Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer 1

up vote 2 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 $\Vert T\Vert=\Vert T(1)\Vert$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.