Is a norm in $\mathbb{R}^d$ increasing in every coordinate?

Question

Consider any norm $\|\cdot\|$ in the $d$ dimensional Euclidean space and consider a fix point there $x$ such that all of its coordinates are positive. Let $e_k$ some vector in the standard basis of this space and consider the function $t \mapsto \|x +te_k\|$ for non negative $t$. Is this function increasing?

Answer 1

It's hopeless: Draw in $\mathbb{R}^2$ a parallelogram symmetric with respect to the origin, where the upper side is slanted going up to the right, like in the following figure:

Since this is a convex set and symmetric with respect to the origin, its interior is the unit ball of a norm of $\mathbb{R}^2$. Now pick a point on the upper side and you'll see that if you move it to the right, it will enter the ball, hence its norm will decrease.

That was for the intuitive and visual part.

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For the formal part:

Define the norm $$\forall u=(u_x,u_y)\in\mathbb{R}^2,\ N(u)=\frac14\lvert u_x+u_y\rvert+\frac12\lvert u_x-u_y\rvert.$$ It is straightforward to check that $N$ is a norm (and is in fact obtained from the $1$-norm and a linear bijection). The ball is shown on the figure above.

Let $u=(1,5/3)$. Clearly $N(u)=1$. Now for $t=1/3$, $u+e_x/3=(4/3,5/3)$ and $N(u+e_x/3)=11/12<1=N(u)$.

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Answer 2

A counterexample for that function being strictly increasing is $\|.\|_{max}:\mathbb{R}^d\rightarrow \mathbb{R}$ given by $\|x\|_{max}=\max_i|x_i|$ where $x_i$ are the components of $x\in\mathbb{R}^d$. Take $x=2e_1+\sum_{i=2}^de_i$. Then $t\mapsto \|x+te_2\|_{max}$ is constant for $t\in[0,1]$.

Answer 3

Consider $\|\cdot\|:\mathbb R^2\to\mathbb R, (x_1,x_2)\mapsto |x_1-x_2|+ |x_1|$. This is a norm:

$\|\cdot\|\geq 0$ is obvious. If $\|x\| = 0$ then $|x_1| = 0$ and $x_1 = x_2$, hence $x_2=0$ aswell.

$\|\lambda x\| = |\lambda|\|x\|$ is obvous.

$\begin{align*}\|x+y\| &= |x_1-x_2+y_1-y_2|+|x_1+y_1| \leq |x_1-x_2|+|y_1-y_2|+|x_1|+|y_1|\\ &= \|x_1\|+\|x_2\|.\end{align*}$

But now consider $x=(1,\frac 12)$. We have $\|x\| = \frac 32$ but $\|x+\frac 12e_2\| = \|(1,1)\| = 1 < \frac 32 = \|x\|$.

Answer 4

It is not too hard to check that $\|(x,y)\|^2 = 4 (x-y)^2 + (x+y)^2$ defines a norm on $\mathbb{R}^2$. (The level sets an ellipse rotated by 45°.)

Let $\phi(t) = \|(t,1)\|^2$ and note that $\phi'(0) = -6$.

If you draw a little picture it will be visually obvious what is happening.