Let $(V,\left\Vert \cdot\right\Vert_V),(W,\left\Vert \cdot\right\Vert_W) $ be norm spaces.
For any linear transformation $\;T:V\rightarrow W\;$ we will define:$\;\left\Vert T\right\Vert _{op}=sup\{\left\Vert Tv\right\Vert _{W}:\left\Vert v\right\Vert _{V}=1\}$
Show that:
$\forall c\geq0$ , $\;\left\Vert T\right\Vert _{op}\leq c\;$ $\iff \left\Vert Tx\right\Vert _{W}\leq c\left\Vert x\right\Vert _{V}\;\; \forall x \in V$
edit: This is what I tried:
For $\impliedby$:
We have that $\left\Vert Tx\right\Vert _{W}\leq c\left\Vert x\right\Vert _{V}\;\; \forall x \in V$, specifically for any $x \in V$ such that $\left\Vert x\right\Vert _{V} = 1$, therefore we get from homogeniety : $c\geq sup\{\left\Vert Tv\right\Vert _{W}:\left\Vert v\right\Vert _{V}=1\} = \;\left\Vert T\right\Vert _{op}$
Since $\;\left\Vert T\right\Vert _{op}\geq 0$ we get from transitivty that $c\geq 0$
For $\implies$: I tried by contradiction, which means that I assume that $\forall c\geq0$ , $\;\left\Vert T\right\Vert _{op}\leq c\;$ but there exists $x_0\in V$ such that $\left\Vert Tx_0\right\Vert _{W}> c\left\Vert x_0\right\Vert _{V}$
We know that $\frac{x_{0}}{\left\Vert x_{0}\right\Vert _{\text{V}}}$ is a unit vector in $V$. Therefore: $\left\Vert T\right\Vert _{op}\geq\left\Vert T(\frac{x_{0}}{\left\Vert x_{0}\right\Vert _{\text{V}}})\right\Vert _{W}=\frac{1}{\left\Vert x_{0}\right\Vert _{V}}\left\Vert Tx_{0}\right\Vert _{W}>c\frac{1}{\left\Vert x_{0}\right\Vert _{V}}\left\Vert x_{0}\right\Vert _{V}=c$
Which gives us a contradiction
Use the fact that $ x/||x||$ has norm 1 and the linearity of T.
EDIT: $||T||=\text{inf}\{ c\geq 0: ||T(x)|| \leq c||x|| \;\text{for all} \; x \in V\} $
Proof: I am going to suppose the above is finite, the second case is easier in my opinion (try it).
Suppose ||T|| is finite then, $$ ||T||= \text{inf}\{c\geq 0: ||T(x)|| \leq c||x||\; \text{for all}\;x \}, \\ = \text{inf}\{ c \geq 0: \left|\left|T\left(\frac{x}{||x||}\right)\right|\right| \leq c,\; ||x|| \neq 0\}\cup \{0\}, \\ = \text{inf}\{c \geq 0: ||T(y)|| \leq c, \; ||y||=1 \}.$$
The above set is the set of upper bounds of $\{ ||T(x) ||: ||x||=1\}$ therefore it’s infinum is the lowest upper bound I.e the supremum.