MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 2 deleted 40 characters in body

One can prove by using repeatedly the Krein-Milman theorem that

  • If $T : C[0,1] \rightarrow X$ is a morphism an operator of normed vector spacesnorm at most one, with $T$ isometric on the $2$-dimensional subspace spanned by $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ increases distances : $\forall f, ||Tf|| \geq ||f||$.is an isometry.
  • If $T$ is an endomorphism of $C[0,1]$ of norm at most one, which fixes the functions $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is the identity operator.
show/hide this revision's text 1 [made Community Wiki]

One can prove by using repeatedly the Krein-Milman theorem that

  • If $T : C[0,1] \rightarrow X$ is a morphism of normed vector spaces, with $T$ isometric on the $2$-dimensional subspace spanned by $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ increases distances : $\forall f, ||Tf|| \geq ||f||$.
  • If $T$ is an endomorphism of $C[0,1]$ of norm at most one, which fixes the functions $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is the identity operator.