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If $T:X\to Y$ is a bounded linear transformation of Banach spaces $X$ and $Y$, then there is an adjoint transformation $Y^*\to X^*$ that satifies $ =$ for all $x\in X$ and $y^*\in Y^*.$

One standard result is that $\left \| T \right \|=\left \| T^{*} \right \|.$ I have seen several proofs of this fact but I have a question about Rudin's.

Rudin writes the following sequence of equalities:

$\left \| T \right \|=\sup \left \{\langle Tx,y^{*} \rangle:\left \| x\leq 1 \right \|,\left \| y^{*}\le 1 \right \| \right \}=\\\sup \left \{\langle x,T^*y^{*} \rangle:\left \| x\leq 1 \right \|,\left \| y^{*}\le 1 \right \| \right \}=\\ \sup \left \{T^*y^{*}:\left \| y^{*}\le 1 \right \| \right \}=\left \| T^{*} \right \|$

which I take to mean:

$\left \| T \right \|=\sup_\left \{ \left \| x \right \|\le 1 \right \}(\sup_\left \{ \left \| y^* \right \|\le 1 \right \}\left \{\langle Tx,y^{*} \rangle \right \})=\\ \left \| T \right \|=\sup_\left \{ \left \| x \right \|\le 1 \right \}(\sup_\left \{ \left \| y^* \right \|\le 1 \right \}\left \{\langle x,T^*y^{*} \rangle \right \})=\\ \left \| T \right \|=\sup_\left \{ \left \| y^* \right \|\le 1 \right \}(\sup_\left \{ \left \| x \right \|\le 1 \right \}\left \{\langle x,T^*y^{*} \rangle \right \})=\left \| T^* \right \|.$

My question is simple: what justifies the interchange of the suprema? The way Rudin writes it, he seems to be claiming in general that that if $f\in \mathscr F, $ and $x\in X$, then $\sup_{\mathscr F}\sup_{X}f(x)=\sup_{X}\sup_{\mathscr F}f(x), $which is intuitive enough, but I have not been able to prove it.

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    Use `\langle x , y \rangle` for $\langle x,y \rangle$ instead of $$ – 2017-02-16

1 Answers 1

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If we have a double-indexed family $\{ u_{\alpha\beta} : \alpha \in A, \beta \in B\}$, then we have

$$\sup \: \{ u_{\alpha\beta} : \alpha \in A, \beta \in B\} = \sup_{\beta\in B} \sup\: \{u_{\alpha\beta} : \alpha \in A\} = \sup_{\alpha\in A} \sup\: \{ u_{\alpha\beta} : \beta \in B\}.$$

Clearly, since $C\subset D \implies \sup C \leqslant \sup D$, we have

$$\sup \:\{u_{\alpha\beta} : \alpha \in A\} \leqslant \sup \: \{ u_{\alpha\beta} : \alpha \in A, \beta \in B\}$$

for every fixed $\beta\in B$, and consequently

$$\sup_{\beta\in B} \sup\: \{u_{\alpha\beta} : \alpha \in A\} \leqslant \sup \: \{ u_{\alpha\beta} : \alpha \in A, \beta \in B\}.$$

Conversely, if $c < \sup \: \{ u_{\alpha\beta} : \alpha \in A, \beta \in B\}$, then there is an $(\alpha_0,\beta_0) \in A\times B$ with $c < u_{\alpha_0\beta_0}$, and hence

$$c < \sup\:\{ u_{\alpha\beta_0} : \alpha \in A\} \leqslant \sup_{\beta\in B} \sup\: \{u_{\alpha\beta} : \alpha \in A\}.$$

Since this holds for all $c < \sup \: \{ u_{\alpha\beta} : \alpha \in A, \beta \in B\}$, the equality follows. The argument for the other order of taking the suprema is analogous.