Machine-translated from Chinese.
定理 1. $X,Y$为赋范线性空间. 若$\phi:X\rightarrow Y$为等距满射, 则$\phi$为仿射变换.
证: 首先由于$\phi$是等距变换, $\phi$是单射, 且$\phi$连续. 同时,
$\phi-\phi(0)$也是等距满射. 故不妨设$\phi(0)=0,$
只需验证$\phi$为线性变换. 为此, 只要有$\phi(x+y)=\phi(x)+\phi(y)$即可.
这等价于总有
$$
\phi\left(\frac{x+y}{2}\right)=\frac{\phi\left(x\right)+\phi\left(y\right)}{2}.
$$
$\,\forall\,x,y\in X,$ 记$z=\frac{x+y}{2},$
取$X$上关于$z$点的中心对称变换$\psi(w):=2z-w.$ 那么
$$
|\phi(w)-z|=|w-z|,\quad |\phi(w)-w|=2|w-z|.
$$
取$G$为保持$x,y$两点不动的等距自变换群,
记$\lambda=\sup\{|gz-z|:g\in G\},$ 那么$\,\forall\,g\in G,$
$$
|gz-x|=|gz-gx|=|z-x|,\quad |gz-z|\le |gz-x|+|x-z|=2|z-x|,\quad \lambda<\infty.
$$
定义$g^\ast :=\psi g^{-1}\psi g,$ 那么$\,\forall\,g\in G,$
$$
2|gz-z|=|\psi gz-gz|=|g^{-1}\psi gz-z|=|\psi g^{-1}\psi gz-z|=|g^\ast z-z|\le \lambda.
$$
关于$g$取上确界, 就得到了$\lambda=0.$
从而所有保持两点不动的等距自变换也保持中点不动.
接下来,
我们取$\psi'$为$Y$中关于$z':=\frac{\phi(x)+\phi(y)}{2}$的中心对称变换.
此时$h:=\psi\phi^{-1}\psi'\phi$保持$x,y$两点不动, 从而由前面的结论,
$hz=z.$ 具体展开并化简移项得到$\psi'\phi z=\phi z.$
然而中心对称变换$\psi'$的唯一不动点为$z',$ 这就说明$\phi z=z',$
从而命题得证.
参考: https://zhuanlan.zhihu.com/p/24978670
文章最后更新于 2022-03-24 21:24:48
Theorem 1. $X,Y$ is a normed linear space. If $\phi:X\rightarrow Y$ is an isometric surjection, then $\phi$ is an affine transformation.
Certificate: First, since $\phi$ is an isometric transformation, $\phi$ is an injective, and $\phi$ is continuous. At the same time,
$\phi-\phi(0)$ is also an equidistant surjection. So we might as well assume $\phi(0)=0,$
Just verify that $\phi$ is a linear transformation. To do this, just have $\phi(x+y)=\phi(x)+\phi(y)$.
This is equivalent to always having
$$
\phi\left(\frac{x+y}{2}\right)=\frac{\phi\left(x\right)+\phi\left(y\right)}{2}.
$$
$\,\forall\,x,y\in X,$ Remember $z=\frac{x+y}{2},$
Take the central symmetry transformation $\psi(w):=2z-w.$ on $X$ about the point $z$, then
$$
|\phi(w)-z|=|w-z|,\quad |\phi(w)-w|=2|w-z|.
$$
Let $G$ be the isometric self-transformation group that keeps the two points $x,y$ stationary,
Remember $\lambda=\sup\{|gz-z|:g\in G\},$ then $\,\forall\,g\in G,$
$$
|gz-x|=|gz-gx|=|z-x|,\quad |gz-z|\le |gz-x|+|x-z|=2|z-x|,\quad \lambda<\infty.
$$
Define $g^\ast :=\psi g^{-1}\psi g,$ then $\,\forall\,g\in G,$
$$
2|gz-z|=|\psi gz-gz|=|g^{-1}\psi gz-z|=|\psi g^{-1}\psi gz-z|=|g^\ast z-z|\le \lambda.
$$
Taking the supremum of $g$, we get $\lambda=0.$
Thus All isometric self-transformations that keep two points stationary also keep the midpoint stationary.
Next,
We take $\psi'$ as the central symmetry transformation of $Y$ with respect to $z':=\frac{\phi(x)+\phi(y)}{2}$.
At this time $h:=\psi\phi^{-1}\psi'\phi$ keeps $x,y$ motionless at two points, so from the previous conclusion,
$hz=z.$ Specifically expand and simplify the transfer terms to get $\psi'\phi z=\phi z.$
However, the only fixed point of the central symmetry transformation $\psi'$ is $z',$, which means $\phi z=z',$
Thus the proposition is proved.
Reference: https://zhuanlan.zhihu.com/p/24978670
The article was last updated on 2022-03-24 21:24:48
