Given a norm, show a map is Affine So I have this problem:
Let $\|v\|_1=|x|+|y|$ be a norm in the plane, if $v=(x,y)$. 

Now I'm asked to show that, if $T: \mathbb R^2 \rightarrow \mathbb R^2 $ satisfies $\|T(v)-T(u)\|_1=\|v-u\|_1$, then $T$ is of the type $T(v)=p+A(v)$ with $A$ a linear map. 

I am not quite sure where to start, I have worked for some time with properties of norms, and I don't arrive anywhere. 
It may be similar in some way to proving that a rigid motion satisfying that norm given is affine... but I can't find the way to prove this. Can you help me with this problem?
 A: (1) Note that $X=(\mathbb{R}^2, d)$ is a metric space where
$d(x,y)=\|x-y\|$. Hence $T$ is an isometry. 
(2) Assume that
$$ \| v-u\| +\|u-x\|=\|v-x\|$$
Then $$ \| Tv-Tu\| +\|Tu-Tx\|=\|Tv-Tx\| $$
(3) If $e_i$ is a canonical basis, then $c(t)=te_1$ is unique
shortest path from $(0,0)$ to $e_1$.
If $T(0,0)=(0,0)$, then $Te_1\in \{\pm e_i\}$.
If $Te_1=e_1$ and $Te_2=-e_1$, then $(0,0)$ is a unique mid point between $e_1$ and $-e_1$.
Hence $T(t,t)=(0,0)$ where $0\leq t\leq 1$. It is a contradiction.
If $Te_2=e_2$, then $TA=A$ where $A=
 \{ (t,t)|0\leq t\leq \}$.
$\| T(0,0)-T(t,t)\| =\|(t,t)\|$ so that $T$ is identity on $A$.
Hence $T$ is identity on $\mathbb{Z}^2$
For $t>0$, $A_t=t\cdot \mathbb{Z}^2$. Since $T(te_i)=te_i$, by
similar argument, $T$ is identity on $A_t$ so that $T$ is
identity.
For general $T$, set $S(x)=T(x)-T(0,0)$ so that $S(0,0)=(0,0)$.
Hence $S$ is identity. That is, $T(x)=x+ T(0,0)$.
A: Recall the Mazur-Ulam theorem: a surjective isometry between normed linear spaces must be an affine map. There is a Wikipedia stub on the subject, but it's not very helpful. Check out this short and rather elegant proof instead.
The problem here is that surjectivity is not assumed, which is the hole I intend to fill.
I first claim that $\operatorname{Range} T$ is orthogonally convex, meaning that its intersections with horizontal or vertical lines are convex. To prove this, note that for $\lambda \in [0, 1]$,
$$S[(x_1, y); \lambda|x_1 - x_2|] \cap S[(x_2, y); (1 - \lambda)|x_1 - x_2|] = \lbrace (\lambda x_1 + (1 - \lambda)x_2, y) \rbrace.$$
(It's not as painful as it looks.) So, if $T(p) = (x_1, y)$ and $T(q) = (x_2, y)$, but
$$(\lambda x_1 + (1 - \lambda)x_2, y) \notin \operatorname{Range} T$$
then
$$S[p; \lambda|x_1 - x_2|] \cap S[q; (1 - \lambda)|x_1 - x_2|] \subseteq T^{-1} \lbrace (\lambda x_1 + (1 - \lambda)x_2, y) \rbrace = \emptyset.$$
But, the only way that two spheres can have non-empty intersection is if the distance between the centres is greater than the sum of their radii, in other words,
$$|x_1 - x_2| = \lambda|x_1 - x_2| + (1 - \lambda)|x_1 - x_2| < \|p - q\| = \|(x_1, y) - (x_2, y)\| = |x_1 - x_2|,$$
which is a contradiction. Hence $(\lambda x_1 + (1 - \lambda)x_2, y) \in \operatorname{Range} T$, proving horizontal convexity. Vertical convexity follows by a similar argument.
Obviously, $\operatorname{Range} T$ is not a singleton. This means, $\operatorname{Range} T$ must contain a non-trivial horizontal or vertical line segment. We will assume it is a vertical line segment, and the other case will be analogous.
Let $y_1, y_2 \in \Bbb{R}$ and $\varepsilon > 0$ such that $(y_1, y_2 \pm \varepsilon) \in \operatorname{Range} T$. Then, for any $r > \varepsilon$,
$$S[(y_1, y_2 + \varepsilon); r] \cap S[(y_1, y_2 - \varepsilon); r] = \lbrace (y_1 + r - \varepsilon, y_2), (y_1 - r + \varepsilon, y_2) \rbrace.$$
Let $p, q \in \Bbb{R}^2$ such that $Tp = (y_1, y_2 + \varepsilon)$ and $Tq = (y_1, y_2 - \varepsilon)$. Then,
$$S[p; r] \cap S[q; r] \subseteq T^{-1}\lbrace (y_1 + r - \varepsilon, y_2), (y_1 - r + \varepsilon, y_2) \rbrace,$$
which is finite. As above, we cannot have $S[p; r] \cap S[q; r] = \emptyset$, as this would lead to $r < \varepsilon$. If it were a singleton, then this would imply $r = \varepsilon$. So, $S[p; r] \cap S[q; r]$ is a doubleton, and so $(y_1 \pm (r - \varepsilon), y_2) \in \operatorname{Range} T$. This is true for all $r > \varepsilon$, so the whole horizontal line $y = y_2$ is contained in $\operatorname{Range} T$.
This now puts $(y_1, y_2)$ on a non-trivial horizontal line segment contained in $\operatorname{Range} T$, so an analogous argument shows that both the vertical line $x = y_1$ and horizontal line $y = y_2$ are contained in $\operatorname{Range} T$. From applying this argument to other points along these lines, we easily see that $\operatorname{Range} T = \Bbb{R}^2$.
