# Converse of Spectral Theorem when initial operator is not bounded

It is easy to show that if a bounded linear operator on a separable Hilbert space $$\mathcal{H}$$ has:

(i) eigenvectors that constitute an orthonormal basis

(ii) eigenvalues that are real that go to zero

Then the operator must be compact and symmetric.

My question is, suppose I only know that the operator is linear. Can I still show that the operator is compact and symmetric?

My conjecture is no. Let $$\mathcal{H}$$ have basis $$\{e_i\}_{i=1}^\infty$$. Define our operator $$T$$ such that $$Te_k = \frac{1}{k^{1/3}}e_k$$. Then the eigenvalues tend to $$0$$.

Consider $$h = \sum_{k=1}^\infty \frac{1}{k^{2/3}}e_k$$. Then $$h \in \mathcal{H}$$ since $$\lVert h \rVert^2 = \sum_{k=1}^\infty k^{-4/3} < \infty$$.

Suppose T is continuous. Then $$\lVert Th \rVert = \sum_{k=1}^\infty k^{-1}$$ which diverges, so $$T$$ is not bounded. Does this argument work? Is $$T$$ a linear operator?

To get unbounded examples you can proceed as follows. Complete $$(e_k)_{k\in\mathbb{N}}$$ to a Hamel basis $$(e_i)_{i\in\mathbb{N}\sqcup I}$$ of $$\mathcal{H}$$ with $$\|e_i\|=1$$ for all $$i\in I$$. If $$\mathcal{H}$$ is infinite-dimensional, then $$I$$ is infinite (in fact, uncountable). Let $$\phi\colon \mathbb{N}\to I$$ be an injective function and define $$Te_k=\lambda_k e_k$$ for $$k\in\mathbb{N}$$, $$Te_{\phi(l)}=l e_1$$ for $$l\in\mathbb{N}$$ and $$Te_i=e_2$$ for $$i\in I\setminus\phi(\mathbb{N})$$. Then $$(e_k)$$ is an orthonormal basis of eigenvectors, we can choose a null sequence $$(\lambda_k)$$ for the eigenvalues and yet the operator $$T$$ is not bounded.
The operator $$T$$ constructed above will not be symmetric. In fact, every symmetric everywhere defined linear operator on a Hilbert space is bounded.