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?



LF or LB space that happens to be finite dimensional
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](https://en.wikipedia.org/wiki/Haar_wavelet).
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 $h\_0,\dots,h\_m$.


