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user25235
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One can prove by using repeatedly the Krein-Milman theorem that

  • If $T : C[0,1] \rightarrow X$ is a morphisman operator of normed vector spacesnorm at most one, with $T$ isometricisometric 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.

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.

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

  • If $T : C[0,1] \rightarrow X$ is an operator of norm 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 $T$ 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.
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user25235
  • 225
  • 2
  • 6

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.