# Prove that if $||Tu-Tv||=||u-v||$ then $T$ is of the form $Tu=p+Au$ with $A$ linear.

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.

• This is known as the Mazur-Ulam Theorem. – Theo Bendit Oct 11 '18 at 12:49
• @JoséCarlosSantos I disagree. The dupe target uses the Euclidean norm, whereas this question uses the $1$-norm. – Theo Bendit Oct 11 '18 at 12:57
• @TheoBendit You are right. I will rectract my closing vote. – José Carlos Santos Oct 11 '18 at 12:57
• Can't we apply Mazur-Ulam Theorem as suggests by @TheoBendit? – Idonknow Jan 4 at 10:06

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.