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?


The operator is not necessarily bounded. The example you give is unclear, specifying the values on an orthonormal basis does not uniquely determine a linear operator (it does for a bounded 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.

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