What is the norm on the completion of a Hilbert space? Let $X$ be a Hilbert space with a norm $|u|_X = |u|_{X_1} + |Gu|_{X_1}$, where $G:X \to X_1$ is linear and continuous, $X_1$ is a Hilbert space.
Define $$|u|_Y = |Gu|_{X_1} + |Tu|_{Z}\quad\text{for $u \in X$}\tag{1}$$ where $T:X \to Z$ is a continuous linear map and $Z$ is a Hilbert space.
Now define $Y$ to be completion of $X$ with respect to $|\cdot|_Y$.

My question is, what is the norm on $Y$? The expression (1) makes sense only for $u \in X$ (and not $u \in Y$) because $T:X \to Z$. Is there a natural way to extend this norm to $Y$? I think it is done using Cauchy sequences and defining a extension of $T$. However, this type of space must have been studied before so I am wondering if someone has a reference or knowledge of this.
 A: Your idea with the Cauchy sequences and the extension of $T$ is correct, and is a fairly typical argument in functional analysis so is how I imagine most people would solve this. The involved theorem is actually more general so I will state it here:

Theorem: If $X,Y$ and $Z$ are metric spaces with $Y,Z$ complete and $X$ dense in $Z$, then any uniformly continuous function $f:X\rightarrow Y$ extends uniquely to a uniformly continuous function $\hat{f}:Z\rightarrow Y$.

If you are unfamiliar with this theorem, I would recommend proving it yourself. A simple corollary is that if $Y,Z$ are Banach spaces and $X\subset Z$ is a dense subspace, then any bounded linear map $T\in\mathcal{L}(X,Y)$ lifts uniquely to a bounded linear map $\tilde{T}\in\mathcal{L}(Z,Y)$. Analysts will often use this without comment, and there are many other applications, such as defining the Bochner integral without any of the laboured set-up of the Lebesgue integral.
So in this case, once you have defined $Y$, note that $|Gu|_{X_1}\leq|u|_Y$ for all $u\in X$, so $G$ lifts to $\tilde{G}\in\mathcal{L}(Y,X_1)$. Similarly, $|Tu|_Z\leq|u|_Y$, so $T$ lifts to $\tilde{T}\in\mathcal{L}(Y,Z)$. Then for any $u\in Y$,
$$|u|_Y=|\tilde{G}u|_{X_1}+|\tilde{T}u|_Z.$$
