Proof. For every nonzero vector $x\in X$ , since f is a min-phase-isometry, we see that

(2.1) $$ \begin{align} (\|f(2x)+f(x)\| \wedge\|f(2x)-f(x)\|) +(\|f(x)+f(0)\|)\wedge\|f(x)-f(0)\|)=2\|x\|. \end{align} $$

However,

(2.2) $$ \begin{align} \|f(2x)+f(x)\|+\|f(x)+f(0)\| & \geq \|f(2x)+f(x)+(-f(x)-f(0))\|\nonumber \\ &\geq\|f(2x)-f(0)\|\geq2\|x\| \end{align} $$

and

$$ \begin{align*} \|f(2x)+f(x)\|+\|f(x)-f(0)\|&\geq \|f(2x)+f(x)+(-f(x)+f(0))\|\\ &\geq\|f(2x)+f(0)\|\geq2\|x\|. \end{align*} $$

Similarly, we can get

$$ \begin{align*} \|f(2x)-f(x)\|+\|f(x)+f(0)\|&\geq2\|x\| \end{align*} $$

and

$$ \begin{align*} \|f(2x)-f(x)\|+\|f(x)-f(0)\|\geq2\|x\|. \end{align*} $$

These inequalities combined with (2.1) imply that at least one equality is achieved. Without loss of generality, we may assume that (2.2) is an equality. Since Y is strictly convex, we conclude that

$$ \begin{align*} 2f(x)= -f(2x)-f(0) \end{align*} $$

and hence $\|f(x)\|\geq \|x\|$ .

Proof. Set $z=\tfrac 12(x+y)$ . Since $\|x-y\|\leq \|x+y\|$ , it follows that

(2.3) $$ \begin{align} \|x+z\|\geq2\|z\|- \|z-x\|=\|x+y\|-\|x-y\|/2 \geq \|x-y\|/2=\|x-z\|. \end{align} $$

Similarly, we get

(2.4) $$ \begin{align} \|z+y\| \geq \|x-y\|/2=\|z-y\|. \end{align} $$

Since f is a min-phase-isometry, we see that

$$ \begin{align*} \min \{\|f(x)+f(y)\|, \|f(x)-f(y)\|\}=\|x-y\|. \end{align*} $$

Obviously, we have

(2.5) $$ \begin{align} \|x-z\|+\|z-y\|=\|x-y\| \end{align} $$

and it is easily checked that

$$ \begin{align*} &\|f(x)-f(z)\|+\|f(z)-f(y)\|\geq \|f(x)-f(y)\| \\ &\|f(x)-f(z)\|+\|f(z)+f(y)\|\geq \|f(x)+f(y)\|\\ &\|f(x)+f(z)\|+\|f(z)-f(y)\|\geq \|f(x)+f(y)\|\\ &\|f(x)+f(z)\|+\|f(z)+f(y)\|\geq \|f(x)-f(y)\|. \end{align*} $$

These combined with (2.3), (2.4) and (2.5) guarantee that at least one of the previous four inequalities is an equality. We may assume that the first one is an equality, that is, $\|f(x)-f(z)\|+\|f(z)-f(y)\|=\|f(x)-f(y)\|$ . Since Y is strictly convex,

$$ \begin{align*} f(z)=\tfrac{1}{2}(f(x)+f(y)) \end{align*} $$

as desired. The proof is complete.

Proof. The conclusion follows immediately from

$$ \begin{align*} \|f(x)\|=\min\{\|f(x)+f(0)\|,\|f(x)-f(0)\|\} =\min\{\|x+0\|,\|x-0\|\}=\|x\|.\\[-36pt] \end{align*} $$

Proof. Let $y=\lambda x$ and $\lambda \geq 0$ . Replacing $\lambda $ by $1/\lambda $ , we may assume that $\lambda \in [0,1]$ . Then

$$ \begin{align*} (1-\lambda)\|x\|&=\min\{\|f(x)+f(y)\|,\|f(x)-f(y)\|\}\\ &\geq\|f(x)\|-\|f(y)\| =\|x\|-\|y\| =\|x\|-\lambda\|x\|=(1-\lambda)\|x\|, \end{align*} $$

so $\|f(x)+f(\lambda x)\|\wedge \|f(x)-f(\lambda x)\|=\|f(x)\|-\|f(\lambda x)\|$ . Strict convexity of Y yields $f(\lambda x)\in \{-\lambda f(x),\lambda f(x)\}$ . Finally, from $\min \{\|f(-z)+f(z)\|, \|f(-z)-f(z)\|\}=0$ for ${z\in X}$ , we deduce that $f(-z)\in \{-f(z),f(z)\}$ . This finishes the proof.

Proof. Assume, in contrast, there are $x,y\in X$ which are linearly independent such that $f(y)=\lambda f(x)$ with $\lambda \neq 0$ . Lemma 2.6 guarantees that

$$ \begin{align*} \|\lambda x+y\| \wedge\|\lambda x-y\|=\|f(\lambda x)+f(y)\| \wedge \|f(\lambda x)-f(y)\|=0. \end{align*} $$

This is a contradiction since $x,y$ are linearly independent. The proof is complete.

Proof. By Proposition 2.4 and Lemma 2.6, we only need to consider those $x,y\in X$ such that $\|x-y\|>\|x+y\|$ . We may assume that $\|x\|\leq \|y\|$ and $x,y$ are linearly independent. Then it follows from Lemma 2.7 that $f(x)$ and $f(y)$ are linearly independent.

We apply Proposition 2.4 and Lemma 2.6 to $-x$ and y to obtain $\alpha _1,\beta _1\in \{-1,1\}$ such that

(2.6) $$ \begin{align} f(y-x)=\alpha_1 f(x)+\beta_1f(y). \end{align} $$

Set $u=y-x$ and $v=y+x$ . Then

$$ \begin{align*} \|u-v\|\leq\|u+v\|. \end{align*} $$

It follows from Proposition 2.4 and Lemma 2.6 that there are $\alpha _2,\beta _2\in \{-1,1\}$ such that

$$ \begin{align*} f(u+v)=\alpha_2f(u)+\beta_2f(v), \end{align*} $$

that is, $f(2y)=\alpha _2f(y-x)+\beta _2f(x+y)$ . This combined with (2.6) yields

$$ \begin{align*} f(x+y)=\alpha f(x)+\theta f(y) \end{align*} $$

for some $\alpha \in \{-1,1\}$ and $\theta \in \{\pm 1,\pm 3\}$ . To complete the proof, it remains to exclude the possibility that $|\theta |=3$ . If $|\theta |=3$ , Lemma 2.5 and the strict convexity of Y imply that

$$ \begin{align*} \|x+y\|=\|f(x+y)\|=\|\alpha f(x)+\theta f(y)\|>3\|f(y)\|-\|f(x)\|\geq2\|y\|. \end{align*} $$

This contradiction gives the desired conclusion.

Proof. We may assume that $x,y$ are linearly independent and $a,b$ are nonzero. Proposition 2.8 yields the existence of $\alpha ,\beta \in \{-1,1\}$ satisfying $f(x+y)=\alpha f(x)+ \beta f(y)$ . We conclude from Proposition 2.8 and Lemma 2.6 that there are $\alpha _1, \beta _1 \in \{-1,1\}$ such that

(2.7) $$ \begin{align} f(ax+by)=\alpha_1 a f(x)+\beta_1 b f(y). \end{align} $$

However, Proposition 2.8 and Lemma 2.6 again imply

$$ \begin{align*} f(ax+by)= f(x+y+(a-1)x+(b-1)y)=\alpha_2f(x+y)+\beta_2(f(a-1)x+(b-1)y) \end{align*} $$

for some $\alpha _2, \beta _2\in \{-1,1\}$ . Continuing in this way produces $\alpha _3, \beta _3\in \{-1,1\}$ such that

(2.8) $$ \begin{align} f(ax+by)&=\alpha_2f(x+y)+\beta_2(f(a-1)x+(b-1)y) \nonumber \\ &=\alpha_2(\alpha f(x)+ \beta f(y))+\beta_2 (\alpha_3 (a-1)f(x)+\beta_3(b-1)f(y)) \nonumber \\ &=(\alpha_2 \alpha + \beta_2 \alpha_3(a-1))f(x)+(\alpha_2\beta+ \beta_2\beta_3(b-1))f(y). \end{align} $$

Since $f(x)$ and $f(y)$ are linearly independent from Lemma 2.7, we obtain from (2.7) and (2.8) that

$$ \begin{align*} \alpha_1=\alpha_2 \alpha \quad \mbox{and} \quad \beta_1 =\alpha_2\beta. \end{align*} $$

The proof is complete.

Proof of Theorem 1.1.

If $\dim X=1$ , choose a norm-one ${x_0\in X}$ . For every ${\lambda \in \mathbb {R}}$ , define $U:X\rightarrow Y$ by $U(\lambda x_0)=\lambda f(x_0)$ . Then U is a linear isometry which is phase-equivalent to f by Lemma 2.6.

For $\dim X=2$ , fix two linearly independent vectors $x, y\in X$ . By Proposition 2.8, there are $\alpha ,\beta \in \{-1,1\}$ such that

$$ \begin{align*} f(x+y)=\alpha f(x)+\beta f(y). \end{align*} $$

Then we can define a map $U:X\rightarrow Y$ by

$$ \begin{align*} U(ax+by)=a\alpha f(x)+b\beta f(y) \end{align*} $$

for all $a,b\in \mathbb {R}$ . It is clear that U is linear and, by Lemma 2.9, U is phase-equivalent to f. Thus, $\|U(ax+by)\|=\|f(ax+by)\|=\|ax+by\|$ .

Let $\dim X\geq 3$ and let $x,y,z\in X$ be linearly independent. We will show that $f(x),f(y)$ and $f(z)$ are also linearly independent. Assume that this is not true. Then there are nonzero $a,b\in \mathbb {R}$ such that $f(z)=af(x)+bf(y)$ . We deduce from this and Lemma 2.9 that

$$ \begin{align*}f(z)=af(x)+bf(y)\in \{\pm f(ax+by),\pm f(ax-by)\}.\end{align*} $$

Since f is a min-phase-isometry, we see that

$$ \begin{align*} z\in \{\pm (ax+by),\pm(ax-by)\}. \end{align*} $$

This is a contradiction. It follows that the dimension of $f(X)$ is at least three. This combined with Proposition 2.8 guarantees that the conditions of the fundamental theorem of projective geometry are satisfied. So we conclude that f is induced by an injective linear mapping $A:X\rightarrow Y$ , that is, for each $x\in X$ , there is a real number $\lambda _x$ such that

(2.9) $$ \begin{align} f(x)=\lambda_x A(x). \end{align} $$

Hence, for any $y\in X$ , we have

$$ \begin{align*} f(x+y)=\lambda_{x+y}A(x+y)=\lambda_{x+y}A(x)+\lambda_{x+y}A(y). \end{align*} $$

It follows from Proposition 2.8 that

$$ \begin{align*} f(x+y)=\alpha f(x)+\beta f(y)= \alpha\lambda_x A(x)+\beta\lambda_y A(y) \end{align*} $$

with $\alpha ,\beta \in \{-1,1\}$ . Thus,

$$ \begin{align*} |\lambda_y|=|\lambda_{x+y}|=|\lambda_x|. \end{align*} $$

Therefore, $|\lambda _x|$ is a constant independent of x and we denote it by $\lambda $ . Let $U=\lambda A$ . Then by (2.9), U is a linear isometry and is phase-equivalent to f. This completes the proof when $\dim X\geq 3$ .