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.

Thank you in advance.

  • 2
    $\begingroup$ At lest all norms of the type $\|\|(v,w)\|\|:= \|| (\|v\|,\|w\|)\||$ with all norms $\||\cdot\||$ in $\mathbb{R}^2$. $\endgroup$
    – Dirk
    Jan 10 '13 at 11:26
  • $\begingroup$ 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. $\endgroup$
    – user53153
    Jan 11 '13 at 0:07
  • $\begingroup$ 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. $\endgroup$
    – James Well
    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.

  • $\begingroup$ 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 $\phi((1,1)+(1,-1))=2\sqrt{2}> 2 =\phi(1,1)+\phi(1,-1)$. $\endgroup$
    – James Well
    Jan 6 '19 at 17:48
  • $\begingroup$ 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. $\endgroup$
    – Jochen
    Jan 6 '19 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.