# How can I compare unbounded linear operators?

Let $X$, $Y$ be Hilbert spaces. Let $S, T : X \rightarrow Y$ be unbounded operator.

Suppose $S$ and $T$ be bounded operators. Then we can compare by their maximum distance on the unit ball of $X$.

Suppose $S$ and $T$ are unbounded but closed operators, with the same set of definition. Then we can compare them locally on the unit ball, i.e. for every point $x$ on unit ball of $X$ which lies in the domain of $S$ and $T$ there exists a neighbourhood $U(x) \in S^1(X)$ such that the distance of $S$ and $T$ is bounded on the intersection of the domain of $S$ and $T$ with $U(x)$. (Can we?). This means we compare their graphs.

Is there something like this for general unbounded operators? I can only imagine comparing them on finite-dimensional subspaces, to get some sense of distance.

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It is a little strange to write "Let $S, T : X \to Y$ be unbounded operator" and then "Suppose $S$ and $T$ be bounded operators." – Qiaochu Yuan Dec 3 '11 at 6:05
Any bounded operator is called unbounded, too, which is completely common, isn't it? – shuhalo Dec 4 '11 at 18:00