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Here I think it is implict that we have to use the metric induced by the norm. Also, an isometry here means a surjective function that preserves distances between points.

Well, I tried to find a isometric embedding that is also surjective. But I couldn't, and I think that the function that solves my problem isn't to intuitive (so I won't find it too easily) and so I came here asking for it... (hope this isn't duplicated).

My bounded creativity suggested me only $\varphi(f)(x)=f(x)b +(1-f(x))a$, but I think $\varphi$ is not surjective (and it doesn't even preserves distances hehe :/)

Help me, please!

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  • $\begingroup$ can't you just precompose functions with any homeomorphism between the intervals? $\endgroup$ – Tim kinsella Sep 22 '18 at 3:57
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    $\begingroup$ Hint: Given $f\in\mathcal{C}([0,1])$ investigate the function $\varphi(f):=f\circ\alpha$ where $ \alpha(t):=a+t(b-a)$. $\endgroup$ – Jens Schwaiger Sep 22 '18 at 3:57
  • $\begingroup$ @JensSchwaiger but couldn't be the case that $\alpha (t) \not \in [0,1]$? $\endgroup$ – Robson Sep 22 '18 at 4:38
  • $\begingroup$ Correction of my comment: $f\in\mathcal{C}([0,1])$ should be replaced by $f\in\mathcal{C}([a,b])$. $\endgroup$ – Jens Schwaiger Sep 22 '18 at 4:51
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If $X$ and $Y$ are homeomorphic spaces, with homeomorphism $h: X \to Y$, define

$F: C(X) \to C(Y)$ by $F(f) = f \circ h^{-1} \in C(Y)$ for $f \in C(X)$.

$\|F(f)\|_\infty$ is just the supremum all its of absolute values over the domain and as the $h$ just shuffles the domains bijectively, the suprema will be the same in either case (these suprema are even assumed as the spaces are compact, but we don't need that even because all functions under consideration are bounded). So $F$ is an isometry (and linear too).

Now note that $[0,1]\simeq [a,b]$.

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You have the right idea, except you should apply it to the domain not the codomain of $f$, i.e., $$ \varphi(f)(x)=f(a+(b-a)x) $$ for each $f\in C([0,1])$.

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  • $\begingroup$ think better on my example (and it applie to yours too), $\varphi (f)$ has domain $[a,b]$, could be the case that $a<0$ and $b>0$ and so $0$ is in it... so $\varphi (f)(0)=f(a)$ but $f$ isn't defined at this point, right? $\endgroup$ – Robson Sep 22 '18 at 4:26

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