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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?

simonzack
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3 Answers3

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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 $X$.

Hagen von Eitzen
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  • In particular, if the topology induced by $|\cdot |_1$ on $X$ does not contain the topology induced by $|\cdot|_2$, addition will not be continuous, since if you restrict to $X\times{0}$ addition is just the projection. – Eric Wofsey Sep 09 '16 at 22:43
  • you are saying $(x,y) \mapsto \phi(x+y)$ can be not continuous where $\phi$ is a linear mapping between different normed vector spaces, which is obvious, and that sometimes $\phi$ is implicit, when it is the canonical embedding of $(X,|.|_1)$ into $(X,|.|_2)$ (for example $\phi : l^1 \to l^2$) – reuns Sep 09 '16 at 22:46
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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 $X \times X$.

Rob Arthan
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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.

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Tsemo Aristide
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