In mathematics, a linear operator f : V → V {\displaystyle f:V\to V} is called locally finite if the space V {\displaystyle V} is the union of a family of finite-dimensional f {\displaystyle f} -invariant subspaces. In other words, there exists a family { V i | i ∈ I } {\displaystyle \{V_{i}\vert i\in I\}} of linear subspaces of V {\displaystyle V} , such that we have the following:
⋃ i ∈ I V i = V {\displaystyle \bigcup _{i\in I}V_{i}=V}
( ∀ i ∈ I ) f [ V i ] ⊆ V i {\displaystyle (\forall i\in I)f[V_{i}]\subseteq V_{i}}
Each V i {\displaystyle V_{i}} is finite-dimensional. An equivalent condition only requires V {\displaystyle V} to be spanned by finite-dimensional f {\displaystyle f} -invariant subspaces. If V {\displaystyle V} is also a Hilbert space, sometimes an operator is called locally finite when the sum of the { V i | i ∈ I } {\displaystyle \{V_{i}\vert i\in I\}} is only dense in V {\displaystyle V} .
Examples Every linear operator on a finite-dimensional space is trivially locally finite. Every diagonalizable (i.e. there exists a basis of V {\displaystyle V} whose elements are all eigenvectors of f {\displaystyle f} ) linear operator is locally finite, because it is the union of subspaces spanned by finitely many eigenvectors of f {\displaystyle f} . The operator on C [ x ] {\displaystyle \mathbb {C} [x]} , the space of polynomials with complex coefficients, defined by T ( f ( x ) ) = x f ( x ) {\displaystyle T(f(x))=xf(x)} , is not locally finite; any T {\displaystyle T} -invariant subspace is of the form C [ x ] f 0 ( x ) {\displaystyle \mathbb {C} [x]f_{0}(x)} for some f 0 ( x ) ∈ C [ x ] {\displaystyle f_{0}(x)\in \mathbb {C} [x]} , and so has infinite (or zero) dimension. The operator on C [ x ] {\displaystyle \mathbb {C} [x]} defined by T ( f ( x ) ) = f ( x ) − f ( 0 ) x {\displaystyle T(f(x))={\frac {f(x)-f(0)}{x}}} is locally finite; for any n {\displaystyle n} , the polynomials of degree at most n {\displaystyle n} form a T {\displaystyle T} -invariant subspace.
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