AxiomOfChoice
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- 1
If f is a continuous functional on a normed space, do you have
<br /> \sup_{\|x\| < 1} |f(x)| = \sup_{\|x\| = 1} |f(x)|<br />
If so, why? If not, can someone provide a counterexample?
<br /> \sup_{\|x\| < 1} |f(x)| = \sup_{\|x\| = 1} |f(x)|<br />
If so, why? If not, can someone provide a counterexample?