TITLE: Is the following function a norm?
QUESTION [7 upvotes]: 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 -$p\in[1,\infty]$-).

REPLY [2 votes]: The answer is no in general:
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:
$$\begin{align*}
f(2,0,2,0)&=\lVert(2,2)\rVert=4,\\
f(0,1,0,1)&=\lVert(1,1)\rVert=2,\\
f(2,1,2,1)&=\lVert(3,2)\rVert=5+2=7\\
&\not\leq f(2,0,2,0)+f(0,1,0,1)=6.
\end{align*}$$