TITLE: On the Equivalent Norms for Subspaces
QUESTION [6 upvotes]: By a function space I mean a vector space of functions.
Assume that $M$ is a function spaces which $\|\cdot\|_{X}$ and $\|\cdot\|_{Y}$ are defined such that
\begin{align*}
\|f+g\|_{X}&\leq\|f\|_{X}+\|g\|_{X}\\
\|af\|_{X}&=|a|\|f\|_{X}
\end{align*}
but with the possibility that $\|f\|_{X}=\infty$ for $f\in M$. Of course $\|\cdot\|_{Y}$ also bears these properties.
Now let $X$ and $Y$ be such that $\|f\|_{X}<\infty$ and $\|f\|_{Y}<\infty$ respectively for $f\in M$, so both $\|\cdot\|_{X}$ and $\|\cdot\|_{Y}$ are norms on $X$ and $Y$ respectively.
Assume further that a subspace $D$ is such that $\overline{D}^{X}=X$ and $\overline{D}^{Y}=Y$ and we have 
\begin{align*}
\|f\|_{X}\approx\|f\|_{Y},~~~~f\in D
\end{align*}
in the sense that 
\begin{align*}
c^{-1}\|f\|_{Y}\leq\|f\|_{X}\leq c\|f\|_{Y},~~~~f\in D
\end{align*}
where the constant $c>0$ is independent of $f\in D$.
Initially I was thinking that if $X=Y$. Note that $f\in X$ only satisfies that $\|f\|_{X}<\infty$ but could be $\|f\|_{Y}=\infty$ and vice versa.
If $X=Y$, then whether it is necessarily that
\begin{align*}
\|f\|_{X}\approx\|f\|_{Y}.
\end{align*}
But actually this seems need no to be the case.
For the second concern, suppose $f\in X$ and $f_{n}\in D$ are such that $\|f_{n}-f\|_{X}\rightarrow 0$, we wish to show that $\|f\|_{Y}\leq C\|f\|_{X}$, this will imply at the same time that $\|f\|_{Y}<\infty$. 
At first we have 
\begin{align*}
\|f\|_{X}=\lim_{n\rightarrow\infty}\|f_{n}\|_{X}\geq c^{-1}\limsup_{n\rightarrow\infty}\|f_{n}\|_{Y}.
\end{align*}
But one cannot assert immediately that 
\begin{align*}
\limsup_{n\rightarrow\infty}\|f_{n}\|_{Y}\geq\|f\|_{Y}.
\end{align*}
And of course $\|f_{n}-f\|_{X}\rightarrow 0$ says nothing about the convergence of $\|f_{n}-f\|_{Y}$.
Philosophy: Initially I have a philosophical view that two norms are equivalent on a dense subspace, then they should also be equivalent on the space, which is just a slightly larger than the dense subspace, but I think that my philosophical view is not accurate.
Therefore I am looking for counterexample.
At least, can we say that $X=Y$ as sets? Note that $X=Y$ need no implies $\|\cdot\|_{X}\approx\|\cdot\|_{Y}$.

REPLY [3 votes]: This answer is a slight variation on the answer given by Slup. Consider $\ell^{2}$, let $e_{i}$ be standard orthonormal basis and consider the space
$$X=\{x\in\ell_{2}:x=\sum^{n}_{i=1}x_{i}e_{i}\text{ where }n\in\mathbb{N},\ x_{i}\in\mathbb{R}\}$$
the set of all finite sums of standard basis vectors. Note that $X$ is a subspace of $\ell^{2}$. Let $C$ be a linear subspace of $\ell_{2}$ such that $\ell_{2}=X\oplus C$, such a space exists due to Zorn's lemma. Consider the norm $\|\cdot\|_{2}$ and the norm $\|\cdot\|$ on $\ell_{2}$ given by
$$\|x+c\|=\|x\|_{2}+\|c\|_{3}\qquad(x\in X,c\in C).$$
Note that $\|\cdot\|_{2}$ and $\|\cdot\|$ are equivalent, indeed equal, on $X$, but $\overline{X}^{\|\cdot\|_{2}}=\ell_{2}$ and $\overline{X}^{\|\cdot\|}=X$.