Spectra of restrictions of bounded operators Suppose $T$ is a bounded operator on a Banach Space $X$ and $Y$ is a non-trivial closed  invariant subspace for $T$. It is fairly easy to show that for the point spectrum  one has $\sigma_p(T_{|Y})\subseteq\sigma_p(T)$ and this is also true for the approximate point spectrum, i.e.
$\sigma_a(T_{|Y})\subseteq\sigma_a(T)$. However I think it is not true in general that $\sigma(T_{|Y})\subseteq\sigma(T)$. We also have
$$
\partial(\sigma(T_{|Y}))\subseteq\sigma_a(T_{|Y})\subseteq\sigma_a(T)
$$
Hence $\sigma(T_{|Y})\cap\sigma(T)\ne\emptyset$. Moreover, if $\sigma(T)$ is discrete then $\partial(\sigma(T_{|Y}))$ is also discrete, which implies that $\partial(\sigma(T_{|Y}))=\sigma(T_{|Y})$, so at least in this case the inclusion  $\sigma(T_{|Y})\subseteq\sigma(T)$ holds true. So for example holds true for compact, strictly singular and quasinilpotent operators.
Question 1: Is it true, as I suspect, that $\sigma(T_{|Y})\subseteq\sigma(T)$  doesn't hold in general? A counterexample will be appreciated. On $l_2$ will do, as I think that on some Banach spaces this holds for any operators. For example, if $X$ is hereditary indecomposable (HI), the spectrum of any operator is discrete. 
Question 2 (imprecise): If the answer to Q1 is 'yes', is there some known result regarding how large the spectrum of the restriction can become? 
Thank you.

For example, consider the right shift operator $R$ on $X = \ell^2({\mathbb Z})$, $Y = \{y \in X: y_j = 0 \ \text{for}\ j < 0\}$.  Then $Y$ is invariant under $R$, and $\sigma(R)$ is the unit circle while $\sigma(R|_Y)$ is the closed unit disk.
