# continuous and residual spectrum of the right shift

Let $$1\le p, $$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.

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.

• 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 ...) – 0xbadf00d Feb 2 at 18:21
• 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 $\sigma_c(R)=\emptyset$. – 0xbadf00d Feb 2 at 18:24
• Yes, you are right. I tend to think about Hilbert spaces. – Martin Argerami Feb 2 at 18:27
• I'm on my phone right now, so I'll get back to this a bit later. – Martin Argerami Feb 2 at 18:31