Proving that this supremum is equal to an infimum I want to prove that for $T\in L(\mathbb{R^l},\mathbb{R^m})$,i.e the space of linear transformation, $$||T||_L = \sup\{ |Tx| : x\in\bar B_1(0)\}$$.
But my defnition of $||T||_L$ is the following:
$$||T||_L = \inf\{ M :|Tx|\le M|x|\}$$
So we know that since $T$ is linear we get that 
$$|Tx|\le M|x|\le M$$
since $|x| \leq 1$ but the thing is how do I get that the supremum of this $M$'s is the same as its infimum? 
Thanks a lot in advance. 
 A: Let $A = \{|Tx| : x\in \bar B_1(0)\}$ and $B = \{M : |Tx| < M|x|, \text{ for } x \ne 0 \}$. We will show $\sup A = \inf B$. As you allready noted, we have 
$$ |Tx| < M|x| \le M $$
for $|Tx| \in A$ and $ M \in B$, hence $\sup A \le \inf B$. Now let $\epsilon > 0$, and $x \ne 0$. Note that we have 
\begin{align*}
  |Tx| &= \left|T \left(\frac{x}{|x|}\right)\right|\cdot |x|\\
       &< (\sup A + \epsilon)|x| 
\end{align*}
as $x/|x| \in \bar B_1(0)$. Therefore $\sup A + \epsilon \in B$ for all $\epsilon > 0$, hence $\inf B \le \sup A + \epsilon$ for all $\epsilon > 0$, givnig $\inf B \le \sup A$.
Together, we have $\inf B = \sup A$.
A: If $c<\inf\{\,M:|Tx|<M|x|\,\}$ then there exists $x_0$ with $|Tx_0|\ge c|x_0|$. We may assume $x_0\ne 0$ (why?). Now let $x_1=\frac1{|x_0|}x_0$ and verify that also $|Tx_1|\ge c|x_1|$. Hence $\sup\{\,|Tx|:x\in \overline B_1(0)\,\}\ge c$. 
If $c>\inf\{\,M:|Tx|<M|x|\,\}$ then $|Tx|<c|x|$ for all $x$, specifically $|Tx|<c$  for all $x\in \overline B_1(0)$, so that $c>\sup\{\,|Tx|:x\in \overline B_1(0)\,\}$.
We conclude $\inf\{\,M:|Tx|<M|x|\,\} = sup\{\,|Tx|:x\in \overline B_1(0)\,\}$.
Remark: I would prefer to write  $\inf\{\,M:\forall x\;|Tx|<M|x|\,\} $ instead of  $\inf\{\,M:|Tx|<M|x|\,\}$.
