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From classical theorem of Mazur-Ulam it follows that if $(X,\|\cdot\|)$ Banach space and $T\colon X\to X$ is surjective isometry (i.e. $\|T(x)-T(y)\|=\|x-y\|$ for all $x,y\in X$) with $T(0)=0$ then $T$ is linear. My question is what if we instead of $\|T(x)-T(y)\|=\|x-y\|$ for all $x,y\in X$ require $\|T(x)+T(y)\|=\|x+y\|$ for all $x,y\in X$. In other words is it true:

if $(X,\|\cdot\|)$ Banach space and map $T\colon X\to X$ continuously bijective map such that $\|T(x)+T(y)\|=\|x+y\|$ for all $x,y\in X$ and $T(0)=0$ then $T$ is linear?

Thank's in advance.

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Nice question.

Suppose $||T(x) + T(y)|| = ||x + y|| \forall x, y \in \mathbb{R}^n$. Then $||T(-x) + T(x)|| = ||-x + x|| = ||0|| = 0$ $\forall x \in \mathbb{R}^n$, so $T(-x) = -T(x) \forall x \in \mathbb{R}^n$. Therefore \begin{align*} ||T(x) - T(y)|| = ||T(x) + T(-y)|| = ||x - y|| \forall x, y \in \mathbb{R}^n \end{align*} and we can apply the Mazur-Ulam Theorem.

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