The usual definitions of a (strong) convergence and a weak convergence of a sequence $(x_n)_{n \geq 1}$ go something like this:
Definition 1: A sequence $(x_n)_{n \geq 1}$ in a normed space $X$ converges to $x \in X$ if
$\hspace{3cm} ||x_n - x||_X\rightarrow 0$ as $n \rightarrow \infty$.
Definition 2: A sequence $(x_n)_{n \geq 1}$ in a normed space $X$ converges weakly to $x \in X$ if for every $f \in X^{*}$, we have that
$\hspace{3cm} f(x_n) \rightarrow f(x)$ as $n \rightarrow \infty$,
where $X^*$ represents a dual space.
My questions are:
If instead of $x \in X$ we need to show that $x \in Y$ where $X \subset Y$, what would change then in Definition 1 and Definition 2 given above? I think that in this case we just understand the sequence $(x_n)_{n \geq 1} \in X$ as a sequence in $Y$ and that's it. But I am not sure are there other possibilities.
What if $X$ is not a subset of $Y$?
If we use that $X \subset Y$ is continuous or compact embedding or that $X$ is dense in $Y$ or etc., I am sure that it would be easier to show convergence or weak convergence of the sequence $(x_n)_{n \geq 1} \in X$ to the $x \in Y$. But how precisely?
Background: I work on a problem where I have a sequence of functions $(x_n)_{n \geq 1}$ with values in some normed space X. I am trying to show weak convergence to a limit $x \in Y$. Proving (strong) convergence is not possible. Sequence $(x_n)_{n \geq 1}$ is smooth (has values in some appropriate Sobolev space such as $H^s$ or similar) and the limit is discontinuous (has values in the spaces such as $BV, L^{\infty}, L^1,...$). I was thinking of trying to show that $X$ is precompact in $Y$ (meaning that every sequence in $X$ has a subsequence that converge in $Y$).
Help with this would be great (especially with the questions 2 and 3). Thanks in advance.
If $\|\cdot\|$ is a norm only on $X$, the expression $\|x_n-x\|$ does not make sense for $x\in Y\setminus X$. Likewise, $f(x)$ is not defined for $f\in X^*$ and $x\notin X$. You can however ask for the convergence in a different norm on $Y$ or weakly in $Y$. If the embedding $X\hookrightarrow Y$ is compact (i.e., the unit ball of $X$ is relatively compact in $Y$) convergence may become easier: It would be sufficient that the sequence $x_n$ is bounded in $X$ and that for any two subsequences which converge in $Y$ the limit is the same. Then the sequence converges in $Y$. The advantage is that you do not have to find limits but that you only have to show uniqueness of limits.