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.