I am trying to show that there exists a nontrivial normed functional on $\mathbb R^2$ invariant by isometries. That is:
If $A$ is any set, let $\mathcal B_{A}=\{f: \mathbb R^2 \rightarrow \mathbb R: f \text{ is bounded}\}$. Let $\|f\|=\sup\{|f(x)|: x \in A\}$. Notice that we are dealing with a vector space.
I am looking for a linear functional satisfying:
- $I$ is normed, that is, $|I(f)|\leq \|f\|$
- $I$ is nontrivial, that is, $I(1)=1$
- If $s \in \mathbb R^2$, let $f_s(x)=f(x+s)$ for every $x \in X$ and let $f^-(x)=f(\bar x)$ (complex conjugation). There are the translations and reflections. If $|t|=1$, let $f^t(x)=f(tx)$ (complex product). We want that $I(f)=I(f^-)=I(f_s)=I(f^t)$. These are the rotations.
I have already proved that there exists $I_0$ satisfies everything but the rotation thing.
Let $T=\{z: |z|=1\}$. I also know that there exists a nontrivial normed linear functional $I_1: \mathcal B_T \rightarrow \mathbb R$ satisfying $I_1(f^t)=I(f)$ for every $t \in \mathbb T$, $f \in \mathcal B_T$.
The book I am following tells me to define $I(f)=I_1(G_f)$, where $G_f: \mathcal B_t \rightarrow \mathbb R$ is given by $G_f(t)=I_0(f^t)$. This is a bounded function since $\|f^t\|=\|f\|$. It's easy to see that $I$ is a nontrivial normed linear functional. I have verified that $I$ is invariant by rotations and by translations, but I'm stuck on showing $I$ is invariant by the reflection.
My attempt to show it was:
Notice that $(f^-)^t(x)=f(\bar t \bar x)=(f^{\bar t})^-(x)$. Therefore, $G_{\bar f}(t)=G_f(\bar t)$ since $I_0$ is invariant by $\,^-$. I don't know how to proceed from here.
I'm tagging this question as measure theory since I need to prove this in order to define a nice measure on $\mathbb R^2$.