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Math Help - Iterated sup

  1. #1
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    Iterated sup

    Let f:A\times B\to\mathbb R bounded above, prove that \underset{(x,y)\in A\times B}{\mathop{\sup }}\,f(x,y)=\underset{x\in A}{\mathop{\sup }}\,\left(\underset{y\in B}{\mathop{\sup }}\,f(x,y)\right).

    How to prove this? I think is not that easy.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    Let f:A\times B\to\mathbb R bounded above, prove that \underset{(x,y)\in A\times B}{\mathop{\sup }}\,f(x,y)=\underset{x\in A}{\mathop{\sup }}\,\left(\underset{y\in B}{\mathop{\sup }}\,f(x,y)\right).

    How to prove this? I think is not that easy.
    Let (x_0,y_0)\in A\times B be fixed. We clearly then that that \displaystyloe f(x_0,y_0)\leqslant \sup_{y\in B}f(x_0,y) and moreover it's clear that \displaystyle \sup_{y\in B}f(x_0,y)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y) putting these two together gives \displaystyle f(x_0,y_0)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y) and since (x_0,y_0)\in A\times B was arbitrary it follows that \displaystyle \sup_{(x,y)\in A\times B}f(x,y)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y). Conversely, fix y_0\in B and x_0\in A then evidently \displaystyle \sup_{(x,y)\in A\times B}f(x,y)\geqslant f(x_0,y_0). Since y_0 was arbitrary we have that \displaystyle \sup_{(x,y)\in A\times B}f(x,y)\geqslant \sup_{y\in B}f(x_0,y) and since x_0 was arbitrary \displaystyle \sup_{(x,y)\in A\times B}f(x,y)\geqslant \sup_{x\in A}\sup_{y\in B}f(x,y). The conclusion follows.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    and moreover it's clear that \displaystyle \sup_{y\in B}f(x_0,y)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y)
    Okay, not much clear for me, haha, wonder why is this true? I can't see it.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    Okay, not much clear for me, haha, wonder why is this true? I can't see it.
    Why? If I said to you consider the function g:\mathbb{R}\to\mathbb{R} one would say that g(x_0)\leqslant \sup_{x\in\mathbb{R}}g(x). Let g(x)=\sup_{y\in B}f(x,y)
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