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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.

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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 $P$.

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  • $\begingroup$ Sorry, it's not very clear to me what $E_\epsilon$ and its range space are. Why is $E_\epsilon$ a projection and why is $P$ bounded from below on the range space $M$ of $Q$. $E_\epsilon$ is a charactersitc function of operators, so its domain is set of operatos, and its range is also set of operators, am i right? $\endgroup$ – Yung Jun 15 '15 at 6:16
  • $\begingroup$ $E_{\epsilon}$ is a projection since it is obtained by applying a characteristic function to $P$. A characteristic function is a projection, and the Borel functional calculus is a $\ast$-homomorphism. $\endgroup$ – Prahlad Vaidyanathan Jun 15 '15 at 8:23
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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.

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  • $\begingroup$ Why is that if $P$ is not compact, then there exists nonzero $\lambda_0 \in \sigma (P)$ such that it is an eigenvalue with infinite multilplicity, or accumilation point for $\sigma (P)$ ? $\endgroup$ – Yung Jun 15 '15 at 7:25
  • $\begingroup$ Because if $P $ is selfadjoint, every nonzero eigenvalue has finite multiplicity, and the only accumulation point of the spectrum is zero, then $ P $ is compact. $\endgroup$ – Martin Argerami Jun 15 '15 at 7:55
  • $\begingroup$ I still don't get, this is just the contrapositive of the previous statement, but why is it true? I've never learned that, could you explain a little more? $\endgroup$ – Yung Jun 15 '15 at 8:02
  • $\begingroup$ I know every nonzero eigenvalue of a compact operator has finite multiplicity, and zero is the only possible accumulation point of the spectrum of a compact operator, but I don't know why if in addition, the compact operator is self-adjoint, then converse true. $\endgroup$ – Yung Jun 15 '15 at 8:06
  • $\begingroup$ It's the Spectral Theorem. In the case as I said, $\int _{|\lambda|\geq\varepsilon }\lambda\,dE (\lambda) $ is finite-rank, and $\int _{|\lambda|<\varepsilon }\lambda\,dE (\lambda) $ has norm at most $\varepsilon $, so $P $ is a limit of finite-rank operators. $\endgroup$ – Martin Argerami Jun 15 '15 at 8:08

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