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 Post subject: Linear isometry
PostPosted: Thu May 18, 2017 9:10 am 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 127
Location: Melbourne, Australia
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$. If:

  • $f(\mathbf{0})=\mathbf{0}$
  • $\left| {f\left( {\bf{u}} \right) - f\left( {\bf{v}} \right)} \right| = \left| {{\bf{u}} - {\bf{v}}} \right|$ for all $ {{\bf{u}},{\bf{v}}}$

then prove that $f$ is linear.

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 Post subject: Re: Linear isometry
PostPosted: Thu May 18, 2017 6:31 pm 
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 423
Let \(\displaystyle{\langle{\,\,,\,\,\rangle}}\) denote the usual inner product of \(\displaystyle{\mathbb{R}^2}\).

If \(\displaystyle{u\in\mathbb{R}^2}\), then

\(\displaystyle{|f(u)-f(0)|=|u-0|\iff |f(u)|=|u|}\), so,

\(\displaystyle{|f(u)|=|u|\,\,,\forall\,u\in\mathbb{R}^2\,,(I)}\).

Now, if \(\displaystyle{u\,,v\in\mathbb{R}^2}\) , then


\(\displaystyle{\begin{aligned}|f(u)-f(v)|^2=|u-v|^2&\iff |f(u)|^2-2\,\langle{f(u),f(v)\rangle}+|f(v)|^2=|u|^2-2\,\langle{u,v\rangle}+|v|^2\\&\iff \langle{f(u),f(v)\rangle}=\langle{u,v\rangle}\end{aligned}}\)

Finally, if \(\displaystyle{u\,,v\in\mathbb{R}^2}\) and \(\displaystyle{k\in\mathbb{R}}\), then

\(\displaystyle\begin{aligned}|f(u+k\,v)-(f(u)+k\,f(v))|^2&=|f(u+k\,v)|^2-2\,\langle{f(u+k\,v),f(u)+k\,f(v)\rangle}+|f(u)+k\,f(v)|^2\\&=|u+k\,v|^2-2\,\langle{f(u+k\,v),f(u)\rangle}-2\,k\,\langle{f(u+k\,v),f(v)\rangle}+|f(u)+k\,f(v)|^2\\&=|u+k\,v|^2-2\,\langle{u+k\,v,u\rangle}-2\,k\,\langle{u+k\,v,v\rangle}+|f(u)+k\,f(v)|^2\\&=|u+k\,v|^2+|f(u)+k\,f(v)|^2-2\,|u|^2-2\,k\,\langle{v,u\rangle}-2\,k\,\langle{u,v\rangle}-2\,k^2\,|v|^2\\&=|u|^2+2\,k\,\langle{u,v\rangle}+k^2\,|v|^2+|f(u)|^2+2\,k\,\langle{f(u),f(v)\rangle}\,+\\
&\qquad\qquad\qquad\qquad k^2\,|f(v)|^2-2\,|u|^2-4\,k\,\langle{u,v\rangle}-2\,k^2\,|v|^2\\&=-|u|^2-k^2\,|v|^2-2\,k\,\langle{u,v\rangle}+|u|^2+2\,k\,\langle{u,v\rangle}+k^2\,|v|^2\\&=0\end{aligned}\)

so \(\displaystyle{f(u+k\,v)=f(u)+k\,f(v)}\)

We deduce that \(\displaystyle{f}\) is \(\displaystyle{\mathbb{R}}\) - linear.


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 Post subject: Re: Linear isometry
PostPosted: Thu May 18, 2017 7:31 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 127
Location: Melbourne, Australia
Here is another solution I've seen ...

For convenience, identify $\mathbb{R}^2$ with $\mathbb{C}$ here. Then note that for any such function $f:\mathbb{C} \to \mathbb{C}$, also $z_1 \cdot f(z)$ a solution for any point $z_1$ on the unit circle. Also $\overline{f(z)}$ is a solution. Note that $\vert f(1)\vert=1$ and hence we can wlog assume that $f(1)=1$. So $f(i)$ is a point on the unit circle with distance $\sqrt{2}$ to $1$. Hence $f(i) =\pm i$, so w.l.o.g. assume that $f(i)=i$. But then for any $z \in \mathbb{C}$, both $z$ and $f(z)$ have the same distance to $0,1$ and $i$. So supposing $z \ne f(z)$, all $0,1,i$ lie on the perpendicular bisector between these points and in particular $0,1$ and $i$ are collinear which clearly is absurd. Hence $f(z)=z$ for all $z$ which proves the claim.

Credits
This solution is due to Tintarn...

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 Post subject: Re: Linear isometry
PostPosted: Sun May 21, 2017 8:48 am 

Joined: Thu Dec 10, 2015 1:58 pm
Posts: 59
Location: India
An alternative approach: It suffices to show that $f$ is surjective and preserves mid-points, i.e., if $x,y \in \mathbb{R}^2$ then, $$f\left(\frac{x+y}{2}\right) = \frac{f(x) + f(y)}{2}$$ then, by simple induction one can extend the above identity to $f(rx+(1-r)y) = rf(x) + (1-r)f(y)$ where, $r$ is a dyadic-rational in $(0,1)$ and then using continuity of $f$ and density of dyadic rationals one concludes $f(tx + (1-t)y) = tf(x) + (1-t)f(y)$, for $t \in (0,1)$, and hence $f$ is linear.

One argument for showing $f$ is surjective is noting that $f$ maps spheres of radius $r \ge 0$ onto itself ($rS^1 \overset{f}{\longrightarrow} rS^1$). Suppose not, then since, $f$ is injective, $S^1$ is homeomorphic to closed connected subset $f(S^1) \subset S^1\setminus \{p_0\} \equiv (0,1)$ for some $p_0 \in S^1$, i.e., homeomorphic to a closed sub-interval $[a,b] \subset (0,1)$ which is absurd. Since, removing mid-point from $[a,b]$ produces two connected components but $S^1\setminus \left\{f^{-1}((a+b)/2)\right\}$ is connected. Hence, $f$ is surjective.

Now to see that $f$ preserves mid-points, choose two arbitrary points $x,y \in \mathbb{R^2}$ and consider the spheres (homeomorphic to $S^1$) centered at $f(x)$ and $f(y)$ with radius $\frac{1}{2}|x-y|$. They intersect only at $\displaystyle \frac{f(x) + f(y)}{2}$, and the intersection of preimage under $f$ is only $\displaystyle \frac{x+y}{2}$. Hence, $\displaystyle f\left(\frac{x+y}{2}\right) = \frac{f(x) + f(y)}{2}$.

N.B.: The proof extends to 'surjective' isometries between real normed linear spaces $(X,\lVert\cdot \rVert_X) \overset{f}{\longrightarrow} (Y,\lVert\cdot \rVert_Y)$, whenever, $Y$ is strictly convex (i.e., the spheres in $Y$ do not contain line-segments). :)


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