Jtdylktuy

Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y.$ Then, we have for arbitrary $Tin B(X,Y), $
begin{align} Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert= suplimits_{Vert x Vert = 1}Vert T x Vert=suplimits_{xneq 0} frac{Vert T x Vert}{Vert x Vert}.end{align}

Proof

Since $T$ is bounded and linear, $exists Kgeq 0$ such that for all $xin X,;Vert T x Vert leq K Vert x Vert.$ If $Vert x Vertleq 1$, then $Vert T x Vert leq K Vert x Vertleq K.$

Thus, begin{align}tag{1}label{1} suplimits_{Vert x Vertleq 1}Vert T x Vertleq inf{ Kgeq 0:Vert T x Vert leq K Vert x Vert,forall;xin K }=Vert T Vertend{align}
By definition of $inf,$ for every $epsilon> 0,exists ;x_{epsilon}in X,x_{epsilon}neq 0$ such that
begin{align} Vert T x_{epsilon} Vert>left(Vert T Vert -epsilonright)Vert x_{epsilon} Vert.end{align}
Let $u_{epsilon}=frac{x_{epsilon}}{Vert x_{epsilon} Vert},$ then $u_{epsilon}=1$ and $ Vert T u_{epsilon} Vert>Vert T Vert -epsilon.$ We obtain from $ eqref{1} $

begin{align} tag{2}label{2}Vert T Vert geqsuplimits_{Vert x Vertleq 1}Vert T x Vertstackrel{text{how?}}{geq} suplimits_{Vert x Vert = 1}Vert T x Vertstackrel{text{how?}}{geq} suplimits_{Vert x_{epsilon} Vertneq 0} Vert Tleft(frac{x_{epsilon} }{big Vert x_{epsilon} Vert}right)bigVertstackrel{text{how?}}{geq}Vert T Vert -epsilon.end{align}
Since $epsilon>0$ was arbitrary, then
begin{align} Vert T Vert =suplimits_{Vert x Vertleq 1}Vert T x Vert= suplimits_{Vert x Vert = 1}Vert T x Vert=suplimits_{xneq 0} frac{Vert T x Vert}{Vert x Vert}.end{align}
Can you please explain the how's in $ eqref{2} ?$

Answering your "how" questions regarding the inequalities:

The first one is because you're taking the sup over a smaller set. When $A subset B$, and $f$ is a real valued map whose domain contains $A$ and $B$, we have $$sup_{x in A} f(x) le sup_{x in B} f(x).$$ Here $B = {x , : , |x| le 1}$ and $A = { x , : , |x | = 1}$.

The second is again just restricting to a smaller set. For each $epsilon$, the vector $x_epsilon/|x_epsilon|$ has norm $1$, so if we only consider the sup over those vectors, we are again considering a smaller set, and thus get the inequality for the same reason as above.

The third is by the definition of $x_epsilon$. We've chosen $x_epsilon$ such that $$left | Tleft(frac{x_epsilon}{| x_epsilon|} right)right | ge |T | - epsilon.$$

1. Since $bigl{lVert TxrVert,|,lVert xleqslant1bigr}supsetbigl{lVert TxrVert,|,lVert x=1bigr}$, $supbigl{lVert TxrVert,|,lVert xleqslant1bigr}leqslantsupbigl{lVert TxrVert,|,lVert x=1bigr}$
   


2. Because $leftlVertdfrac{x_varepsilon}{lVert x_varepsilonlVert}rightrVert=1$.


3. It was proved before that $leftlVert Tleft(dfrac{x_varepsilon}{lVert x_varepsilonlVert}right)rightrVertgeqslantlVert TrVert-varepsilon$.