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Math Help - Norm of a Bounded Linear Operator - another question

  1. #1
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    I have one final enquiry, and that is the relation,

    ||B|| = \sup_{||x||\leq 1}||Bx|| = \sup_{||x|| = 1}||Bx||.


    I don't exactly know what this is saying. I mean, I could interpret it as the least upper bound on the sphere, ||x|| \leq 1, and that that is the same as the least upper bound on the surface of the sphere. Thinking in that way, I can see kind of why this is the case. Is there any way of proving it formally?

    Further thoughts:

    The norm of a bounded linear operator is given by,

    \displaystyle ||A|| = \sup_{||x||\leq 1}||Ax|| = \sup_{||x|| = 1} ||Ax||.


    Now, the way I think about this is that this is the least upper bound on the closed unit sphere, and that this can be viewed as the same as the least upper bound on the surface of the unit sphere, I think. However, how can this be proved formally?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by HTale View Post
    I also have one final (related) enquiry, and that is the relation,

    ||B|| = \sup_{||x||\leq 1}||Bx|| = \sup_{||x|| = 1}||Bx||.

    I don't exactly know what this is saying. I mean, I could interpret it as the least upper bound on the sphere, ||x|| \leq 1, and that that is the same as the least upper bound on the surface of the sphere. Thinking in that way, I can see kind of why this is the case. Is there any way of proving it formally?
    My terminology is a bit different. I'll call the set \{x:\|x\| \leqslant 1\} the unit ball, and the set \{x:\|x\| = 1\} the unit sphere.

    The ball is bigger than the pshere, so the sup over the ball is greater than or equal to the sup over the sphere: \sup_{\|x\|\leqslant 1}\|Bx\| \geqslant \sup_{\|x\| = 1}\|Bx\|.

    For the reverse inclusion, use the same sort of argument as in my previous comment. If 0<\|y\|<1 then \left\|\tfrac{\textstyle y}{\|y\|}\right\| = 1, and so  \|B\tfrac{\textstyle y}{\|y\|}\|\leqslant \sup_{\|x\|=1}\|Bx\|. Therefore \|By\| \leqslant \sup_{\|x\|=1}\|Bx\|\,\|y\| < \sup_{\|x\|=1}\|Bx\|. Hence \sup_{\|y\|\leqslant 1}\|By\| \leqslant \sup_{\|x\| = 1}\|Bx\|.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    My terminology is a bit different. I'll call the set \{x:\|x\| \leqslant 1\} the unit ball, and the set \{x:\|x\| = 1\} the unit sphere.

    The ball is bigger than the pshere, so the sup over the ball is greater than or equal to the sup over the sphere: \sup_{\|x\|\leqslant 1}\|Bx\| \geqslant \sup_{\|x\| = 1}\|Bx\|.

    For the reverse inclusion, use the same sort of argument as in my previous comment. If 0<\|y\|<1 then \left\|\tfrac{\textstyle y}{\|y\|}\right\| = 1, and so  \|B\tfrac{\textstyle y}{\|y\|}\|\leqslant \sup_{\|x\|=1}\|Bx\|. Therefore \|By\| \leqslant \sup_{\|x\|=1}\|Bx\|\,\|y\| < \sup_{\|x\|=1}\|Bx\|. Hence \sup_{\|y\|\leqslant 1}\|By\| \leqslant \sup_{\|x\| = 1}\|Bx\|.
    Thanks a lot Opalg, and thank you mr fantastic for sorting out the two posts!
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