= Solution
\b[No.] On the complex <l2 sequence space>, let $L(x_n)=(x_n/n)$. This is a bounded <self-adjoint operator>. At $\lambda=0$, $\ker L=\{0\}$, so zero is not an <eigenvalue> (an eigenvector must be nonzero). But $y=(1/n)\in\ell^2$ has no preimage in $\ell^2$: its only formal preimage is the constant-one sequence. Thus $L$ is not surjective and
$$
\boxed{0\in\Sigma(L)\quad\hbox{but}\quad0\notin\sigma_{\mathrm p}(L).}
$$
Alternatively $\|Le_n\|=1/n\to0$ excludes a bounded inverse. This example separates the <spectrum of a bounded operator> from its <point spectrum>.
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