如何正式书写赋范向量空间按$d(x,y)=\Vert x-y\Vert$成度量空间的证明？

Got it, let's break this down properly—first the formal proof that a normed space gives a metric space, then we'll circle back to why the reverse isn't true, tying in the translation invariance and homogeneity differences you noted.

First, let's recap the core definitions we need to work with:

Key Definitions #

• A normed vector space is a pair $(V, \Vert \cdot \Vert)$ where $V$ is a vector space over a field $\mathbb{F}$ (either $\mathbb{R}$ or $\mathbb{C}$), and the norm $\Vert \cdot \Vert: V \to [0, \infty)$ satisfies three axioms:

  1. Non-negativity & Definiteness: $\Vert x \Vert \geq 0$ for all $x \in V$, and $\Vert x \Vert = 0$ if and only if $x = 0$.
  2. Homogeneity: $\Vert \alpha x \Vert = |\alpha| \Vert x \Vert$ for all $x \in V$, $\alpha \in \mathbb{F}$.
  3. Triangle Inequality: $\Vert x + y \Vert \leq \Vert x \Vert + \Vert y \Vert$ for all $x, y \in V$.

• A metric space is a pair $(X, d)$ where $X$ is a set, and the metric $d: X \times X \to [0, \infty)$ satisfies three axioms:

  1. Non-negativity & Definiteness: $d(x, y) \geq 0$ for all $x, y \in X$, and $d(x, y) = 0$ if and only if $x = y$.
  2. Symmetry: $d(x, y) = d(y, x)$ for all $x, y \in X$.
  3. Triangle Inequality: $d(x, z) \leq d(x, y) + d(y, z)$ for all $x, y, z \in X$.

We’ll prove that defining $d(x, y) = \Vert x - y \Vert$ turns $(V, \Vert \cdot \Vert)$ into a valid metric space by verifying each metric axiom using the norm’s properties.

1. Non-negativity & Definiteness #

By the norm’s non-negativity axiom, $\Vert x - y \Vert \geq 0$, so $d(x, y) \geq 0$ holds for all $x, y \in V$.
For definiteness: $d(x, y) = 0$ if and only if $\Vert x - y \Vert = 0$. By the norm’s definiteness rule, this is true exactly when $x - y = 0$, i.e., $x = y$. This checks out perfectly.

2. Symmetry #

We need to show $d(x, y) = d(y, x)$. Substitute the metric definition:
$d(x, y) = \Vert x - y \Vert = \Vert (-1)(y - x) \Vert$.
Using the norm’s homogeneity axiom with $\alpha = -1$, this simplifies to $|-1| \cdot \Vert y - x \Vert = 1 \cdot \Vert y - x \Vert = d(y, x)$. Symmetry is confirmed.

3. Triangle Inequality #

We need to verify $d(x, z) \leq d(x, y) + d(y, z)$ for all $x, y, z \in V$.
Start with the left-hand side: $d(x, z) = \Vert x - z \Vert$. Notice we can rewrite $x - z$ as $(x - y) + (y - z)$.
Applying the norm’s triangle inequality to this sum:
$\Vert (x - y) + (y - z) \Vert \leq \Vert x - y \Vert + \Vert y - z \Vert$.
Substituting back the metric definition, this translates directly to $d(x, z) \leq d(x, y) + d(y, z)$. The triangle inequality holds.

As you pointed out, not every metric space can be turned into a normed vector space. The key properties that norm-induced metrics have (which arbitrary metrics lack) are:

• Translation Invariance: A norm-induced metric will always satisfy $d(x + z, y + z) = d(x, y)$ for all $x, y, z \in V$. This comes directly from the definition: $d(x+z, y+z) = \Vert (x+z)-(y+z) \Vert = \Vert x - y \Vert = d(x, y)$.
• Homogeneity: A norm-induced metric will satisfy $d(\alpha x, \alpha y) = |\alpha| d(x, y)$ for all $x, y \in V$, $\alpha \in \mathbb{F}$. Again, from the definition: $d(\alpha x, \alpha y) = \Vert \alpha x - \alpha y \Vert = \Vert \alpha(x - y) \Vert = |\alpha| \Vert x - y \Vert = |\alpha| d(x, y)$.

Example of a Non-Norm-Induced Metric #

Take $\mathbb{R}$ with the metric $d(x, y) = |e^x - e^y|$. Test translation invariance:
$d(x+1, y+1) = |e^{x+1} - e^{y+1}| = e \cdot |e^x - e^y| = e \cdot d(x, y)$. Since $e \neq 1$ (unless $d(x,y)=0$), this metric isn’t translation-invariant, so it can’t be induced by any norm.

Another example: the discrete metric on any vector space, where $d(x,y)=0$ if $x=y$, else $d(x,y)=1$. Check homogeneity: $d(2x, 0) = 1$ if $x≠0$, but $|2|d(x,0)=2\cdot1=2≠1$. It fails homogeneity, so it can’t come from a norm.

内容的提问来源于stack exchange，提问作者stefano ferrari