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.

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Again? Where is your previous try? – Chris Eagle Oct 23 '12 at 14:44
And what does "nondecreasing function in each of its variables" mean? – Chris Eagle Oct 23 '12 at 14:45
@ChrisEagle. im gonna edit and explain better – Tomás Oct 23 '12 at 14:49
@ThomasAndrews, your example is wrong. – Tomás Oct 23 '12 at 14:54
@Tomas Are you saying that $f(-1,0,0,...0)$ is not positive, or that $f(1,0,..,0)$ is not positive, or that $f(0,0,...,0)$ is not zero? – Thomas Andrews Oct 23 '12 at 14:57

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.

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Interesting @MartinSleziak. So is this question wrong: <math.stackexchange.com/questions/218937/…; ? – Tomás Oct 23 '12 at 15:22
Well, either that or there is a mistake in my answer. – Martin Sleziak Oct 23 '12 at 15:28
@MartinSleizak, i ploted the function and it is not nondecreasing. Maybe it is not a norm, but i think it is. – Tomás Oct 23 '12 at 15:32
I think you answer is good, you can give this counter example there. – Tomás Oct 23 '12 at 15:33

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 $\mathbb{R}^n$.

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The question did not say "Euclidean norm". – Karolis Juodelė Oct 23 '12 at 14:59
@KarolisJuodelė Notice the last line of my answer. As pointed out by Martin Sleziak, it doesn't work in general. I was thinking the OP meant the standard norm and gave a proof for that case. – Joe Johnson 126 Oct 23 '12 at 15:01
I didn't notice that line. Although I am still afraid that this answer could have been misleading to someone (were it the only answer). Though, in case you were wondering, that downvote isn't mine... – Karolis Juodelė Oct 23 '12 at 15:21

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.

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