Task:
Let $(V,\langle\cdot,\cdot\rangle)$ be a Euclidean vector space, $U\subset V$ a $\mathbb{R}$-subspace. Show that there exists an $f\in\operatorname{End}_{\mathbb{R}}(V)$ with the following properties:
- $\|f(v)\|\leq\|v\|$ for all $v\in V$, and
- $\|f(v)\|=\|v\|$ if and only if $v\in U$.
Problem/Approach:
For 1. Let $f:V\rightarrow U$ with $v\mapsto f(v)=u$ be the orthogonal projection onto the subspace $U$. We know that this mapping is linear.
Since $U\subseteq V$, we have $V=U^{\perp}\oplus U$ and every $v\in V$ is uniquely determined by $v=u+w$ with $u\in U$, $w\in U^{\perp}$. Then $f(u)=u$ since $u\in U$ and $\|f(v)\|=\|u\|$ and $f(w)=0$ since $w$ is orthogonal to $U$.
It follows that $\|u\|\leqslant\|v\|$, since for $u^{\prime}\in U$, we have $\|v-u\|\leq\|v-u^{\prime}\|$. Therefore, $\|f(v)\|=\|u\|\leq\|v\|$, which proves the claim.
For 2. "$\Leftarrow$"
Let $v\in U$, then $f(v)=v$ and thus $\|f(v)\|=\|v\|$.
"$\Rightarrow$"
From 1. , we know that $\|f(v)\|\leq\|v\|$. Now, since $v\in U$, we have $\|v\|=\|u\|$ for all $u\in U$, so $\|f(v)\|=\|u\|=\|v\|$.
My problem:
I am using "It follows that $\|u\|\leqslant\|v\|$, since for $u^{\prime}\in U$, we have $\|v-u\|\leq\|v-u^ {\prime}\|$. Therefore, $\|f(v)\|=\|u\|\leq\|v\|$" not entirely satisfied. After a while it doesn't look right to me anymore. Do you have another idea how I can justify $\|u\|\leqslant\|v\|$? Maybe with pythagoras theorem? Thanks for any help.