# Linear operator norm

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?

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 $$||T||_*\leq \sup\{|T(x)|:\ ||x||\leq1\}.$$