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Let $1\le p<q\le\infty$, $L:\ell^p\to\ell^p,x\mapsto(x_2,x_3,\ldots)$ and $R:\ell^q\to\ell^q,y\mapsto(0,x_1,x_2,\ldots)$.

It's easy to show $L'=R$, $\sigma(L)=\sigma(R)=\overline B_1(0)$, $\sigma_p(L)=B_1(0)$, $\sigma_r(L)=\emptyset$, $\sigma_c(L)=\partial B_1(0)$, $\sigma_p(R)=\emptyset$ and $B_1(0)\subseteq\sigma_r(R)$.

But I'm failing to show that $\sigma_c(R)=\emptyset$. How can we do that?

Clearly, if $\lambda\in\sigma_c(R)$, then $\mathcal R(\lambda-R)$ is dense and hence $\lambda\not\in\sigma_p(L)$, i.e. $\lambda\in\partial B_1(0)$ ... But I'm not able to derive a contradiction from that.

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Since $q>1$, you have $R=L'$, considering $L$ acting on $L^p$ for $p$ conjugate to $q$. From $$ \|(R-\lambda I)x\|_p=\|Rx-\lambda x\|_p\geq\|Rx\|_p-|\lambda|\,\|x\|_p=(1-|\lambda|)\,\|x\|_p, $$ we get that $R-\lambda I$ is bounded below for all $\lambda\in B_1(0)$. This precludes $\lambda$ from being an approximate eigenvalue. Combined with the fact that the boundary of the spectrum consists of approximate eigenvalues shows that $\sigma_{\rm ap}(R)=\partial B_1(0)$.

From $\sigma(R)=\sigma_r(R)\cup\sigma_{\rm ap}(R)$ we conclude that $B_1(0)\subset\sigma_r(R)$.

If $\lambda\in\partial B_1(0)$, then $ \lambda\not\in\sigma_p(L)$ (here we use $p<\infty$). This means that $\ker(L-\lambda I)=\{0\}$. It is well-known (proof at the end) that if an operator is injective, then its adjoint has dense range. So $\overline{\operatorname{ran}(R-\lambda I)}=\ell^q$, showing that $\lambda\in\sigma_c(R)$. That is, $\partial B_1(0)\subset \sigma_c(R)$. We have $$ \overline{B_1(0)}=\sigma(R)=\sigma_r(R)\cup\sigma_c(R) $$ (since $\sigma_p(R)=\emptyset$). As $\sigma_r(R)$ and $\sigma_c(R)$ are disjoint, with $B_1(0)\subset\sigma_r(R)$ and $\partial B_1(0)\subset\sigma_c(R)$, it follows that $$ \sigma_r(R)=B_1(0),\qquad \sigma_c(R)=\partial B_1(0). $$


Proof that $T:X\to Y$ is a bounded operator and $T'$ is injective, then $T$ has dense range.

If $\operatorname{ran}T$ were not dense, there exists $y\in Y\setminus\overline{\operatorname{ran}T}$. By Hahn-Banach there exists $g\in Y'$ with $g(y)=1$ and $g(Tx)=0$ for all $x\in X$. This means that $$ (T'g)(x)=g(Tx)=0 $$ for all $x$, so $T'g=0$. As $T'$ is injective, we get that $g=0$ a contradiction.

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  • $\begingroup$ Thank you for your answer. It's clear to me that $B_1(0)\subseteq\sigma_r(R)$ (missed that in the question). But how do we prove $\overline{\operatorname{ran}(R-\lambda I)}=\ker(T-\overline\lambda I)^\perp$ (and what is $T$? $T=L$ I guess)? (I'm curious, by the way, why you're considering $\overline\lambda$. I mean, we know that if $\lambda\in\partial B_1(0)$, then $\lambda\not\in B_1(0)=\sigma_p(L)$ and clearly $\overline\lambda\not\in\sigma_p(L)$ as well. So, I guess the point is the former identity ...) $\endgroup$ – 0xbadf00d Feb 2 at 18:21
  • $\begingroup$ The formula that I know, for a general bounded linear operator $T$ on a complex Banach space $X$, is $\overline{\mathcal R(T)}=(\mathcal N(T'))_\perp$. Now, if $\lambda\in\mathbb C$, then $(\lambda-T)'=\lambda-T'$ (not $\overline\lambda-T'$). Did you confuse that with the Hilbert space adjoint? My expectation would be $\sigma_c(R)=\emptyset$. $\endgroup$ – 0xbadf00d Feb 2 at 18:24
  • $\begingroup$ Yes, you are right. I tend to think about Hilbert spaces. $\endgroup$ – Martin Argerami Feb 2 at 18:27
  • $\begingroup$ I'm on my phone right now, so I'll get back to this a bit later. $\endgroup$ – Martin Argerami Feb 2 at 18:31
  • $\begingroup$ Thank you in advance for your effort. $\endgroup$ – 0xbadf00d Feb 2 at 18:32

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