Question about the " pseudo-gradient vector fields"

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I'm working on this theorem, but i don't understand why there exists $x$ with $$||x||=1$$

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Well, $F:=h(v)\in X'$ is a functional and as such its norm is defined as $\Vert F \Vert = \sup\{(F,x)\vert x\in X, \Vert x\Vert =1\}$. The number $2/3 \Vert F\Vert$ is less then that supremum, hence it is dominated by some number within the set, in other words, there is an $x\in X$ with $\Vert x\Vert=1$ and $(F,x)> 2/3 \Vert F\Vert$.