Let $K$ and $L$ be two compact set and $T$ is an linear onto isometric from $C(K)$ to $C(L)$. My question is that $T(1)$ is the identity map in $C(L)$, where 1 is the identity map in $C(K)$ . give me some hint.
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2Are you asking if $T(1)$ must be the identity map? By identity do you mean $f(x)=x$? I think your setup and question is not clear. – Seth May 09 '14 at 19:09
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I am asking about$f(x)=1$, for all $ x$ in $X $ . – user149010 May 09 '14 at 19:15
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$C(K)$ is a ring of continuous functions? – rschwieb May 09 '14 at 19:23
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Actually $C(K)$ is collection of all continuous function. I am consider the linear isometric $T$ considering $C(K)$ as a vector space . – user149010 May 09 '14 at 19:29
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@user149010 It is worth to read – Norbert May 09 '14 at 21:54
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I guess the simplest example could be something like this : Let $K = L = [0,1]\sqcup [2,3]$ and let $T : C(K) \to C(L)$ be given by $$ T(f) = (f\vert_{[0,1]}, -f\vert_{[2,3]}) $$ However, I do think that with sufficiently stringent conditions on $T, K,$ and $L$, the Banach-Stone theorem should help you ensure that $T(1) = 1$ (For instance, if $T$ is an algebra isomorphism, this is true, but that is obviously asking for too much)
Prahlad Vaidyanathan
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OP said that $T$ is linear isometric and onto so by Banach-Stne theorem $T(f)=f\circ\phi$ for some homeomorphism $\phi$. In this case we obviously have $T(1)=1$. The Question is how to prove this without appealing to such strong theorem. – Norbert May 10 '14 at 09:20