Using the completion of $ T:PC[0,1]\to PC[0,1]$ to $ \hat{T}:L^2[0,1]\to L^2[0,1]$ in order to find the image of $\hat{T}$ Given a bounded linear operator: $$ T:PC[0,1]\to PC[0,1]$$ and its completion $$ \hat{T}:L^2[0,1] \to L^2[0,1]$$ (defined by $\hat{T}x:=\lim Tx_n$ where $x_n \to x$).
If $T$'s image is $C[0,1]$ (all continuous functions), how can I show that the image of $\hat{T}$ is all of $L^2[0,1]$? (note, I may be wrong about this. Also - if it helps, it's a statement based on this question).
(note, $PC$ stands for piecewise-continuous functions, which are dense in $L^2[0,1]$. Also $C[0,1]$ is dense in $L^2[0,1]$).
I have tried to claim the following:
Let $f\in L^2[0,1]$. We will show that there exists a $g\in L^2[0,1]$ such that $\hat{T}g=f$.
Since $C$ is dense in $L^2$, there is a sequence $f_n \in C[0,1]$ such that $f_n \to f$. Since $f_n \in Img(T)$, there exist $g_n \in PC[0,1]$ such that $Tg_n = f_n$ for all $n$. Taking the limit we obtain that $\lim Tg_n = \lim f_n$.
Since the RHS is $f$, if I could only claim that $g_n$ must a converging sequence, converging to some $g \in L^2$ then I would be able to say, by the definition of the completion $\hat{T}$, that: $\hat{T}g=\lim Tg_n=\lim f_n=f$ so that we have found our $g$.    
I am unable to prove this last claim, i.e. that $g_n$ converges to $g$. So - either this is the wrong way to go about this, or I'm missing something.
EDIT:
I believe the claim holds if we replace $C[0,1]$ with $PC[0,1]$, since then we have that $T$ is a bijection and then it is possible to claim that each $f \in L^2$ has a distinct image in $L^2$ as such: $f,h \in L^2[0,1]$, with sequences $f_n, h_n \in PC[0,1]$ such that $f_n \to f$ and $h_n \to h$ then $\hat{T}f = \lim Tf_n \neq \lim Th_n = \hat{T}h$
EDIT 2:
This last comment on $PC$ is actually also incorrect since I'm basically stating that injective means surjective here, and that's not necessarily true.
Any ideas?
Thanks.
 A: Here is an example of a bounded operator $T$ on $L^2[0,1]$ which

*

*maps $C[0,1]$ onto itself,


*maps $PC[0,1]$ onto itself,
and yet it does not map $L^2[0,1]$ onto itself.
For $f$ in $L^2[0,1]$, set
$$
  T(f)(x) = f(\sqrt x), \quad \forall x\in  [0,1].
  $$
Then $T$ is bounded because
$$
  \|T(f)\|^2 =
  \int_0^1|f(\sqrt x)|^2\, dx =
  \int_0^12y|f(y)|^2\, dy \leq
  2\int_0^1|f(y)|^2\, dy =  2 \|f\|^2.
  $$
Points (1) and (2)  above are easy to verify since the correspondence $x\mapsto \sqrt x$ is an increasing  homeomorphism
on $[0,1]$.
To prove that  $T$ is not onto $L^2[0,1]$, pick  a real number $\beta $ in the interval $(-\frac 1 2,  -\frac 14]$, and let
$$
  g(x) = x^\beta , \quad \forall x\in  [0,1].
  $$
The clearly $g$ lies in $L^2[0,1]$, because $\beta >-\frac12$,  but I claim that $g$ is not in the range of $T$.  Indeed, if $g=T(f)$, for some
$f$, then
$$
  x^\beta  = g(x)  = T(f)(x) = f(\sqrt x),
  $$
whence
$$
  f(x) = f(\sqrt {x^2}) = x^{2\beta },
  $$
which is not in $L^2[0,1]$, because $2\beta \leq -\frac12$.
