This thread is about extending a dense domain $D \subset L^{2}$ into $L^{2}$. I do not understand what Deyton means in his comment about getting contraction map when doing this.
I cannot see any theorem there, you can define a contraction map (norm less than 1) from $W: L2(ℝ)→L2(ℝ×ℝ)$, but you will loose the sense since you want to deal with signals and have $L^{2}$-energy of them.
Let $W: L^{2}$(ℝ) → $L^{2}$(ℝ×ℝ) be such that for all x ∈ $L^{2}$(ℝ) we have ∥Wx∥$_{L2(ℝ)}$ ≤ ∥x∥$_{L2(ℝ×ℝ)}$. Then ∥W∥ ≤ 1, so W is a contraction.
where $W$ is like Wigner-Ville distribution but Quadratic. It is quadratic nonlinear Time-Frequency Representation.
Why should the value of the corresponding norm estimate stay in one? There is no exponent in the contraction map normally as $\leq (\cdot)^{2}$.
Original situation
Take the signal $x \in L^{2}(\mathbf R \times \mathbf R)$. Assume the domain can be extended to $L^{2}(\mathbf{R} \times \mathbf R)$ (from initially dense subset $D \subset L^{2}(\mathbf R \times \mathbf R)$ as domain) \begin{equation} W : L^{2}(\mathbf{R} \times \mathbf R) \rightarrow L^{2}(\mathbf{R} \times \mathbf{R}), \end{equation} and its corresponding norm estimate \begin{equation} \lVert Wx \rVert_{L^{2}( \mathbf{R} \times \mathbf R ) } \leq \lVert x \rVert^{2}_{L^{2}(\mathbf{R} \times \mathbf{R})}, \end{equation} which is a contraction map an the corresponding norm estimate less than one.
I wanted to comment on your previous question, but I gave a 150 bounty and I lost the privilege to comment the questions.
First of all, I answered your doubts, that is, I said that, indeed, \begin{equation}(1) \ \quad \lVert Wx \rVert^{2}_{L^{2}( \mathbb{R} \times \mathbb{R} )} \leq \lVert x \rVert^{2}_{L^{2}(\mathbb{R} )} \ \ \forall_{x \in D}, \end{equation}
is a correct condition since the domain of $W$ is $D \subset L^2(\mathbb{R})$.
Then you asked me whether the above inequality is preserved if we take $x \in L^2(\mathbb{R})$ instead taking it from $D$.
I don't know the references from which you wrote it up. I don't understand why you want to take $x \in L^2(\mathbb{R})$, as you can see the domain of the map $W$ is $D$ so it may not be defined on the whole $L^2(\mathbb{R})$. Thus, there might be such $x \in L^2(\mathbb{R})$ for which $Wx$ does not make sense.
I mentioned contractions, but I didn't notice that there is no linearity assumption for $W$, the notion contraction for operators is rather used when we consider linear maps.
Further, $D$ is a set of all signals, so I guessed that in the context it is important to use them instead of ones which might not be signals.
Possibility
But, if you use a linear map (consider $W$ to be linear) then the following theorem might be useful:
http://en.wikipedia.org/wiki/Continuous_linear_extension
Theorem
Every bounded linear transformation $\mathsf{T}$ from a normed vector space $X$ to a complete, normed vector space $Y$ can be uniquely extended to a bounded linear transformation $\tilde{\mathsf{T}}$ from the completion of $X$ to $Y$. In addition, the operator norm of $\mathsf{T}$ is $C$ if and only if the the operator norm of $\tilde{\mathsf{T}}$ is $C$.
Thus, if your $W$ is a linear map and the inequality (1) is true, we can extend it on the whole $L^2(\mathbb{R})$ and the norm of the extension will be still less than $1$, so the inequality (1) will be also valid.
I hope that I was clear this time.