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Let $M_1$ and $M_2$ be compact metric spaces.

Denote $C(M_i)$ as the space of continuous functions from $M_i$ to $\Bbb C$ with supremum-norm, $i=1,2$.

A linear function $T:C(M_1)\to C(M_2)$ is said to be positive if $T(f) \ge 0$ whenever $f \ge 0$ (ie. the ranges of $f$ and $T(f)$ are both subsets of $[0, \infty )$ ).

Question: Show that $T$ is bounded and evaluate $||T||$ provided $T$ is positive.

I have proof that it is sufficient to consider the case where $f\ge 0$ by the following argument:

  1. $||\;|f|\;||_{C(M_1)}=||f||_{C(M_1)}$ (here $|f|$ is the complex modulus of $f$). Proof: Obvious by definition of supremum-norm.

  2. $T(|f|)=|T(f)|$. Proof: $T(|f|)=(T(f)+T(\bar f))/2$ as $T$ is linear, $T(\bar f) = \overline {T(f)}$ by writing $f=h+ik$. (Note $|f|\in C(M_1)$ whenever $f\in C(M_1)$ so "$T(|f|)=|T(f)|$" makes sence.)

  3. $||\;T(|f|)\;||_{C(M_2)}=||\;T(f)\;||_{C(M_2)}$ . Proof: Obvious followed by 2 above an by definition of supremum-norm.

So for any $f\in C(M_1)$, we can take $|f| \ge 0$, and by $T $ is positive, $T(|f|) \ge 0$, so we may only need to prove:

"$\exists M \gt 0 \;s.t.\;||\;T(f)\;|| \le M||\;f\;||$, for the case $f \ge 0$ ."(I ignore the subscripts here.)

But then I cannot go further more. Would anyone like to suggest me a way construct that $M$? Thank you!

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up vote 3 down vote accepted

Take $f\in C(M_1)$ such that $f\geq 0$. Consider $\varphi=\Vert f\Vert^{-1} f$, then $\varphi\geq 0$ and $\Vert\varphi\Vert=1$. Consequently $-1\leq\varphi\leq 1$, so positivity of $T$ gives $-T(1)\leq T(\varphi)\leq T(1)$. Hence $\Vert T(\varphi)\Vert\leq \Vert T(1)\Vert$ and $$ \Vert T(f)\Vert=\Vert T(\Vert f\Vert\varphi)\Vert=\Vert f\Vert\Vert T(\varphi)\Vert\leq \Vert T(1)\Vert\Vert f\Vert\tag{1} $$ As you showed later since this inequality holds for $f\geq 0$ it holds for all $f\in C(M_1)$. So $\Vert T\Vert\leq\Vert T(1)\Vert$. After substitution of $f=1$ into $(1)$ we get that $\Vert T\Vert=\Vert T(1)\Vert$.

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Thank you very much! – Lawrence Mar 3 '13 at 9:55
But I want to ask about how the prositivity of $T$ gives $-T(1) \le T( φ ) \le T(1)$ ? – Lawrence Mar 3 '13 at 9:56
OK! I get it! Is that $ 1-φ \ge 0$ so $T(1-φ) \ge 0$, then $T(φ) \le T(1)$? – Lawrence Mar 3 '13 at 10:01
Yes and for the second inequality use $$1+\varphi\geq 0\implies T(1+\varphi)\geq 0\implies T(\varphi)\geq -T(1)$$ – Norbert Mar 3 '13 at 10:08

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