$\sigma_p(f(A))=f(\sigma_p(A))$? Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $\sigma_p(A)$ be the point spectrum of $A$ (the set of $\lambda$ for which $ker(A-\lambda I)\neq 0$). I am trying to find a counter example for

$\sigma_p(f(A))=f(\sigma_p(A))$.

I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $\sigma_p(f(A))$ and $\sigma_p(A)$ are both empty...
Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?
 A: Let $f$ be a constant polynomial.  Then $f(A) = c I$ for some scalar $c$, so $\sigma_p(f(A)) = \{c\}$.  But
$\sigma_p(A)$ might be empty.
But these are the only counterexamples.
If $\lambda \in \sigma_p(A)$, there is nonzero $v \in H$ such that
$A v = \lambda v$, and then $f(A) v = f(\lambda) v$, so $p(\lambda) \in \sigma_p(f(A))$.  Thus $f(\sigma_p(A)) \subseteq \sigma_p(f(A))$.
On the other hand, suppose $\lambda \in \sigma_p(f(A))$.
Thus there is nonzero $v \in H$ such that $f(A) v = \lambda v$.
If $f$ has degree $d > 0$, the polynomial $f(z) - \lambda$ can be factored over the complex numbers as
$$f(z) - \lambda = \prod_{j=1}^d (z - \alpha_j)$$
so that
$$ 0 = (f(A) - \lambda) v = \prod_{j=1}^d (A - \alpha_j I) v $$
For some $k$, $1 \le k \le d$, we must have $\prod_{j=k}^d (A - \alpha_j I)v = 0$ but $\prod_{j=k+1}^d (A - \alpha_j I) v \ne 0$, i.e. with $w = \prod_{j=k+1}^d (A - \alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $w\ne 0$ and
$(A - \alpha_k I) w = 0$, so $\alpha_k \in \sigma_p(A)$, and 
$f(\alpha_k) = \lambda$.  So $\sigma_p(f(A)) \subseteq f(\sigma_p(A))$.
