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Let $ f: \mathbb{R}^n \to \mathbb{R}^m $ be a function, that preserves distances. Prove that there exist a linear transformation $T$, and a vector $\mathbf{j} \in \mathbb{R}^m $ such that $ f(\mathbf{x}) = T\mathbf{x}+\mathbf{j}$ for every $\mathbf{x} \in \mathbb{R}^n $.

First I suppose that $f(\mathbf{0})=\mathbf{0}$, since I can translate, without losing the property of preserving distances. So I need to prove that $f$ is linear.

If I simply prove that $ f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y})$ then it´s done, because obviously I can deduce that this implies $ f(r\mathbf{x}) = rf(\mathbf{x}) $ with $r$ rational. But since I know that $f$ preserves distances, then in particular $f$ is continuous, and it´s easy to prove that this implies that $ f(c\mathbf{x}) = cf(\mathbf{x})$ for every real number $c$.

But I don´t know How can I prove that $f$ respects the sum.

3 Answers 3

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Consider the parallelogram determined by $\mathbf{0}$, $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{x}+\mathbf{y}$. Because $f$ respects distances, the points $f(\mathbf{0}) = \mathbf{0}$, $f(\mathbf{x})$, $f(\mathbf{y})$, and $f(\mathbf{x}+\mathbf{y})$ must also give a parallelogram (since the distance from $\mathbf{0}$ to $f(\mathbf{x})$ equals the distance from $f(\mathbf{y})$ to $f(\mathbf{x}+\mathbf{y})$, and the distance from $\mathbf{0}$ to $f(\mathbf{y})$ is equal to the distance from $f(\mathbf{x})$ to $f(\mathbf{x}+\mathbf{y})$).

Therefore, $f(\mathbf{x}+\mathbf{y})$ must be the fourth vertex of the parallelogram determined by $\mathbf{0}$, $f(\mathbf{x})$ and $f(\mathbf{y})$, namely $f(\mathbf{x})+f(\mathbf{y})$.

Added. You can verify that $f(\mathbf{x}+\mathbf{y})$ lies in the same plane as $\mathbf{0}$, $f(\mathbf{x})$, and $f(\mathbf{y})$ by using the fact below that $f$ respects midpoints: the diagonals of the original parallelogram bisect each other, hence the line joining $f(\mathbf{x})$ and $f(\mathbf{y})$ and the line joining $\mathbf{0}$ and $f(\mathbf{x}+\mathbf{y})$ bisect each other.

You can also prove homogeneity by showing that $f(\frac{1}{2}(\mathbf{x}+\mathbf{y})) = \frac{1}{2}f(\mathbf{x}) + \frac{1}{2}f(\mathbf{y})$, by considering the spheres of radius $\frac{1}{2}\lVert \mathbf{x}-\mathbf{y}\rVert$ around $\mathbf{x}$, $\mathbf{y}$, $f(\mathbf{x})$ and $f(\mathbf{y})$; that is, $f$ respects midpoints of line segments. From this it follows that $f(t\mathbf{x}+(1-t)\mathbf{y}) = tf(\mathbf{x}) + (1-t)f(\mathbf{y})$ for all dyadic rationals $t\in (0,1)$, hence by continuity for all $t\in [0,1]$. From there, it follows that $f(n\mathbf{x}) = nf(\mathbf{x})$ for all integers $n$, and then homogeneity follows.

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    @ArturoMagidin your proposition follows from the fact that in any inner product space the "triangular equality" holds if and only the three points are in the same convex linear combination. You just replace by |x|^2 and use linear properties of dot product. Can you help me with http://math.stackexchange.com/questions/$1$48414/extending-f-x-subset-mathbbrm-to-mathbbrn-an-isometric-immersion similar problem?2012-05-22
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One trick is to use the polarization identity. This allows us to prove linearity directly, without approximation arguments.

Preserving distances means that $ \|f(x)-f(y)\|=\|x-y\| $ for all $x,y$. The assumption $f(0)=0$ allows us to also get $\|f(x)\|=\|x\|$ for all $x$.

Now, using the polarization identity, $ \langle f(x),f(y)\rangle=\frac{\|f(x)\|^2+\|f(y)\|^2-\|f(x)-f(y)\|^2}4 =\frac{\|x\|^2+\|y\|^2-\|x-y\|^2}4=\langle x,y\rangle. $ Now we get, for any $x,y,z\in\mathbb{R}^n$, $\lambda\in\mathbb{R}$, $ \langle f(\lambda x+y),f(z)\rangle=\langle\lambda x+y,z\rangle=\lambda \langle x,z\rangle+\langle y,z\rangle=\langle\lambda f(x)+f(y),f(z)\rangle, $ so $ \langle f(\lambda x+y)-\lambda f(x)-f(y),f(z)\rangle=0 $ for any $z$, in particular $z=x$, $z=y$, $z=x+y$. But then $ \|f(\lambda x+y)-\lambda f(x)-f(y)\|^2=\langle f(\lambda x+y)-\lambda f(x)-f(y),f(\lambda x+y)-\lambda f(x)-f(y)\rangle=0 $ after distributing on the second term of the inner product.

So $f(\lambda x+y)=\lambda f(x)+f(y)$ for all $x,y\in\mathbb{R}^n$, $\lambda\in\mathbb{R}$.

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    Yes, as long as the distance is the one coming from the norm induced by the inner product.2013-12-13
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I think the second property is easier to prove, and you can probably get the first one from here...

If $f(x)=a$ and $f(2x)=b$, then $||b||=2||a|| (*)$ and $||b-a||=||a|| \,.$

From here you can deduce that $b=2a$. Indeed

$||a||^2= ||b-a||^2=||a||^2-2 a \cdot b + \|b\|^2 \,.$

Thus $\|b^2\|=2a \cdot b (**)$

Then by $(*)$ and $(**)$

$\|b-2a\|^2=\|b\|^2-4a \cdot b+4\|a\|^2=0 $

Now, you can prove by induction exactly the same way that $f(nx)=nf(x)$.

Also, $f(-x)=-f(x)$ follows the same way. $\|f(-x)\|=\|f(x)\|$ and $\|f(x)-f(-x)\|=2\|f(x)\|$.

From here you can deduce exactly as you mentioned that $f(cx)=cf(x)$.