Locally finite operator

In mathematics, a linear operator ${\displaystyle f:V\to V}$ is called locally finite if the space ${\displaystyle V}$ is the union of a family of finite-dimensional ${\displaystyle f}$-invariant subspaces.^[1][2]: 40

In other words, there exists a family ${\displaystyle \{V_{i}\vert i\in I\}}$ of linear subspaces of ${\displaystyle V}$, such that we have the following:

• ${\displaystyle \bigcup _{i\in I}V_{i}=V}$
• ${\displaystyle (\forall i\in I)f[V_{i}]\subseteq V_{i}}$
• Each ${\displaystyle V_{i}}$ is finite-dimensional.

An equivalent condition only requires ${\displaystyle V}$ to be the spanned by finite-dimensional ${\displaystyle f}$-invariant subspaces.^[3][4] If ${\displaystyle V}$ is also a Hilbert space, sometimes an operator is called locally finite when the sum of the ${\displaystyle \{V_{i}\vert i\in I\}}$ is only dense in ${\displaystyle V}$.^{[2]: 78–79}

Examples

• Every linear operator on a finite-dimensional space is trivially locally finite.
• Every diagonalizable (i.e. there exists a basis of ${\displaystyle V}$ whose elements are all eigenvectors of ${\displaystyle f}$) linear operator is locally finite, because it is the union of subspaces spanned by finitely many eigenvectors of ${\displaystyle f}$.
• The operator on ${\displaystyle \mathbb {C} [x]}$, the space of polynomials with complex coefficients, defined by ${\displaystyle T(f(x))=xf(x)}$, is not locally finite; any ${\displaystyle T}$-invariant subspace is of the form ${\displaystyle \mathbb {C} [x]f_{0}(x)}$ for some ${\displaystyle f_{0}(x)\in \mathbb {C} [x]}$, and so has infinite dimension.
• The operator on ${\displaystyle \mathbb {C} [x]}$ defined by ${\displaystyle T(f(x))={\frac {f(x)-f(0)}{x}}}$ is locally finite; for any ${\displaystyle n}$, the polynomials of degree at most ${\displaystyle n}$ form a ${\displaystyle T}$-invariant subspace.^[5]