Completion of normed space Let $(X,\|\cdot\|_X)$ be a Banach space and $S$ a subspace of $X$ (not necessarily closed).
Consider a linear (not neccesarily bounded) operator $T:S \rightarrow X$ and equip $S$ with the norm $\|u\|_S := \|u\|_X + \|T(u)\|_X$, which makes $T$ continuous.
Denote $S'$ the completion of $S$ with respect to the norm $\|\cdot\|_S$.
1) Can we assure that $S'$ i contained in $X$ (obviously, the notion of contained is in the sense of the Cauchy classes of equivalence). I think so, because $\|\cdot \|_X \leq \|\cdot\|_S$ in $S$ and therefore every pair of equivalent Cauchy sequences in $(S,\|\cdot\|_S)$ are also equivalent Cauchy sequences in $(X,\|\cdot\|_X)$.
2) Can we extend $T$ in some way to $T': S' \rightarrow X$? Would it be something like $T'(s') = \lim_n T(s_n)$, where $s_n$ is a sequence in $S$ converging to $s'$ in the extension of the norm $\|\cdot\|_S$?
3) In case 2) is true, does it hold that $\operatorname{Ker}(T') = \operatorname{Ker}(T)$?
In case you're wondering why I'm asking this, the answer is that I'm trying to know if what holds in Sobolev Spaces can also be extended in some way for more general settings. I mean, if we consider $\Omega \subset \mathbb R$ an open interval and $X = L^2(\Omega)$, $S = C_c^\infty(\overline\Omega)$ and $T$ the derivative operator we know that 1)-2)-3) hold, being $S' = H^1(\Omega)$ and the Kernels the space of constant functions.
 A: I'll answer the second of your questions first. Yes, we can extend $T$ to $S'$, and what you wrote is one of the standard ways of describing the extension. That is a special case of a more general theorem. If $E$ is a normed space, $F$ a Banach space, and $\tilde{E}$ the completion of $E$, then every continuous linear map $L \colon E \to F$ has a unique continuous linear extension $\tilde{L} \colon \tilde{E} \to F$. Also this is a special case of a more general theorem, we don't need the norms, the same (mutatis mutandis) holds if $E$ is a topological vector space, $\tilde{E}$ its completion and $F$ a complete topological vector space. And the vector space structure is also irrelevant, we can take $E$ as a uniform space with completion $\tilde{E}$, $F$ a complete uniform space, and $L \colon E \to F$ a uniformly continuous function. (If you don't know uniform spaces yet, take metric spaces.)
The answer to your first question is more involved. While indeed equivalent Cauchy sequences in $S$ correspond to equivalent Cauchy sequences in $X$, it may be the case that classes of non-equivalent Cauchy sequences in $S$ correspond to classes of equivalent Cauchy sequences in $X$.
I find it more transparent to identify $S$ with the graph of $T$, that is to consider
$$Y = \{(u, T(u)) : u \in S\} \subset X \times X.$$
Endowing $X\times X$ with the norm $\lVert (x,y)\rVert_{X\times X} = \lVert x\rVert_X + \lVert y\rVert_X$ makes it a Banach space (this norm induces the product topology) and the projection $\pi_1 \colon (x,y) \mapsto x$ induces an isometry $\iota \colon Y \to S$ when $Y$ is normed by the restriction of $\lVert\,\cdot\,\rVert_{X\times X}$ and $S$ is normed by $\lVert\,\cdot\,\rVert_S$. This gives an identification of $S'$ with $\overline{Y}\subset X \times X$. And the natural interpretation of "$S'$ is contained in $X$" is that the restriction of $\pi_1$ to $\overline{Y}$ provides that inclusion, i.e. that $\pi_1\lvert_{\overline{Y}}$ is injective. That is easily seen to be equivalent to saying that $\overline{Y}$ is the graph of a linear map (with domain $\pi_1(\overline{Y}) \subset X$), and to the condition $(0,v) \in \overline{Y} \iff v = 0$. Thus in this sense $S'$ is contained in $X$ if and only if $T$ is a closable operator. There may be continuous injections $S' \hookrightarrow X$ also in other cases, but the extension of the natural inclusion $S \hookrightarrow X$ to $S'$ is injective if and only if $T$ is closable.
To sum up: in general we cannot assure that $S'$ is contained in $X$, but modest niceness conditions on $T$ suffice.
The short answer to the third question is also "in general, no". If $\ker T$ is not a closed subspace of $X$, then we trivially have $\ker T \subsetneqq \overline{\ker T} \subset \ker T'$ (well, if $T$ isn't closable the inclusions need to be interpreted appropriately). But we can also have $\ker T \subsetneqq \ker T'$ if $\ker T$ is closed in $X$, even if $T$ is continuous. Consider in $X = \ell^2(\mathbb{N})$ the subspace $S$ of sequences with only finitely many nonzero terms, and $T(u) = u - \langle u \mathop{\vert} \xi\rangle\cdot \xi$ where $\xi$ is a unit vector with infinitely many nonzero terms. Then $T$ is injective, $S'$ can be naturally identified with $X$, and $\ker T' = \mathbb{C}\cdot \xi$.
