Is vector addition continuous under arbitrary norms Suppose $X$ is a normed space, and $X\times X$ is the vector space with vector addition and such defined component-wise, equipped with some norm. Then is $(x, y) \mapsto x + y$ continuous?
I know that if the norm is just $(x, y) \mapsto |x| + |y|$ then this is satisfied, but what if theres some other norm?
 A: No, not if you pick an unrelated norm on the product.
Assume we already have two not topologically equivalent norms $|\cdot|_1$ and $|\cdot|_2$ on $X$. Then addition may not be continuous if we take the norm induced by $|\cdot|_1$ on $X\times X$ and $|\cdot|_2$ on $X$.
A: If $X$ is a finite-dimensional vector space over $\Bbb{R}$ then any two norms on $X$ determine the same topology and vector addition will be continuous with respect to that that topology.
If $X$ is infinite-dimensional then different norms on $X$ can give different topologies on $X$ and you can't expect vector addition to be continuous if you choose an arbitrary norm on $X \times X$.
A: It is not true in general, suppose it was true, consider two norms $\| \|_i, i=1,2$ defined on $X$,
 and endow $X\times X$ with $\|\|_2\times \|\|_2$. Set $f:X\times X\rightarrow (X,\|\|_1): f(x,y)=x+y$, consider $i:(X,\|\|_2)\rightarrow X\times X$ defined by $i(x)=(x,0)$, $i$ is continue and $f\circ i=Id_X:(X,\|\|_2)\rightarrow (X,\|\|_1)$ is continue. A similar argument implies that $Id_X:(X,\|\|_1)\rightarrow (X,\|\|_2)$ is continue. This is equivalent to saying that $\|\|_1$ and $\|\|_2$ are equivalent, 
(see the answer here Equivalent norms )a fact which is not always true.
