# Convergence in norm independent of the choice of the norm.

When I read the proof of the Lie product formula in Reed Simon's book on functional analysis (which essentially reduces to showing $\left\Vert X_n - Y_n\right\Vert\rightarrow 0$ as $n\rightarrow 0$ for some specific sequences $\left\lbrace X_n \right\rbrace_{n=0}^{\infty}$, $\left\lbrace Y_n \right\rbrace_{n=0}^{\infty}$ in $\operatorname{Mat}_n(\mathbf{C})$), I thought that the proof depends crucially on the choice of the norm, for the proof requires that $\left\Vert \cdot\right\Vert$ be submultiplicative (which is satisfied by any cool matrix norm, e.g. the Hilbert–Schmidt norm or the operator norm, anyway), but then I realized that this is not case. For if $\left\Vert \cdot\right\Vert'$ is not submultiplicative, we have $\left\Vert X_n - Y_n\right\Vert'\leqslant C\left\Vert X_n - Y_n\right\Vert \rightarrow 0$ for some constant $C>0$ (since any two norms on a finite-dimensional vector space are equivalent), and hence convergence in norm is independent of the choice of the norm (which seems to be true for every finite-dimensional vector space). What happens if the underlying vector space is no longer finite-dimensional? I know that equivalence of norms no longer holds on infinite-dimensional vector spaces, but are there restrictions which guarantee equivalence of norms on infinite-dimensional vector spaces?

• it does, indeed. thanks. – user55315 Jan 16 '13 at 18:37

What is asked is whether we can find two norms $N_1$ and $N_2$, and a sequence $\{x_n\}$ such that $N_1(x_n-x)\to 0$ and $N_2(x_n-y)\to 0$ with $x\neq y$.