# 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 $V$? (I can prove everything but the triangle inequality. As a matter of fact, any norm in $\mathbb{R}^d$ whose entries are non decreasing will make $f$ a norm; in particular, any $p$-norm make $f$ a norm -$p\in[1,\infty]$-).

Let $\lVert\cdot\rVert$ be the norm defined on $\mathbb{R}^2$ by: $$\lVert(x,y)\rVert=\lvert x+y\rvert+2\lvert x-y\rvert.$$
Then consider the norms $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_\infty$ on $\mathbb{R}^2$ and define $$\forall v\in\mathbb{R}^4,\ f(v_1,v_2,v_3,v_4)=\Bigl\lVert\bigl(\lVert (v_1,v_2)\rVert_1,\lVert(v_3,v_4)\rVert_\infty\bigr)\Bigr\rVert.$$
Then: \begin{align*} f(2,0,2,0)&=\lVert(2,2)\rVert=4,\\ f(0,1,0,1)&=\lVert(1,1)\rVert=2,\\ f(2,1,2,1)&=\lVert(3,2)\rVert=5+2=7\\ &\not\leq f(2,0,2,0)+f(0,1,0,1)=6. \end{align*}
• There is a condition for the function $f$ to be a norm, that is a sufficient condition. I asked here if it was also necessary: mathoverflow.net/questions/257807/… Commented Dec 22, 2016 at 18:40