[proofplan]
We compare three quantities with the defining [operator norm](/page/Operator%20Norm), taken as the supremum of $\|Tx\|_Y$ over the closed unit ball of $X$. The nonzero hypothesis supplies at least one unit vector, so the unit-sphere supremum is not an empty supremum. The quotient formula follows by normalizing each nonzero vector, and the infimum formula follows by observing that admissible constants are exactly upper bounds for all quotients.
[/proofplan]
[step:Define the three comparison quantities and produce a unit vector]
Define the closed unit ball and unit sphere of $X$ by
\begin{align*}
B:=\{x\in X:\|x\|_X\le 1\}
\end{align*}
and
\begin{align*}
S_X:=\{x\in X:\|x\|_X=1\}.
\end{align*}
Define
\begin{align*}
M:=\sup\{\|Tx\|_Y:x\in S_X\},
\end{align*}
define
\begin{align*}
Q:=\sup\left\{\frac{\|Tx\|_Y}{\|x\|_X}:x\in X,\ x\ne 0\right\},
\end{align*}
and define
\begin{align*}
I:=\inf\{C\ge 0:\|Tx\|_Y\le C\|x\|_X\text{ for every }x\in X\}.
\end{align*}
By the definition of the operator norm,
\begin{align*}
\|T\|_{\mathcal L(X,Y)}=\sup\{\|Tx\|_Y:x\in B\}.
\end{align*}
Since $X\ne\{0\}$, choose $x_0\in X$ with $x_0\ne 0$. Define
\begin{align*}
u_0:=\frac{1}{\|x_0\|_X}x_0\in X.
\end{align*}
The norm homogeneity over $\mathbb F$ gives
\begin{align*}
\|u_0\|_X=\left|\frac{1}{\|x_0\|_X}\right|\|x_0\|_X=1.
\end{align*}
Thus $S_X$ is nonempty. In particular, $M$ is the supremum of a nonempty set of nonnegative [real numbers](/page/Real%20Numbers).
[/step]
[step:Identify the unit-sphere supremum with the closed-unit-ball supremum]
Because $S_X\subset B$, every value appearing in the definition of $M$ also appears in the definition of $\|T\|_{\mathcal L(X,Y)}$. Hence
\begin{align*}
M\le \|T\|_{\mathcal L(X,Y)}.
\end{align*}
Conversely, let $x\in B$. If $x=0$, then linearity gives $Tx=T0=0$, so
\begin{align*}
\|Tx\|_Y=0\le M.
\end{align*}
If $x\ne 0$, define
\begin{align*}
u:=\frac{1}{\|x\|_X}x\in X.
\end{align*}
Then $\|u\|_X=1$, so $u\in S_X$. By linearity of $T$ and homogeneity of the norm on $Y$,
\begin{align*}
\|Tx\|_Y=\|x\|_X\|Tu\|_Y.
\end{align*}
Since $x\in B$, we have $\|x\|_X\le 1$, and since $u\in S_X$, we have $\|Tu\|_Y\le M$. Therefore
\begin{align*}
\|Tx\|_Y\le M.
\end{align*}
Taking the supremum over all $x\in B$ gives
\begin{align*}
\|T\|_{\mathcal L(X,Y)}\le M.
\end{align*}
Combining the two inequalities yields
\begin{align*}
\|T\|_{\mathcal L(X,Y)}=M.
\end{align*}
[guided]
The definition of the operator norm uses all vectors in the closed unit ball, while the first desired formula uses only vectors of norm exactly $1$. We first compare the two sets. Since every unit vector has norm at most $1$, we have $S_X\subset B$. Therefore every number $\|Tx\|_Y$ with $x\in S_X$ is included among the numbers $\|Tx\|_Y$ with $x\in B$, so
\begin{align*}
M\le \|T\|_{\mathcal L(X,Y)}.
\end{align*}
For the reverse inequality, we must show that allowing vectors of norm less than $1$ does not increase the supremum. Let $x\in B$. If $x=0$, then linearity of $T:X\to Y$ gives $Tx=T0=0$, so
\begin{align*}
\|Tx\|_Y=0\le M.
\end{align*}
The inequality $0\le M$ is valid because $S_X$ is nonempty and every norm is nonnegative.
Now suppose $x\ne 0$. The way to compare $x$ with the unit sphere is to normalize it. Define
\begin{align*}
u:=\frac{1}{\|x\|_X}x\in X.
\end{align*}
Norm homogeneity gives
\begin{align*}
\|u\|_X=\left|\frac{1}{\|x\|_X}\right|\|x\|_X=1,
\end{align*}
so $u\in S_X$. Also $x=\|x\|_X u$, and linearity of $T$ gives
\begin{align*}
Tx=T(\|x\|_X u)=\|x\|_X Tu.
\end{align*}
Taking the norm in $Y$ and using homogeneity again,
\begin{align*}
\|Tx\|_Y=\|x\|_X\|Tu\|_Y.
\end{align*}
Since $x\in B$, $\|x\|_X\le 1$, and since $u\in S_X$, $\|Tu\|_Y\le M$. Hence
\begin{align*}
\|Tx\|_Y\le M.
\end{align*}
This holds for every $x\in B$, so the supremum over $B$ is bounded above by $M$:
\begin{align*}
\|T\|_{\mathcal L(X,Y)}\le M.
\end{align*}
Together with the earlier inequality, this proves
\begin{align*}
\|T\|_{\mathcal L(X,Y)}=M.
\end{align*}
[/guided]
[/step]
[step:Normalize nonzero vectors to obtain the quotient formula]
Let $x\in X$ satisfy $x\ne 0$, and define
\begin{align*}
u:=\frac{1}{\|x\|_X}x\in X.
\end{align*}
Then $\|u\|_X=1$, and linearity gives
\begin{align*}
\frac{\|Tx\|_Y}{\|x\|_X}=\|Tu\|_Y\le M.
\end{align*}
Taking the supremum over all nonzero $x\in X$ gives
\begin{align*}
Q\le M.
\end{align*}
Conversely, if $u\in S_X$, then $u\ne 0$ and
\begin{align*}
\|Tu\|_Y=\frac{\|Tu\|_Y}{\|u\|_X}\le Q.
\end{align*}
Taking the supremum over all $u\in S_X$ gives
\begin{align*}
M\le Q.
\end{align*}
Thus $Q=M$. From the previous step,
\begin{align*}
Q=\|T\|_{\mathcal L(X,Y)}.
\end{align*}
[/step]
[step:Identify the infimum of admissible constants]
Let
\begin{align*}
\mathcal A:=\{C\ge 0:\|Tx\|_Y\le C\|x\|_X\text{ for every }x\in X\}.
\end{align*}
Since $T\in\mathcal L(X,Y)$, the map $T$ is bounded, so $\mathcal A$ is nonempty.
If $C\in\mathcal A$ and $x\in X$ with $x\ne 0$, then dividing the inequality $\|Tx\|_Y\le C\|x\|_X$ by the positive number $\|x\|_X$ gives
\begin{align*}
\frac{\|Tx\|_Y}{\|x\|_X}\le C.
\end{align*}
Thus $C$ is an upper bound for the quotient set, and hence $Q\le C$. Taking the infimum over $C\in\mathcal A$ gives
\begin{align*}
Q\le I.
\end{align*}
Conversely, let $x\in X$. If $x=0$, then $Tx=0$, so
\begin{align*}
\|Tx\|_Y=0=Q\|x\|_X.
\end{align*}
If $x\ne 0$, the definition of $Q$ as the supremum of the quotient set gives
\begin{align*}
\frac{\|Tx\|_Y}{\|x\|_X}\le Q.
\end{align*}
Multiplying by $\|x\|_X>0$ yields
\begin{align*}
\|Tx\|_Y\le Q\|x\|_X.
\end{align*}
Therefore $Q\in\mathcal A$, so
\begin{align*}
I\le Q.
\end{align*}
Combining this with $Q\le I$ gives $I=Q$. Since $Q=\|T\|_{\mathcal L(X,Y)}$, we obtain
\begin{align*}
\|T\|_{\mathcal L(X,Y)}=I.
\end{align*}
Together with the previous equalities, this proves all three formulas.
[/step]
