LF or LB space that happens to be finite dimensional

Question

Let $\{V_n\}_{n=1}^\infty$ be a collection of finite dimensional vector subspaces of $L^2[0,1]$ such that $V_n \subset V_{n+1}$ and $\bigcup_{n=1}^\infty V_n$ is dense in $L^2[0,1]$. Suppose further that each $V_n$ consists of smooth functions.

Consider a collection of invertible linear mappings $\{T_n\}_{n=1}^\infty$ defined on each $V_n$ such that $T_n(A) \subset T_{n+1}(A)$ for all $A \subset V_n$.

Now, if we fix some $m \in \mathbb{N}$, then the image $T_n(V_m)$ has the same dimension as $V_m$ for all $n \geq m$. Now, I wonder what the vector space \begin{equation} \bigcup_{n=1}^\infty T_n(V_m) \end{equation} looks like. It certainly looks like finite-dimensional, but is it possible to say more about this space? For example, does it still consist of smooth functions?

Also, if $v \in V_m$, in what sense does $T_n(v)$ converge in $\bigcup_{n=1}^\infty T_n(V_m)$ as $n \to \infty$?

I think LF or LB spaces are relevant topics, but cannot find some concrete information to apply to this case. Could anyone please help me?

Edit : OK, I need to change the condition $T_n(A) \subset T_{n+1}(A)$ by the one that

"there exists a projection $P_{n,n'} : T_n(V_m) \to T_{n'}(V_m)$ for all $n \geq n' \geq m$ such that $P_{n', k} \circ P_{n,n'}= P_{n,k}$ for $n \geq n' \geq k$ and $P_{n,n}=Id$." Also, I further asssume that the image of each $T_n$ consists of smooth functions as well. In this case, Is the above union a finite-dimensional space consisting of smooth functions, so that I can use "any" $L^p$ norm on it?

Answer 1

The expression \begin{equation} \tilde W_m:=\bigcup_{n=1}^\infty T_n(V_m) \end{equation} is undefined in general for $m\ge2$, because $T_n$ is defined (and is invertible) only on $V_n$ and hence $T_n(V_m)$ may be undefined for $n<m$.

So, instead of $\tilde W_m$, one may want to consider \begin{equation} W_m:=\bigcup_{n=m}^\infty T_n(V_m). \end{equation} However, the condition that $T_n(A)\subset T_{n+1}(A)$ for all $A\subset V_n$ implies that $T_m(V_m)\subset T_n(V_m)$ for all $n\ge m$. It is also given that $T_n(V_m)$ has the same dimension as $V_m$ for all $n\ge m$. So, $T_n(V_m)=T_m(V_m)$ for all $n\ge m$, and hence we simply have \begin{equation} W_m=T_m(V_m). \end{equation}

As to whether $W_m$ will still consist of smooth functions, the answer is: not in general. E.g., let $V_n$ be the space of all polynomials of degree $\le n$. Let $(h_n)_{n=0}^\infty$ be an enumeration of (say) the Haar basis. For a polynomial $p\in V_n$ such that $p(t)=a_0+a_1 t+\cdots+a_n t^n$ for some scalars $a_0,\dots,a_n$ and all $t\in[0,1]$, let \begin{equation} T_n(p):=a_0 h_0+a_1 h_1+\cdots+a_n h_n. \end{equation} Then all your conditions on the $V_n$'s and the $T_n$'s hold, whereas $W_m$ contains the (discontinuous) Haar functions $h_0,\dots,h_m$.

Answer 2

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.