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 $\mathbb{C}$.
Best Answer
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:
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 $T$.