Let $X$ be a topological vector space, with $d$ an invariant metric compatible with the metric. Let $f:X\to X$ be an involutive linear isomorphism.
How do you show that $f$ is an isometry?
I encountered this question while trying to prove item 1.41 in Rudin's Functional Analysis, where the claim is that if $X$ is an $F$-space and $N$ is a closed subspace of $X$ then $X/N$ is also an $F$ space. Then a metric is defined on $X/N$ by $$\rho(\pi(x),\pi(y)) := \inf\left(\{d(x-y,z)\,|\,z\in N\}\right)$$
I'm trying to prove this is indeed a metric and got stuck at $\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))$.
If $f$ is known to be an involution, all you need to prove that it's an isometry is the Lipschitz property $d(f(x),f(y))\le d(x,y)$ for all $x,y$. Since $f\circ f=\textrm{id}$, it then follows that $d(f(x),f(y)) = d(x,y)$ for all $x,y$, and that $f$ is onto.
I somehow don't see the connection of the above with the second question; the symmetry $\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))$ follows from the fact that $x-y$ and $y-x$ are at equal distance from $N$. Indeed, $$ \operatorname{dist}(v,N) = \inf_{z\in N}d(v,z)=\inf_{z\in N}d(v,-z) = \inf_{z\in N}d(-v,z) = \operatorname{dist}(-v,N) $$