On the Equivalent Norms for Subspaces

Question

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 $\|\cdot\|_{X}\approx\|\cdot\|_{Y}$.

Answer 1

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 $\overline{X}^{\|\cdot\|}=X$.

Answer 2

I don't now what is a precise meaning of function space. I think that what I have written below gives a counterexample you are seeking for $M$ being some abstract vector space (possibly algebraically isomorphic with some space of functions).

Let $D$ be any vector space with norm $||-||$ such that $(D,||-||)$ is not a Banach space. Let $(\overline{D}, ||-||')$ be a completion of $(D, ||-||)$. Then by Zorn's lemma we may write

$$\overline{D} = D \oplus C$$

for some linear subspace $C$ of $\overline{D}$ - decomposition as vector spaces but not as normed spaces. Now let $C_1$ and $C_2$ be two distinct vector spaces with linear isomorphisms $\phi_1:C_1\rightarrow C$ and $\phi_2:C_2\rightarrow C$, respectively. Consider vector space

$$M = C_1 \oplus D \oplus C_2$$

Now we define two functions $||-||_1,||-||_2$ on $M$. For this pick $c_1\in C_1,\,d\in D, c_2\in C_2$ and define

$$||(c_1,d,c_2)||_1 = \begin{cases}||\phi_1(c_1) + d||' &\mbox{ if }c_2 = 0\\ +\infty &\mbox{ if } c_2\neq 0 \end{cases}$$

and

$$||(c_1,d,c_2)||_2 = \begin{cases}||d + \phi_2(c_2)||' &\mbox{ if }c_1 = 0\\ +\infty &\mbox{ if } c_1\neq 0 \end{cases}$$

Then $X =\{(c_1,d,0)\in M|c_1\in C_1,d\in D\} \subseteq M$ and $Y = \{(0,d,c_2)\in M|d\in D,c_2\in C_2\}\subseteq M$. Hence clearly $X \neq Y$.

Edit (thanks to indication of Floris Claassens in the comments below this answer).

If you want $||-||_1,||-||_2$ to be norms on $M$, then you can modify definitions as follows:

$$||(c_1,d,c_2)||_1 = ||\phi_1(c_1) + d||' + ||\phi_2(c_2)||'$$

$$||(c_1,d,c_2)||_2 = ||\phi_1(c_1)||' + ||d + \phi_2(c_2)||'$$

It does not change descriptions of $X, Y$ from the original, unedited answer.

Further edit.

In general if $X$ and $Y$ are complete, then they are both completions of $D$ with respect to the same (uniform structure) topology given by equivalent norms ${||-||_1}_{\mid D}$, ${||-||_2}_{\mid D}$ and hence by uniqueness of completion they must be abstractly isomorphic as normed spaces.