# Inequivalent norms

On an infinite dimensional, do there exist two norms which are not equivalent?

I actually know that on a infinite dimensional space, the number of inequivalent norms are $2^{dimX}$ However, I want to know if we can explicitly construct two inequivalent norms on any given space.

• This is actually not easy to do in general, but I think the idea you should keep in mind is pretty much always like it is in the sequence spaces. If you look at $c_{00}$ (the space of real sequences of which all but finitely many elements are zero), then you can give it any $\ell^p$ norm that you like, but they are non-equivalent because $c_{00}$ is dense in $\ell^p$ for all $p$.
– Ian
Commented Aug 29, 2016 at 19:30
• Commented Aug 29, 2016 at 20:01

If $\{ x_{\alpha} \}_{\alpha\in\Lambda}$ is a Hamel basis for your space $X$, then every $x\in X$ can be written in a unique way as $\sum_{\alpha\in\Lambda}\beta_{\alpha}x_{\alpha}$, where only finitely many $\beta_{\alpha}$ are non-zero. You can check that $$\|x\|_{\Gamma} = \sum_{\alpha\in\Lambda}\Gamma_{\alpha}|\beta_{\alpha}|$$ defines a norm $\|\cdot\|_{\Gamma}$ for any choice function $\Gamma : \Lambda\rightarrow(0,\infty)$. Such norms will not be equivalent in general. For example, if $\{ x_{\alpha_n}\}_{n=1}^{\infty}$ is a sequence of the Hamel basis, then you can arrange it so that $\|x_{\alpha_n}\|_{\Gamma}=1$ for all $n$ in the one norm $\|\cdot\|_{\Gamma}$, and arrange it so that $\|x_{\alpha_n}\|_{\Gamma'}=n$ for another norm $\|\cdot\|_{\Gamma'}$; in this case the norms $\|\cdot\|_{\Gamma}$, $\|\cdot\|_{\Gamma'}$ cannot be equivalent.