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Consider the following function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2} $:

$$f(x):=\alpha\left( |x-y|^{\alpha-2}(x-y)-|x-z|^{\alpha-2}(x-z) \right)$$

Where $\alpha >0,\alpha \neq 1$. I'm trying to show that $|f|>0$ as long as $y\neq z$, $x\neq y$, $x\neq z$.

We have $$|f (x)|^2=$$

$$|x-y|^{2\alpha-2} -2|x-y|^{\alpha-2}|x-z|^{\alpha-2}\langle x-y,x-z\rangle +|x-z|^{2\alpha-2}$$

To simplify, denote $a=x-y$ and $b=x-z$. Then $$|f (x)|^2=|a|^{2\alpha-2} -2|a|^{\alpha-2}|b|^{\alpha-2}\langle a,b\rangle +|b|^{2\alpha-2}$$

To minimize this we need maximize $\langle a,b\rangle$ if its positive (and minimize $\langle a,b\rangle$ if it is negative). From Cauchy's inequality $$|\langle a,b\rangle|\leq |a||b|$$ and equality happens if and only if $$b=c a,$$ where $c$ is a scalar. In this case $$\langle a,b\rangle=c|a|^2.$$

The problem then reduces to minimize for $c>0$ the function $$c\mapsto |f (x)|^2=|a|^{2\alpha-2} -2c^{\alpha-1}|a|^{2\alpha-2} +c^{2\alpha-2}|a|^{2\alpha-2} =|a|^{2\alpha-2}(1-2c^{\alpha-1} +c^{2\alpha-2} ) =|a|^{2\alpha-2}(1-c^{\alpha-1})^2$$ which is strictly positive if and only if $a\neq 0 \Longleftrightarrow x\neq y$ and $c\neq 1 \Longleftrightarrow a\neq b\Longleftrightarrow x-y\neq x-z\Longleftrightarrow y\neq z$.

Is my argument correct?

Medo
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  • Hello :) I don't get it. Is $f(x,y,z)=-f(x,z,y)$? – Jochen Jul 24 '21 at 08:29
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    $y$, $z$ are fixed points. $f$ is a function of $x$ only. – Medo Jul 24 '21 at 08:32
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    @Medo In that case, maybe you shouldn't call it a "function of two variables" in the title. You probably mean "parametric". – user3733558 Jul 24 '21 at 09:16
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    It is a function on $\mathbb{R}^2$. So, it can be seen as function of one variable $x\in \mathbb{R}^2$ or two variables $x_1,x_2\in\mathbb{R}$ where $x=(x_1,x_2)$. In any case it is not parametric in any way. – Medo Jul 24 '21 at 09:31
  • It would be good to clarify that $x$ is a vector in $\mathbb{R}^2$! – FShrike Jul 24 '21 at 10:10
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    Your reasoning looks fine to me. If you wanted to avoid the last part you could have used $|\langle a,b\rangle|\le |a||b|$ to conclude $$|f(x)|^2\ge |a|^{2\alpha-2}-2|a|^{\alpha-1}|b|^{\alpha-1}+ |b|^{2\alpha-2} =(|a|^{\alpha-1} + |b|^{\alpha-1})^2\ge 0.$$ – Adam T Jul 24 '21 at 10:43
  • @ Idiotic Shrike. Do not both the title and first line clarify that ! – Medo Jul 24 '21 at 14:32
  • @ Adam T. The question is show that $|f|>0$ under the given restrictions. – Medo Jul 24 '21 at 14:34

1 Answers1

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Your argument looks good. For fun and posterity, here's a geometric/physical argument:

Let $\alpha \neq 1$ be an arbitrary positive real. For a fixed vector $y$, the vector field $$ F_{y}(x) = \alpha |x - y|^{\alpha-2}(x - y) $$ is non-vanishing away from $y$, points radially away from $y$, and its magnitude is strictly monotone along each ray (decreasing if $0 < \alpha < 1$ and increasing if $1 < \alpha$).

By definition, $f(x) = F_{y}(x) - F_{z}(x)$. Let's call the terms on the right ($F_{y}$ and $-F_{z}$) the summands.

If $x$ does not lie on the line $\overline{yz}$, then $f(x) \neq 0$ because the summands are non-proportional, hence linearly independent.

To handle the remaining cases, partition the line $\overline{yz}$ into

  • The open ray away from $y$ and not containing $z$;
  • The open ray away from $z$ and not containing $y$;
  • The open segment from $y$ to $z$.

On each unbounded ray, the summands are oppositely proportional, but have different magnitude because $x$ is farther from one point ($y$ or $z$) than from the other, so $f(x) \neq 0$.

On the segment, the summands are positively proportional (away from $y$ is towards $z$), so their sum is non-zero.

In all cases, the field $f$ is non-vanishing away from $y$ and $z$, so its magnitude is positive: $|f| > 0$.

Andrew D. Hwang
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