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Consider the norm $||u||=|x|+|y|$ if $u=(x,y)$. Prove that if $T:\mathbb{R^2}\longrightarrow{\mathbb{R^2}}$ satisfies $||Tu-Tv||=||u-v||$ then $T$ is of the form $Tu=p+Au$ with $A$ linear.

I've been stuck with this problem for a while now, I know what everything is: a norm, a linear application... And their properties, however I just don't know what to do.

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    $\begingroup$ This is known as the Mazur-Ulam Theorem. $\endgroup$ – Theo Bendit Oct 11 '18 at 12:49
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    $\begingroup$ @JoséCarlosSantos I disagree. The dupe target uses the Euclidean norm, whereas this question uses the $1$-norm. $\endgroup$ – Theo Bendit Oct 11 '18 at 12:57
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    $\begingroup$ @TheoBendit You are right. I will rectract my closing vote. $\endgroup$ – José Carlos Santos Oct 11 '18 at 12:57
  • $\begingroup$ Can't we apply Mazur-Ulam Theorem as suggests by @TheoBendit? $\endgroup$ – Idonknow Jan 4 at 10:06
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An isometry like $T$ must preserve spheres and balls. In this case, the sphere is diamond-shaped. Try plotting it now if you don't know what it looks like.

If $T$ is an isometry, then so is $S \circ T$ where $S$ is another isometry. If we consider $S$ to be a translation that maps $T(0)$ to $0$, then $S \circ T$ is an isometry that fixes $0$. If we can show that $S \circ T$ must be linear, then this will confirm that $T$ must be of the appropriate form, hence we assume without loss of generality that $T(0) = 0$.

Consider the unit sphere (centred at $0$, with radius $1$). Note that the four corners, $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$, are the only four points on the sphere that are all exactly $2$ distance apart. Since $T$ fixes $0$, the unit sphere must map onto itself, and the four corners must all map to $4$ distinct points that are precisely $2$ apart. That is, the four corners must be stabilised.

Moreover, $T$ must preserve opposite corner pairs (e.g. $(1, 0)$ and $(-1, 0)$), as only opposite corners have the property that only a single point is distance $1$ from each of them (the point $(0, 0)$). When considering non-opposite corners, there are infinitely many such points!

There are known linear isometries: rotation by $\pi/2$, $\pi$, and $3\pi/2$ radians, as well as reflections in the $x$-axis, $y$-axis, and the lines $x = y$ and $x = -y$. Using similar techniques to the second paragraph, we may assume without loss of generality that $T(1, 0) = (1, 0)$ using rotations. Then, since opposite corners are preserved, $(0, 1)$ and $(0, -1)$ must be preserved. By using reflections, we may further assume $(0, 1)$ and $(0, -1)$ are fixed!

Now, we have an isometry $T$ that maps $(0, 0)$ to $(0, 0)$ and fixes all four corners. It falls to us to show that $T$ is the identity map.

First, we show it fixes the axes. If we have a point $(\lambda, 0)$ on the $x$-axis with $\lambda > 0$, then it is a corner on the sphere of radius $\lambda$ centred at $(0, 0)$. By similar arguments, it must map to another corner of the sphere. But, we have $$\|T(\lambda, 0) - (1, 0)\| = \|T(\lambda, 0) - T(1, 0)\| = \|(\lambda, 0) - (1, 0)\| = |\lambda - 1|.$$ If $T(\lambda, 0) = (0, \pm \lambda)$, or indeed if $T(\lambda, 0) = (-\lambda, 0)$, then $\|T(\lambda, 0) - (1, 0)\| = 1 + \lambda$. In either case, we must have $$1 + \lambda = |\lambda - 1| \implies (1 + \lambda)^2 = (\lambda - 1)^2 \implies \lambda = 0,$$ which is a contradiction. Thus, $T(\lambda, 0) = (\lambda, 0)$, so the positive $x$-axis is fixed. Similar arguments work for the negative $x$-axis, and the $y$-axis.

Now, consider a point $(x, y) \in \mathbb{R}^2$, with $x, y \neq 0$. Note that the point of smallest distance from $(x, y)$ to the $x$-axis is $(x, 0)$. The ball of radius $|x|$, centred at $(x, y)$, intersects the $x$-axis at only one point: $(x, 0)$, and the point is on the sphere.

This ball must map to a ball of radius $|x|$, centred at the point $T(x, y)$. It too will intersect the $x$ axis at precisely one point: $T(x, 0) = (x, 0)$. Note that $(x, 0)$ is a corner point on both spheres (centred at $(x, y)$ and at $T(x, y)$). It can't be either of the corners to the left or right of $T(x, y)$, as that would imply $T(x, y)$ lies on the $x$-axis at some point $(x_0, 0)$, and since $T$ is injective, this implies $y = 0$, which is a contradiction. Thus, it must be the top or bottom corner. Hence, we must have $T(x, y)$ must lie on the vertical line passing through $(x, 0)$.

Similar reasoning shows that $T(x, y)$ must lie on the horizontal line passing through $(0, y)$. The only such point is $(x, y)$, so $T(x, y) = (x, y)$, completing the proof.

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