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Let $T\in \ell_\infty(\mathbb{Z,\mathbb{C}})^*$ such that:

  • $T(1_{\ell_\infty})=1$ where $1_{\ell_\infty}$ denotes the constant function $1$;

  • $T(u)\geq 0$ whenever $u$ is real positive.

How to prove $\lVert T\rVert\leq 1$?

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up vote 3 down vote accepted

Let $u\in L_\infty (X,\mathbb{C})$ with $\Vert u\Vert_\infty\leq 1$, then for almost all $x\in X$ we have $|u(x)|\leq 1$. Obviously there exist $\alpha\in\mathbb{C}$ such that $|\alpha|=1$ and $|T(u)|=\alpha T(u)=T(\alpha u)$. Let $\alpha u=u_1+iu_2$, where $u_1,u_2\in L_\infty(X,\mathbb{R})$. From definition of $T$ it follows that $T(u_1), T(u_2)\in\mathbb{R}$, so $$ T(u_1)+iT(u_2)=T(\alpha u)=|T(u)|\in\mathbb{R} $$ Hence $T(u_2)=0$ and $T(\alpha u)=T(u_1)$. For almost all $x\in X$ we have $$ u_1(x)\leq|u_1(x)|\leq |\alpha u(x)|=|u(x)|\leq 1 $$ so from positivity of $T$ we get $$ |T(u)|=T(\alpha u)=T(u_1)\leq T(|u|)\leq T(1_{L_\infty(X,\mathbb{C})})=1 $$ Thus for all $u\in L_\infty(X,\mathbb{C})$ with $\Vert u\Vert_\infty\leq 1$ we have $|T(u)|\leq 1$. Hence $\Vert T\Vert\leq 1$

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Nice proof, I meant l_{\infty}, not L_{\infty}...but it can be easily adapted. – Lola Oct 30 '12 at 16:27
I knew that, I just slightly generalized my answer. – Norbert Oct 30 '12 at 16:28

Let $u\in\ell_{\infty}$ real. Then $-\lVert u\rVert_{\infty}\leq u\leq\lVert u\rVert_{\infty}$, so $\lVert u\rVert_{\infty}\mathbf 1_{\ell_{\infty}}-u\geq 0$ and $\lVert u\rVert_{\infty}\geq T(u)$. This gives $|T(u)|\leq \lVert u\rVert_{\infty}$.

For $u$ which is not real, write $u=a+ib$, where $a$ and $b$ are real bounded sequences. Then $|T(u)|^2= |T(a)|^2+|T(b)|^2\leq \lVert a \rVert^2+\lVert b \rVert^2\leq 2\lVert u \rVert^2$, proving that $u$ is continuous.

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Yes but what if $u(x)\in \mathbb{C}$? – Lola Oct 30 '12 at 15:14
I need to think more in order to prove that the norm is $1$. – Davide Giraudo Oct 30 '12 at 15:36

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