If two norms are equivalent on a dense subspace of a normed space, are they equivalent?

Question

Given a vector space $V$ equipped with two norms $|\cdot|$ and $\|\cdot\|$ which are equivalent on a subspace $W$ which is $\|\cdot\|$-dense in $V$, are the two norms necessarily equivalent?

The statement seems like a relatively straightforward thing to show, but I can't manage it. Having played around with a few proof strategies and not getting anywhere, I'm starting to think that that the statement isn't true, but I'm not yet familiar with many normed spaces and can't think of a counter-example. Any help or hints would be much appreciated.

EDIT: To clear things up, in the book that I took this question from (Linear Analysis by Béla Bollobás) a normed space is defined to be a real or complex vector space, so I think that the intention is for $V$ to be over $\mathbb{R}$ or $\mathbb{C}$.

Answer 1

I submit this counterexample which, in my opinion, proves that the statement is false.

In the vein of this MathOverflow post by Gerald Edgar, let $V$ denote the real vector space of all polynomials of one variable and let $$\lVert P\rVert=\max_{x\in[0, 1]} \lvert P(x)\rvert,\qquad \forall P\in V.$$ Moreover, let $$W=\{a_0+a_2x^2+a_4x^4+\dots+a_{2k}x^{2k}\ :\ a_j\in \mathbb{R}\ k\in \mathbb{N}\}. $$ This is a dense subspace of $V$ (cfr. linked post).

Now consider the following linear operator: $$T(x^n)=\begin{cases} x^n & n\ \text{even} \\ nx^n& n\ \text{odd}\end{cases}$$ Its peculiarities are that:

1. $T\equiv I$ (identity) on the dense subspace $W$;
2. $T$ is not bounded.
3. $T$ is $1:1$.

Define $$\lvert P \rvert=\lVert T(P)\rVert.$$ Since $T$ is linear and $1:1$, this defines a norm on $V$. Moreover, this norms agrees with $\lVert\cdot\rVert$ on $W$. Nevertheless, the two norms are not equivalent, because this would imply boundedness of the operator $T$.

Answer 2

$\mathbb Q(\sqrt 2)$ has two norms which are equivalent on $\mathbb Q$ derived from the two embeddings into $\mathbb R$

Answer 3

This is not true in general; and in fact this is "never" true.

First, take for example $V=\mathcal C([0,1])$, the space of all continuous functions on $[0,1]$, and let $\Vert\;\Vert=\Vert\,\Vert_\infty$. Let also $W\subset V$ be the space of all polynomial functions on $[0,1]$. Then $W$ is dense in $(V,\Vert\;\Vert)$. Let $E\subset V$ be a linear subspace such that $E\oplus W=V$ (you can find this $E$ with some amount of "choice"). Then $E$ certainly has infinite dimension; so one can find (again with some amount of "choice") a norm $\vert \;\vert_E$ on $E$ which is not equivalent to (the restriction of) $\Vert\;\Vert$ to $E$. Now define the norm $\vert\;\vert$ on $V$ by $\vert e\oplus w\vert=\vert e\vert_E+\Vert w\Vert$ for every $(e,w)\in E\times W$. Then $\vert\;\vert$ are equal on $W$ but not equivalent.

This "construction" works in fact on any (infinite-dimensional) vector space $V$ with any given norm $\Vert\;\Vert$ on $V$. That is, the following holds true: For any infinite-dimensional vector space $V$ and any nor $\Vert\;\Vert$, one can find another norm $\vert \;\vert$ on $V$ such that $\vert\;\vert$ and $\Vert\; \Vert$ coincide on a dense (with respect to $\Vert\;\Vert$) linear subspace $W\subset V$ and yet are not equivalent.

To prove this, all what you need is to find a dense linear subspace $W\subset V$ such that the quotient space $V/W$ is infinite-dimensional; equivalently, for which you can write $V=E\oplus W$ with $\dim E=\infty$.

Now, it is "well known" that you can find a dense set $D\subset W$ (with respect to $\Vert\;\Vert$) which is also linearly independent; se e.g. here: Does there exist a linearly independent and dense subset? Take any sequence $(x_n)\subset D$ such that $\Vert x_n\Vert\to 0$. Then $D\setminus\{ x_n;\; n\in\mathbb N\}$ is again dense in $V$ (wrt $\Vert\;\Vert$). (Indeed, since the set $A=\{ 0\}\cup\{ x_n;\; n\in\mathbb N\}$ is countable, it has empty interior wrt $\Vert\;\Vert$ because nonempty open sets in a normed vector space are uncountable, and moreover $A$ is $\Vert\;\Vert$-closed in $V$; so $V\setminus A$ is dense and open in $V$ wrt $\Vert\;\Vert$, and hence $D\setminus A$ is also $\Vert\;\Vert$-dense in $V$). If you denote by $W$ the linear span of $D\setminus\{ x_n;\; n\in\mathbb N\}$ and by $E$ the linear span of $\{ x_n;\; n\in\mathbb N\}$, then everything is OK.

Note also that the "construction" does not contradict Giuseppe's statement since there is no reason for $\vert\;\vert$ to be complete; but perhaps it could be suitably modified?

Answer 4

WARNING: This answer contains a (probably fatal) error, see comments.

The statement is true with an additional assumption:

$V$ is complete with respect to $\lvert \cdot \rvert$ (recall that $W$ is dense with respect to the other norm $\lVert\cdot\rVert$).

Proof. By assumption, constants $C_1, C_2$ exist such that the following inequalities hold: \begin{equation}\tag{1} \begin{array}{ccc} \lvert y\rvert\le C_1\lVert y\rVert, & \lVert y\rVert\le C_2\lvert y\rvert & \forall y\in W. \end{array} \end{equation} So take $x\in V$. By assumption, we can find a sequence $y_n\in W$ converging to $x$ with respect to $\lVert\cdot\lVert$. Because of the first inequality in (1), $y_n$ is Cauchy with respect to both norms and so, using completeness, we see that a vector $y\in V$ exists such that $$\lvert y_n-y\rvert\to 0.$$ Actually, $x=y$ because $$\lvert x-y\rvert\le \lvert x-y_n\rvert+\lvert y_n-y\rvert\le C_1\lVert x-y_n\rVert + \lvert y_n-y\rvert, $$ and the right hand side of this inequality tends to $0$ as $n\to \infty$. Therefore, by letting $n\to \infty$ in the inequalities $$\lvert y_n\rvert\le C_1\lVert y_n\rVert, \qquad \lVert y_n\rVert\le C_2\lvert y_n\rvert$$ we see that the two norms are equivalent on the whole of $V$.