My purpose is to provide a proof, without using the spectral theorem.
Let $T: \scr H \to H$ be a linear compact self-adjoint operator.
$T$ self-adjoint implies:
- $ \langle \phi, T \psi \rangle = \langle T\phi, \psi \rangle = \overline{\langle \phi, T \psi \rangle}$, which implies $\langle \phi, T \psi \rangle \in \mathbb R$.
- $- \| \phi\| \|T\| \|\psi\| \le \langle \phi, T \psi \rangle \le \| \phi\| \|T\| \|\psi\|$
If in addition $T$ is compact, then its point spectrum $\sigma_p(T)\equiv \{ \lambda \in \mathbb C \ |\ T\psi = \lambda \psi$, $ \ $ for some $\lambda$-eigenvector $\psi \in \scr H$$\}$ is real, non-void.
In particular $\exists\ \Lambda \in \sigma_p(T)$ that verifies $|\Lambda| = \|T\| = \underset {\sigma_p(T)}{\max}|\lambda|$. Let's call the corresponding non-null $\Lambda$-eigenvector $\psi_\Lambda$.
remark. 1. and 2. hold $\forall\ \phi, \psi \in \scr H$. If we take $\phi \in \ker(T)$ and $\psi = \psi_\Lambda$:
$- \| \phi\| \|T\| \|\psi_\Lambda\| \le \langle \phi, T \psi_\Lambda \rangle = \langle T\phi, \psi_\Lambda \rangle = 0
$
Since $\|T\|$ and $\|\psi_\Lambda\|$ are not zero, the only possibility left is $\|\phi\| =0 \implies\phi ={\bf 0} \implies \ker(T) \equiv \{\bf 0\} $.
$T$ is hence injective. CVD(?)
Is this proof correct?
Best Answer
The last step is wrong : since $\|\phi\|,\|T\|$ and $\|\psi_\Lambda\|$ are positive, the inequation : $$-\|\phi\|\|T\|\|\psi_\Lambda\| \leq 0$$ is a triviality.
The result itself is wrong : a finite rank orthogonal projector is compact, bounded and self-adjoint, but is not injective.