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I am trying to show that these two definitions for a bounded linear operator norm on the normed linear space $X$ are equivalent: $$ \sup\{T(x)\,:\, \|x\|\le 1\}=\|T\|_*=\inf\{M>0\,:\, T(x)\le M\cdot\|x\|, \forall f\in X\}. $$


So here is my attempt: Let $\|x\|\le 1$, then using the infimum definition of $\|T\|_*$ we have $$ T(x)\le \|T\|_*\cdot \|x\|\le \|T\|_*, $$ then taking supremum over both sides gives $$ \sup\{T(x)\,:\, \|x\|\le 1\}\le\|T\|_*. $$

I would like to reverse this inequality but I am not seeing how to proceed. Any hints?

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Let $\epsilon>0$. Then, from the definition of the $\inf$, there is $y\in X$ such that

$$(||T||_*-\epsilon)||y||<|T(y)|.$$

Hence for $x:=\frac{y}{||y||}$, we have $$(||T||_*-\epsilon)<|T(x)|\leq \sup\{|T(x)|:\ ||x||\leq1\}.$$

Since this is true for all $\epsilon$ then $$||T||_*\leq \sup\{|T(x)|:\ ||x||\leq1\}.$$

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