# Mazur-Ulam theorem (bijective isometries are affine maps)

1. Jun 16, 2013

### Fredrik

Staff Emeritus
I've been studying the proof of the Mazur-Ulam theorem in the pdf linked to at the end of this Wikipedia article. I'm struggling with some details in that pdf.

Theorem: Let E and F be arbitrary normed spaces. If $f:E\to F$ is a bijective isometry, then f is an affine map.

Some of the things I don't get:

1. Their definition of "affine map". How does $f(tx+(1-t)y)=tf(x)+(1-t)f(y)$ for all $x,y\in X$ and all $t\in[0,1]$ imply that f-f(0) is linear? I don't see how to deal with (f-f(0))(ax) where $a\in\mathbb R$ is arbitrary.

2. The claim that $\psi$ is an isometry. $\psi:E\to E$ is defined by $\psi(x)=2z-x$ for all $x\in E$. This map sends an arbitrary point in E to the point that's "on the opposite side of z", i.e. the point y such that y=z+(z-x). They claim that $\psi$ is an isometry, but consider e.g. $E=\mathbb R$, z=2, x=1. We have $\psi(1)=2\cdot 2-1=3$, but $\|\psi(1)\|=3\neq 1=\|1\|$.

3. If $\psi$ isn't an isometry, then I don't see a reason to think that the map the author denotes by g* should be an isometry either. It's defined by $g^*=\psi\circ g^{-1}\circ\psi\circ g$, where g is a bijective isometry. The step $\|g^*(z)-z\|\leq\lambda$ relies on $g^*$ being an isometry.

4. All they're proving is that for all $a,b\in E$, we have $f\left(\frac{a+b}{2}\right)=\frac 1 2 f(a)+\frac 1 2 f(b)$. It's not obvious that this implies that f is affine.

Last edited: Jun 16, 2013
2. Jun 16, 2013

### micromass

Staff Emeritus
So the claim is that if $f$ is an affine map such that $f(0)=0$, then $f$ is linear. First, take $t\in [0,1]$ arbitrary, then
$$f(tx) = f(tx + (1-t)0) = tf(x) + (1-t)f(0) = tf(x)$$

Now it follows that
$$\frac{1}{2}f(x+y) = f(\frac{1}{2} x+ \frac{1}{2}y) = \frac{1}{2} f(x) + \frac{1}{2}f(y)$$
so $f(x+y) = f(x) + f(y)$ and $f$ is additive.

Now take $\lambda\geq 0$ arbitrary. Then we can write $\lambda = nt$ for some positive integer $n$ and some $t\in [0,1]$. Then
$$f(\lambda x) = f(ntx) = f(tx + ...+ tx) = f(tx) + ... + f(tx) = tf(x) + ... + tf(x) = ntf(x) = \lambda f(x)$$

Now take $\lambda<0$, then

$$0 = f(0) = f(\lambda x - \lambda x) = f(\lambda x) -\lambda f(x)$$
Thus $f(\lambda x) = \lambda f(x)$ and $f$ is linear.

You are verifying the property $\|\psi(x)\| = \|x\|$, but this is not the property that you want to verify. The property is $\|\psi(x) - \psi(y)\| =\|x-y\|$. This is the one that holds for $\psi$. The two properties are equivalent, but only for linear maps.

So, it suffices to show that if $f\left(\frac{a+b}{2}\right)=\frac 1 2 f(a)+\frac 1 2 f(b)$ and $f(0) = 0$, then $f$ is linear.

Note that
$$f\left(\frac{a+0}{2}\right)=\frac 1 2 f(a)+\frac 1 2 f(0)$$
implies that $f(a/2) = f(a)/2$. And as above, we can prove $f(a+b) = f(a) + f(b)$ now.
By induction, it follows easily that
$$f\left(\frac{1}{2^n}x\right)=\frac{1}{2^n} f(x)$$
Now if $t= c/2^n$ for some $c\in \{0,...,2^n\}$ then it follows easily (since the map is additive) that
$$f(tx) = tf(x)$$

Now, fix $x$ and define $g:[0,1]\rightarrow E:t\rightarrow f(tx) - tf(x)$. This function is continuous and vanishes on the dense set $\{c/2^n~\vert~c\in \{0,...,2^n\}\}$. So $g$ vanishes everywhere. Thus $f(tx) = tf(x)$ for all $t\in [0,1]$. As above, we can now prove that $f$ is linear.

3. Jun 16, 2013

### Fredrik

Staff Emeritus
Awesome reply. I don't know how you manage to explain everything I'm stuck on so fast, but I'm glad that you do.

4. Jun 16, 2013

### micromass

Staff Emeritus
You might be interested that if $E$ is a (real) Hilbert space, then things simplify considerably.

Let $\varphi$ be an surjective isometry, then . Thus $\|\varphi(x) - \varphi(y)\| = \|x-y\|$.

Let $\psi(x) = \varphi(x) - \varphi(0)$, then $\psi$ is a surjective isometry which satisfies $\psi(0) = 0$.

Note that in a Hilber space, we have $\|x - y\|^2 = \|x\|^2 + \|y\|^2 - 2<x,y>$. Thus

$$\begin{eqnarray*} 2<\psi(x),\psi(y)> & = & \|\psi(x) - \psi(0)\|^2 + \|\psi(y) - \psi(0)\|^2 -\|\psi(x) - \psi(y)\|^2\\ & = & \|x\|^2 -\|y\|^2 - \|x-y\|^2\\ & = & 2<x,y> \end{eqnarray*}$$

Now let $z$ be arbitrary. Then $z=\psi(y)$ for some $y$. Then

$$\begin{eqnarray*} <\psi(\alpha a +\beta b),z> & = & <\psi(\alpha a + \beta b),\psi(y)>\\ & = & <\alpha a + \beta b, y>\\ & = & \alpha <a,y> + \beta <b,y>\\ & = & \alpha <\psi(a),\psi(y)> + \beta <\psi(b),\psi(y)>\\ & = & <\alpha \psi(a) + \beta \psi(b), z> \end{eqnarray*}$$

This holds for all $z$, thus $\psi(\alpha a + \beta b) = \alpha \psi(a) + \beta \psi(b)$ and $\psi$ is linear.

5. Jun 17, 2013

### Fredrik

Staff Emeritus
Yes, I am interested in that. Thanks for posting it.

I have a followup about the first issue I brought up in post #1. After reading your reply, I see that if X and Y are normed spaces over ℝ, the following conditions on a function $f:X\to Y$ are equivalent.

(a) $f-f(0)$ is linear.
(b) There's a linear $L:X\to Y$ and a $y\in Y$ such that $f(x)=Lx+y$ for all $x\in X$.
(c) For all $t\in[0,1]$ and all $x,y\in X$, we have $f(tx+(1-t)y)=tf(x)+(1-t)f(y)$.

My question is, is this still true if X and Y are normed spaces over ℂ? Define F=f-f(0). Your method shows that (c) implies that F is ℝ-linear. But is it ℂ-linear? I think I see how to deal with arbitrary complex numbers if it's true that F(ix)=iF(x) for all x. But I don't see how to deal with F(ix).

As far as I can tell, this has no relevance to the validity of the proof of the Mazur-Ulam theorem that we've been discussing, since we prove (a) directly, not (c). I'm just curious if (c) works as a definition of "affine map" even in the complex case.

6. Jun 17, 2013

### micromass

Staff Emeritus
I don't think it holds. Take $\mathbb{C}\rightarrow \mathbb{C}:z\rightarrow \overline{z}$. Then this map is easily checked to be affine, but it's not linear. So (a) and (b) are not equivalent anymore.

Note that the same example shows that Mazur-Ulam fails in complex vector spaces.

7. Jun 17, 2013

### Fredrik

Staff Emeritus
Ah, I should have thought of that example. It satisfies (c), but not (a) or (b). (a) and (b) are so trivially equivalent that you probably forgot that I wrote them down as two separate statements.

(a) and (b) are equivalent, and imply (c). But (c) doesn't imply (a) or (b), if "linear" now means "ℂ-linear".

I'm not entirely clear on which condition is the appropriate definition of "affine map" in the context of complex normed spaces. I suspect that it's (a). In that case, Mazur-Ulam doesn't hold for complex normed spaces. But if it's (c), then it does hold.

I have typed up my version of the proof from the pdf (completed by your insights) for my notes. It seems to me that that what we actually prove is that if $f:X\to Y$ is a bijective isometry between normed spaces, then f-f(0) is ℝ-linear. This implies that f-f(0) satisfies (c). But your example shows that in general, it doesn't satisfy (a) (with "linear" meaning "ℂ-linear").

8. Jun 17, 2013

### micromass

Staff Emeritus
I think the right notion of of affine map is the following. Let k be a field. Then an affine map of a k-vector space E to a k-vector space F must satisfy

$$f(\lambda_1 x_1 + ... + \lambda_n x_n) = \lambda_1 f(x_1) + ... + \lambda_n f(x_n)$$

for any $\lambda_1 + ... + \lambda_n = 1$

This is again a new definition that didn't show up yet. However, it can be shown that it's always equivalent to (a).