# Equivalence of Norms Defined on a Cartesian Product

While studying some notes on normed vector spaces, I have come upon the proof that addition $+:V \times V\to V$ of vectors in a normed vector space $V$ is a continuous operation.

The proof of this fact is quite easy except of (in my opinion) one step: the choice of the norm on $V \times V$. The proof has been done for a norm defined as $\|(v_1,v_2)\| = \|v_1\| + \|v_2\|$ and a comment has been made that certain other norms (e.g. $\|(v_1,v_2)\| = \max\{\|v_1\|,\|v_2\|\}$) can be used as well, since they generate the same topology (the product topology, I suppose).

However, I struggle with the question what is the precise set of all norms that can be used in this proof. I suppose that the theorem has to be interpreted in the way that $+$ is continuous with respect to the product topology. Thus, my question can be restated as follows: given a norm $\|\cdot\|$ on $V$ generating a topology $\tau$, which norms can be used on $V \times V$ to generate the product topology with respect to $\tau$?

I do not find this question to be straightforward, since in infinite dimensional spaces, norms need not be equivalent.

• At lest all norms of the type $\|\|(v,w)\|\|:= \|| (\|v\|,\|w\|)\||$ with all norms $\||\cdot\||$ in $\mathbb{R}^2$.
– Dirk
Jan 10 '13 at 11:26
• The (tautological) answer is: all norms on $V\times V$ that are equivalent to $\|v_1\|+\|v_2\|$. There is no description of all norms that are equivalent to a given one.
– user53153
Jan 11 '13 at 0:07
• I don't have the answer to your question, but to comment on what brought it up, you don't need to norm the topology on $V \times V$ to prove that addition is continuous since continuity is a purely topological notion. Jan 6 '19 at 18:07

Any Norm $p$ on $\mathbb R^2$ (you know that the are all equivalent) gives you a norm $(v,w)\mapsto p(\|v\|,\|w\|)$ on $V\times V$ which is equivalent to $\|v\| +\|w\|$ and induces the product topology. Your question could be whether all equivalent norms are of that form.
• I may be mistaken, but I think this is incorrect. Let $V=(\mathbb R, | \cdot |)$ and consider the norm $\rho$ on $\mathbb R^2$ defined by $\rho (a,b)=\sqrt{(a-b)^2+a^2}$. Then $\phi : (v,w) \mapsto p(|v|,|w|)$ is not a norm on $V\times V$ since $\phi((1,1)+(1,-1))=2\sqrt{2}> 2 =\phi(1,1)+\phi(1,-1)$. Jan 6 '19 at 17:48
• You are right. For the triangle inequality one needs that $p$ is increasing in each argument. Perhaps you want to add another answer to the question. Jan 6 '19 at 19:49