Showing properties of a space using dense subsets (soft)

Question

I'm noticing a lot of times during my functional analysis course, that I'm missing some calculus basics (2 years passed since my last class covering this stuff): Especially when working with Lebesgue- and Sobolev-Spaces, one often withdraws to dense subsets of those spaces in order to prove properties of the spaces itself. I try to formulate my problem as abstract as possible:

Imagine there is an abstract normed space $(X,\|\cdot\|)$ and some dense subset $Y\subset X$. First of all, dense means, that I can approximate each object $x\in X$ by a sequence $\{x_n\}$ of objects in $Y$. This means that: $$x_n\overset{n\to\infty}\to x\,\text{in }\|\cdot\|\quad\Leftrightarrow\quad\lim_{n\to\infty}\|x_n-x\| = 0$$

Imagine further, that we want to show a certain property for all objects $x\in X$. Often we would then just consider the approximating sequence of objects in $Y$, and for the objects in $Y$ the property is easily shown (most of the times).

In an exam I would now just write down that the property - because of density - also holds in the limit. Anyway, I'd be cheating on myself If I claim understand why exactly density is enough for this. Another point that's unclear to me is, if this procedure (of saying that it has to hold in the limit because of density) is always valid in the above abstract setting, or if there is some other assumption that has to hold.

Maybe there is someone who can enlighten me a bit.

EDIT: A concrete example

When proving the Poincaré-inequality $\|u\|_{L^p(U)}\leq C\|Du\|_{L^p(U)}$ for $U\subset\mathbb{R}^n$, and e.g. $u\in W^{1,p}_0(U)$, one just restricts oneself to a function $v\in C^\infty_0(U)$, for which the argument can be shown. Then I have here in my notes that after taking a series of $C^\infty_0$ functions $\varphi_n(x)$ that converge to $u$ in $W^{1,1}_0(U)$:

[...] We showed the claim for every $\varphi_n$. By density it has to hold also in the limit.

Answer 1

In my opinion, the keyword is "continuity". That is, if you want to prove that $$P(x) \le Q(x)$$ holds for all $x \in X$ and both function $P,Q : X \to \mathbb{R}$ are continuous, then it is sufficient to prove the inequality for $x$ in the dense subspace $Y$.

If you have shown the inequality on $Y$, then you can show that it holds on $X$. For an arbitrary point $x \in X$, you take a sequence $\{x_k\} \subset Y$ with $x_k \to x$. Then, $$P(x_k) \le Q(x_k)$$ for all $k \in \mathbb{N}$. Passing to the limit shows $$P(x) \le Q(x).$$

However, you have to be careful. It is, e.g., not possible to show the strict inequality $P(x) < Q(x)$, if this inequality merely holds on the dense subspace.