Every non-compact Hermitian operator P has an infinite dimensional invariant subspace on which P is bounded from below I want an explanation of the following statement.
If $P$ is a Hermitian operator on Hilbert space and not compact, there exists an infinite-dimensional subspace $M$, invariant under $P$, on which $P$ is bounded from below.
I've thought about it for 2 days and still can't figure out why, I've thought about the definition of compact operator, spectral theorem, etc..
The above statement comes from Halmos' problem book of Hilbert space, solution 176.
Could anyone understand this explain this to me?
Thank you in advance.
 A: One can begin by proving the following lemma :

For each $\epsilon > 0$, define
$$
\Delta_{\epsilon} := \{z\in \sigma(P) : |z| > \epsilon\}
$$
and set
$$
E_{\epsilon} := \chi_{\Delta_{\epsilon}}(P)
$$

Claim: $\exists \epsilon > 0$ such that $E_{\epsilon}$ is not compact.

Proof: Suppose $E_{\epsilon}$ is compact for all $\epsilon > 0$, then consider
$$
P - PE_{\epsilon} = f_{\epsilon}(P) \text{ where } f_{\epsilon}(z) = z-z\chi_{\Delta_{\epsilon}}(z) = z\chi_{\sigma(P)\setminus \Delta_{\epsilon}}(z)
$$
Hence
$$
\|P - PE_{\epsilon}\| \leq \epsilon
$$
and so $P$ is a limit of compact operators which is compact. This is a contradiction.

By the above lemma, $\exists \epsilon > 0$ such that
$$
Q := \chi_{\Delta_{\epsilon}}(P)
$$
is non-compact, and hence an infinite rank projection. Note that $PQ = QP$, so the range space $M$ of $Q$ is an invariant subspace under $P$.
A: Since $P$ is Hermitian, it satisfies the Spectral Theorem:
$$
P=\int_{\sigma(P)}\lambda\,dE(\lambda),
$$
for a spectral measure $E$ on the Borel $\sigma$-algebra of $\sigma(P)$. 
The fact that $P$ is not compact implies that there exists $\lambda_0\in\sigma(P)$, $\lambda_0\ne0$, such that it is an eigenvalue with infinite multiplicity, or accumulation point for $\sigma(P)$. 
The easy case occurs when $\lambda_0$ is an eigenvalue with infinite multiplicity. In this case $E(\{\lambda_0\})$ is an infinite-dimensional subspace, and
$$
P\,E(\{\lambda_0\})=\int_{\sigma(P)}\lambda\,1_{\{\lambda_0\}}\,dE(\lambda)=\lambda_0\,E(\{\lambda_0\}).
$$
So, if $\xi$ is in the range of $E(\{\lambda_0\})$, we have
$$
\|P\xi\|=\|P\,E(\{\lambda_0\})\xi\|=\|\lambda_0\,E(\{\lambda_0\})\xi\|=|\lambda_0|\,\|\xi\|
$$
and $P$ is bounded below on the range of $E(\{\lambda_0\})$.
When $\lambda_0$ is not an eigenvalue but an accumulation point of $\sigma(P)$, we repeat the argument above replacing the set $\{\lambda_0\}$ with  $(\lambda_0-\delta,\lambda_0+\delta)$ for $\delta $ small enough so that  the interval does $\int _{|\lambda|\geq\varepsilon }\lambda\,dE (\lambda) $ not touch zero. 
