How do I prove the equivalence of two definitions of operator norm without using the equivalnce between the sup versions? $$\|A\|_1=\inf\{c\ge 0: \forall x\in V,\|Ax\|\le c, \|x\| \le 1\}=\inf\ S_1$$
$$\|A\|_2=\inf\{c\ge 0: \forall x\in V,\|Ax\|\le c, \|x\| = 1\}=\inf S_2$$
I am trying to prove these two expresions are equal DIRECTLY, I know there are other equivalente definitions using the sup, I know how to deal with those, so let's not use them.
Similarly to what is done to prove the equivalence between the sup version of these two expresions ( here Why are different definitions of the operator norm equivalent?), I notice the inclusion between sets: $S_2 \subset S_1$ , so
$\|A\|_1=\inf\ S_1 \le \inf\ S_2 =\|A\|_2$  ...(*)
If I can prove the opposite inequality I'd be done, that is where I am having trouble.
I know that for  $ x \in V, \|x\|\le 1 , \|Ax\| \le \|Ax\|/\|x\| $. I don't want to take the sup over $ x$ here since that is the reasoning for the sup version of this problem, so how do I go about reasoning in terms of the inf of the constants that bound each side of the inequality? Let $C_1$ represent any value of $C$ in $S_1$ and $C_2$ represent any value of $C $ in $S_2 $.
I know  $\|Ax\| \le \|A\|_1  \le C_1$ $\forall \|x\| \le 1$
and $\|Ax\| \le  \|A\|_2 \le C_2$ $\forall \|x\| = 1$ that is  $ \|Ax\|/\|x\|  \le \|A\|_2 \le C_2 \forall x \in V$
How can I conclude  $\|A\|_2\le \|A\|_1$ ?
 A: Let us do it very formally. We write $D = \{ x \in V \mid \lVert x \rVert \le 1 \}$ and $B = \{ x \in V \mid \lVert x \rVert = 1 \}$. Your sets $S_1, S_2$ are then given as
$$S_1 = \{ c \ge 0 \mid \forall x \in D : \lVert Ax \rVert \le c \}, \\ S_2 = \{ c \ge 0 \mid \forall x \in B : \lVert Ax \rVert \le c \} .$$
Note that for an infinite-dimensional $V$ these sets may be empty.
By definition we have  $S_1 \subset S_2$ simply because $B \subset D$. In fact, if $\lVert Ax \rVert \le c$ for all $x \in D$, then trivially $\lVert Ax \rVert \le c$ for all $x \in B$.
Note that this argument is valid for arbitrary functions $A$, we did not use that $A$ is linear. Let us prove that for linear $A$ we have $S_2 \subset S_1$. This implies $S_1 = S_2$ and therefore $\lVert A \rVert_1 =  \lVert A \rVert_2$.
So let $c \in S_2$ and $\lVert x \rVert \le 1$. If $\lVert x \rVert = 0$, then $x = 0$ and $\lVert Ax \rVert = 0 \le c$. If $\lVert x \rVert \ne 0$, then $y = x/\lVert x \rVert$ is defined and $\lVert y \rVert = 1$. Thus $\lVert Ay \rVert \le c$. But we have $$\lVert Ax \rVert = \lVert A(\lVert x \rVert \cdot y) \rVert = \lVert \lVert x \rVert Ay \rVert = \lVert x \rVert \cdot \lVert Ax \rVert \le 1 \cdot c = c .$$
A: Since $\inf S_1 = \|A\|_1$ is an upper bound of $\|Ax\|$,
$\forall x ,\|x\| \le 1$ ,$\|Ax\| \le \|A\|_1 $.
In particular
$\forall x ,\|x\| = 1$, $\|Ax\| \le \|A\|_1 $,
so $\|A\|_1 \in S_2$, but  since $\inf S_2 = \|A\|_2$ , it follows that $\|A\|_2 \le \|A\|_1$
