clearly note that $\lambda-T_a=$T_{\lambda-a}$, where$\lambda-a=(\lambda-a_n)_{n\in\mathbb N}$, for all$\lambda\in\mathbb C$. So,$\lambda-T_a$is of the same form. Am I missing something? – 0xbadf00d Feb 21 at 17:08 • (c) I think what you are using the second part is that if$T\in\mathfrak L(X,Y)$, then$\overline{\mathcal R(T)}={\mathcal N(T')}_\perp$. So, don't we get that$T$has dense range if and only if$T'$is injective? For clarity, denote the operator defined in the question by$T_{p,\:a}$. Then we've clearly got$T_{p,\:a}'=T_{q,\:a}$, where$q\in[1,\infty]$with$\frac1p+\frac1q=1$. Then we would obtain that$T_{p,\:a}'$has dense range if and only if$T_{q,\:a}$is injective. But, clearly,$T_{q,\:a}$is injective iff$T_{p,\:a}\$ is injective. Would be great if you reply to that. – 0xbadf00d Feb 21 at 18:47
• I am sorry if I was not sufficiently clear: by "original post" I meant simply the first post in this thread, namely the question itself. I will soon edit my answer to try to better explain my point. – Ruy Feb 21 at 20:14