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Michael Hardy
Michael Hardy
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For the edited question:

Let $\{f_k\}$ be a basis of $L^2$ using smooth functions.
Let $V_n = \mathop{span}(f_1, \ldots, f_n)$.
Let $T_m$ cyclically permute the $f_1, \ldots, f_m$ with $f_1 \mapsto f_m$.

Then all the requirements you gave are satisfied. But $$ \cup_{n = m}^\infty T_n(V_m) = \cup_{n = 1}^\infty V_n = \mathop{span}(\{f_k\})$$$$ \bigcup_{n = m}^\infty T_n(V_m) = \bigcup_{n = 1}^\infty V_n = \mathop{span}(\{f_k\})$$ is not finite dimensional.

For the edited question:

Let $\{f_k\}$ be a basis of $L^2$ using smooth functions.
Let $V_n = \mathop{span}(f_1, \ldots, f_n)$.
Let $T_m$ cyclically permute the $f_1, \ldots, f_m$ with $f_1 \mapsto f_m$.

Then all the requirements you gave are satisfied. But $$ \cup_{n = m}^\infty T_n(V_m) = \cup_{n = 1}^\infty V_n = \mathop{span}(\{f_k\})$$ is not finite dimensional.

For the edited question:

Let $\{f_k\}$ be a basis of $L^2$ using smooth functions.
Let $V_n = \mathop{span}(f_1, \ldots, f_n)$.
Let $T_m$ cyclically permute the $f_1, \ldots, f_m$ with $f_1 \mapsto f_m$.

Then all the requirements you gave are satisfied. But $$ \bigcup_{n = m}^\infty T_n(V_m) = \bigcup_{n = 1}^\infty V_n = \mathop{span}(\{f_k\})$$ is not finite dimensional.

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Willie Wong
Willie Wong
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For the edited question:

Let $\{f_k\}$ be a basis of $L^2$ using smooth functions.
Let $V_n = \mathop{span}(f_1, \ldots, f_n)$.
Let $T_m$ cyclically permute the $f_1, \ldots, f_m$ with $f_1 \mapsto f_m$.

Then all the requirements you gave are satisfied. But $$ \cup_{n = m}^\infty T_n(V_m) = \cup_{n = 1}^\infty V_n = \mathop{span}(\{f_k\})$$ is not finite dimensional.