Open unit ball in integral norm is open in supremum norm
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I have to show the following I know I have to find an $epsilon$ such that $B_infty(f,epsilon)$ is in $B_1(0,1) forall f$ . I just cannot figure out what this epsilon could be. Also I don't know how to use that hint given in the question. I think I am missing something. Kindly help thanks and regards
real-analysis functional-analysis normed-spaces
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edited Nov 17 at 13:06
asked Nov 17 at 13:01
Devendra Singh Rana
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