Let $X:=C^0([0,2],\mathbb C)$,$\phi\in X$ and $T\in L(X)$ defined as:

$$(Tf)(t):=\phi(t)f(t),t\in [0,2]$$

Compute: $\sigma_p(T),\sigma_c(T),\sigma_r(T),\sigma(T)$ and $\rho(T)$

I am quite new to this subject and I have no idea how to compute them. But I just need to compute the first 3 values, since $\sigma(T)=\sigma_p(T) \cup \sigma_c(T) \cup \sigma_r(T)$ (disjoint union) and $\rho(T)=\mathbb C$\ $\sigma(T)$

I want to start with the first one:

By definition $\sigma_p(T)=\{\lambda\in \mathbb C: T-\lambda$ is not injective }, so I need to find all $\lambda$ such that :

$(T-\lambda)f(t)=\phi(t)f(t)-\lambda f(t)$ has a non-trivial solution (in terms of $f(t)$)

Isn't that only possible if $\phi(t)$ is constant in $[0,2]$?

I am thankful for any kinds of tips for the first and of course for the other computations, also for useful references, tipps and tricks for the computation of the spectrum.

  • 1
    $\begingroup$ You are almost correct, $\phi(t)$ has to be $\lambda$ when $f(t) \neq 0$. $\endgroup$ – user99914 Dec 3 '14 at 13:16
  • $\begingroup$ Okay, thanks. And how about the other spectrums? $\endgroup$ – Epsilondelta Dec 3 '14 at 14:09
  • $\begingroup$ If $\phi-\lambda$ is non-zero on $[0,2]$, then $\lambda\in\rho(T)$. If $\phi-\lambda$ is zero at some point of $[0,2]$, then the range of $T-\lambda I$ is not dense, which means $\lambda\in\sigma_{p}(T)$ or $\lambda\in\sigma_{r}(T)$; the first case occurs iff the zero set of $\phi-\lambda$ is such that a non-zero $f \in C[0,2]$ vanishes on the zero set of $\phi-\lambda$. $\endgroup$ – Disintegrating By Parts Dec 3 '14 at 19:10
  • $\begingroup$ Thanks for the answer! Two questions: why the range of $I-\lambda$ is not dense, if $\phi-\lambda$ is zero at some point? And the other question: Should not you say at the end, that f has to vanish in the complement of the zero set of $\phi-\lambda$ ? $\endgroup$ – Epsilondelta Dec 4 '14 at 8:24
  • $\begingroup$ Minor note: the plural of spectrum is spectra. $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '15 at 0:24

We know $\lambda\in\sigma_p(T)$ if and only if there exists $f\neq0$ such that $Tf=\lambda f$. Observe that given such an $f$, the set $\{t\in[0,2]\,:\,f(t)\ne2\}$ is open and non-empty, and so on this set we must have $\phi=\lambda$. Conversely, if there is an open set $U\subseteq[0,2]$ on which $\phi=\lambda$ then we may find a continuous function $f:[0,2]\rightarrow\mathbb{C}$ such that $f=0$ on $[0,2]\setminus U$. Hence we have found $$\sigma_p(T)=\{\lambda\in\mathbb{C}\,:\,[\phi=\lambda]\ \text{has non-empty interior}\}$$ Next we look at the residual spectrum. Suppose $\lambda\in\mathbb{C}\setminus\sigma_p(T)$ is such that $\lambda-\phi$ has a zero in $[0,2]$. I claim $\lambda\in\sigma_r(T)$. Indeed, if $\lambda-\phi(t_0)=0$, note that for all $f\in X$ we have $(\lambda I-T)f(t_0)=0$, so letting $g(t)=1$ for all $t\in[0,2]$ we have $\|g-(\lambda I-T)f\|_\infty\ge1$ for all $f\in X$. Since $g\in X$, this shows $\lambda I-T$ does not have dense range, so $\lambda\in\sigma_r(T)$. Conversely, if $\lambda-\phi$ is never zero then for all $f\in X$, $g:=\frac{f}{\lambda-\phi}$ is continuous, so $f=(\lambda I-T)g$ and hence $\lambda I-T$ has dense range. This shows $$\sigma_r(T)=\{\lambda\in\mathbb{C}\,:\,[\phi=\lambda]\ \text{is non-empty with empty interior}\}$$ Finally we look at $\sigma_c(T)$. However note that if $\lambda\in\mathbb{C}\setminus(\sigma_p(T)\cup\sigma_r(T))$ then $\lambda-\phi$ is never zero on $[0,2]$, so as noted above if $f\in X$ then $(\lambda I-T)g=f$ where $g=\frac1{\lambda-\phi}f$. Moreover, $\frac1{\lambda-\phi}(\lambda I-T)f=f$ for any $f\in X$. Hence $\lambda I-T$ is invertible with inverse $(\lambda I-T)^{-1}f=\frac1{\lambda-\phi}f$. We know $(\lambda I-T)^{-1}$ is bounded by the bounded inverse theorem, but you can also show it directly. Since $\lambda-\phi$ is never zero, there exists $\delta>0$ such that $|\lambda-\phi|\ge\delta$ by compactness of $[0,2]$. This implies $\|(\lambda I-T)^{-1}f\|_\infty\le\frac1\delta\|f\|_\infty$ for all $f\in X$. Hence we have shown $$\sigma_c(T)=\emptyset$$ You can finish off your problem by noting $\sigma(T)=\{\lambda\in\mathbb{C}\,:\,\lambda=\phi(t)\text{ for some }t\in[0,2]\}$ and $\rho(T)=\{\lambda\in\mathbb{C}\,:\,\lambda\neq\phi(t)\text{ for all }t\in[0,2]\}$.

  • $\begingroup$ Wow. I understood most of the things. I will read it some more times, but one question: In the second part (continuous spectrum) you showed that $(\lambda I-T) $ does not have dense range. But the continuous spectrum is defined as the set of all $\lambda \in \mathbb C$ such that $T-\lambda$ is injective and $(T-\lambda)$ has dense range. So where is the mistake? $\endgroup$ – Epsilondelta Feb 8 '15 at 17:06
  • $\begingroup$ Right, I mixed up continuous and residual spectrum, haha. I will edit, thanks. $\endgroup$ – Jason Feb 9 '15 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.