3
$\begingroup$

I posted this question originally here (nobody answered there): https://math.stackexchange.com/questions/2066318/is-the-following-function-a-norm

Let $\| \|$ be any norm in $\mathbb{R}^d$. Consider now $d$ normed vector spaces $(V_i, \|\|_i)$ and let $V$ be the cartesian product vector space. Is the function $f$, given according to the rule $f(v) = \|(\|v_1\|_1, \ldots, \|v_d\|_d)\|$, a norm in $V$?

According to the comments here: Pathological product space norm there is a condition for the function $f$ to define a norm and it is that the norm in $\mathbb{R}^d$ be increasing in each coordinate (case I know how to prove as stated in the first link). Hence, the condition is sufficient. Is it necessary?

$\endgroup$
  • $\begingroup$ Necessary and sufficient condition is that $\|(c_1,\dots,c_d)\|\leqslant \|(a_1,\dots,a_d)\|+\|(b_1,\dots,b_d)\|$ whenever $a_i,b_i\geqslant 0$ and $c_i\in [|a_i-b_i|,a_i+b_i]$. It looks to be strictly weaker than monotonicity in all coordinates, though I do not have an immediate example approving this guess. $\endgroup$ – Fedor Petrov Dec 22 '16 at 8:19
  • $\begingroup$ I think it is necessary. If that condition fails, you can just exploit @Ilmari Karonen's alternative formulation in the comments of that post to show that the ball wrt $f$ fails to be convex. $\endgroup$ – Delio Mugnolo Dec 22 '16 at 9:31
  • $\begingroup$ As a general rule, people don't appreciate when you post the same question across several StackExchange sites (this is called "crossposting"). Also, waiting barely 9 hours between crossposting is a lack of patience that is hardly excusable (notice that, in the meantime, you have received an answer on MSE, so you should have waited for a bit longer). Finally, notice that your question is not exactly research-level, so it might get closed. $\endgroup$ – Alex M. Dec 22 '16 at 14:13
  • 1
    $\begingroup$ @AlexM. I actually waited 2 days. Anyhow, I didn't want to repeat the question, so the question here is about the necessity of certain condition. How much time is it then prudent to wait (a week, a month)? $\endgroup$ – Will M. Dec 22 '16 at 18:37

Your Answer

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

Browse other questions tagged or ask your own question.