Let $X=L^1(\mathbb{R})$ be the space of Lebesgue integrable functions $f:\mathbb{R}\rightarrow \mathbb{C}$ with the usual norm.
Let $T\in B(X)$ be defined by
$$(Tf)(t)= f(t+1)$$
I need to find the point spectrum, the approximate point spectrum and the spectrum itself.
I can show that $T$ defines a linear bijective isometry, so its spectrum is inside the set $\{ \lambda \in \mathbb{C} : | \lambda |=1 \}.$
Thanks to Daniel Fisher's help I can go further:
The point spectrum $\sigma_p(T) = \varnothing.$
Since $f(t) = \lambda f(t+1) \implies |f(t)|=|f(t+1)|\ \ \ \ \ \ \forall t $
But the only periodic integrable function is zero.
Now, the adjoint of $T$ is the "left shift" and similarly we show that its point spectrum is empty.
But the spectrum of $T$ is non empty and $\sigma(T)=\sigma_{ap}(T) \cup \sigma_p(T')$
Therefore there exist $\lambda$ s.t. $|\lambda| =1$ and a sequence $f_n$ of unit norm elements in $X$ s.t. $||\lambda f_n-Tf_n|| \rightarrow 0$
Hence taking $g_n=e^{i\rho} f_n$ for suitable $\rho$ we conclude that
$\sigma(T)=\sigma_{ap}(T)=\{ \lambda \in \mathbb{C} : | \lambda |=1 \}$
Could anyone please confirm whether or not my reasoning makes sense?
I will accept any answer explaining if it is right or what I did wrong!
This is not homework, thank you.
$(Tf)(t)=f(t+1)$ defines a unitary operator with inverse $(T^{\star}f)(t)=f(t-1)$. So the spectrum of $T$ is contained in the unit circle. Neither $T$ nor $T^{\star}$ has any point spectrum, which you have discovered. So $(T-e^{i\theta}I)$ and $(T^{\star}-e^{i\theta}I)$ have dense ranges for all $0 \le \theta < 2\pi$ because $$ \mathcal{R}(T-e^{i\theta}I)^{\perp}= \mathcal{N}(T^{\star}-e^{-i\theta }I)=\{0\}. $$ Nothing above excludes any of these ranges from being all of $X$ so far. However, that is not the case because every $e^{i\theta}$ is in the continuous spectrum of $T$ (and $T^{\star}$). To see this, consider $$ f_{\rho,\theta}(t) = \exp\{-\rho|t|+i\theta t\},\;\;\; \rho > 0,\; 0\le \theta < 2\pi. $$ Notice that $$ Tf_{\rho,\theta}-e^{-i\theta}f_{\rho,\theta} =(e^{-\rho|t|}-e^{-\rho|t+1|})e^{i\theta t}. $$ A direct calculation shows that $$ \lim_{\rho\downarrow 0}\frac{\|(T-e^{-i\theta}I)f_{\rho,\theta}\|}{\|f_{\rho,\theta}\|}=0. $$