Normable topology determined by its restriction to a finite number of factors?

Question

Is it generally true that all norms $\|\cdot\|$ on a finite product of normed spaces $E_1\times\dots\times E_n$ with $\|(0,\dots,0,x,0,\dots,0)\|=\|x\|_i$ where $\|\cdot\|_i$ denotes the norm on $E_i$ are compatible? Or does this only hold for finite dimensional $E_i$?

Answer 1

Seems like the result is not true for normed spaces.

For any linear space $E$ with base vector set $B$ we have $\|\cdot\|_{B,\infty}:=\sum_{b\in B}\lambda_b b\mapsto\max_{b\in B}|\lambda_b|$ is a valid norm on $E$.

If we look at the space $c_{00}\times c_{00}$ with the standard base $B:=\{(e_i,0)\}_i\cup\{(0,e_i)\}_i$ as well as the non-standard base $C:=\{(e_i,0)\}_i\cup\{(ie_i,ie_i)\}_i$ then $\|\cdot\|_{B,\infty}$ coincides with $\|\cdot\|_{C,\infty}$ on $c_{00}\times\{0\}\cup\{0\}\times c_{00}$ but the two norms are not compatible.

I'll try to see if when we restrict ourselves to Banach spaces a positive result could be gained.

Answer 2

The result is true if we assume the $E_i$ to be Banach spaces and restrict ourselves to complete norms. Let $\|\cdot\|_{\rm alt}$ be a complete norm on $E:=E_0\times\dots\times E_{n-1}$ with $\|(0,\dots,0,x_i,0,\dots,0)\|_{\rm alt}=\|x_i\|$ for $x_i\in E_i$. Then $\|(x_0,\dots,x_{n-1})\|_{\rm alt}\le\|(x_0,\dots,x_{n-1})\|$ where the latter denotes the usual norm on product spaces defined as $\sum_{i<n}\|x_i\|$. So ${\rm id}:(E,\|\cdot\|)\rightarrow (E,\|\cdot\|_{\rm alt})$ is a bijective bounded linear map between Banach spaces, hence by the bounded inverse theorem the mapping is a homeomorphism so the norms are equivalent.