Is norm non-decreasing in each variable? Let me try again. Suppose $\|\cdot\|$ is a norm in $\mathbb{R}^n$ and let $$f(x_1,...,x_n)=\|(x_1,...,x_n)\|$$
where $x_i\geq 0, \forall i$. I want to prove or disprove that $f$ is an nondecreasing function in each of its variables.
Thanks
Note: Suppose we vary $x_i$ and fix the other variables. Then I want the function $g(x_i)=f((x_1,...,x_i,...,x_n))$ to be nondecreasing.
 A: We know that $\|x\|_1=|x_1|+|x_2|$ is norm on $\mathbb R^2$. (It is called $\ell_1$-norm or taxicab norm.) It is easy to see that rotation does not change properties of norm.
So for any angle $\varphi$ the function
$$\|x\|=|x_1\cos\varphi-x_2\sin\varphi|+|x_1\sin\varphi+x_2\cos\varphi|$$
is a norm on $\mathbb R^2$.
For $\varphi=\frac\pi 6$ we have
$$\|x\|=\frac{|\sqrt3x_1-x_2|+|x_1+\sqrt3x_2|}2.$$
If we fix $x_2=1$, then this function is not monotone in $x_1$, as we can check by plotting
|sqrt(3)t-1|+|t+sqrt(3)| in WA.
A: I'm guessing you mean that if we hold all variables fixed, except one, then it is nondecreasing.  In that case, calculus can come to the rescue.  Take the derivative of $g$:
$$
\frac{dg}{dx_i}=\frac{d}{dx_i}\sqrt{x_1^2+x_2^2+\cdots x_i^2+\cdots x_n^2}
$$
Since all the $x_j$ that are not equal to $x_i$ are constant, this is
$$
\frac{x_i}{\sqrt{x_1^2+x_2^2+\cdots x_i^2+\cdots x_n^2}}
$$
If $x_i\geq 0$, then the derivative is nonnegative and $g$ is nondecreasing.
Of course, here I am assuming that you are using the standard norm on $\mathbb{R}^n$.
A: I found this post when I had the same question. It seems to me that the answer is no.
Let's look at the $\ell_2$-norm: $\|\boldsymbol{x}\|_2 = \sqrt{x_1^2 + \cdots + x_n^2}$ and recall the definition of a non-decreasing function:

A function, $f(x)$, is non-decreasing on an interval $I$ if $f(b) \geq f(a)$ for all $b > a$ where $a,b \in I$.

Without loss of generality, let $x_2,\ldots,x_n$ be fixed, $x_2+\cdots+x_n=C$ and write $g(x) = \sqrt{x^2 + C}$, where $g:\mathbb{R}\rightarrow\mathbb{R}$. We note that $g(-2)>g(-1)$, but $-2 \ngtr -1$. Hence, the $\ell_2$-norm is not a non-decreasing function in each of its variables.
A: The answer is yes if $f(x)=f(|x|)$
