Isometry on a dense sub-space of a Banach space? Let $X$ be a Banach space and let $D$ be a dense sub-space of $X$. I don't know if the following fact is true:

Fact: For every (linear) isometry $T\in\operatorname{Iso}(X)$ and for every $\varepsilon > 0$ there is an isometry (linear) $S\in\operatorname{Iso}(D)$ such that: $\|S-T\|<\varepsilon$.

Thank for any hint.
 A: I've been thinking about this all week, and I swear I have a counterexample. Consider $L^1([0,1])$ with the dense subset $C^{\infty}$. The isometry on the main space is multiplying each function by a piecewise function which is 1 on the interval $[0,1/2)$ and $-1$ on the rest of the unit interval. This is an isometry, but it does not fix $C^{\infty}$, because it introduces large gaps. 
Now, there are continuous functions that bridge such gaps in short time, but the steeper a function of norm 1 is at $1/2$, the faster its image under a hypothetical isometry of $C^{\infty}$ would have to bridge the gap to stay a fixed distance away. Then take two functions of norm 1, one not so steep, one very steep; the image of 
their difference is the difference of their images, which would bridge the gap a little to slowly and thus not be an isometry.
I'm sorry this isn't fleshed out more, and I'd be interested in hearing your thoughts.
A: I really don't understand your counterexample. To be sure that we speak the same language, i will resume what i understand for your counterexample:
For $X$ you take $L^{1}([0,1])$ and for $D$ you take $C^{\infty}([0,1])$. After you consider the following map: 
$g(x)=\left\{
    \begin{array}{ccccc}
      1& if  &  x\in [0,\frac{1}{2}) \\
      -1& if  &  x \in (\frac{1}{2},1] \\
    \end{array}\right.$
Now for $T$ you take 
$\begin{array}{lll}
T:& L^{1}([0,1]) &\longrightarrow L^{1}([0,1])\\
&f &\longmapsto gf\\
\end{array}$
But i don't understand why you manage to have a contradiction with this.
Please feel free to correct me if my interpretation of you counterexample is not true.
Thank one more
