Can I please get feedback on my proof below? It is a problem from Bartles' Analysis book. Thank you!
$\def\x{{\mathbf x}} \def\0{{\mathbf 0}} \def\N{{\mathbb N}} \def\R{{\mathbb R}}$
The norm of a transformation $L\in{\mathcal L}(\R^n,\R^m)$ is defined by $$\|L\| = \inf\{C\colon \|L(\x)\|\le C\|\x\|\ \mbox{for all}\ \x\in\R^n\}.$$ Prove that $$\|L\| = \sup\left\{\frac{\|L(\x)\|}{\|\x\|}\colon \x\ne\0\right\}.$$
$\textbf{Solution:}$ Given $\|L\| = \inf\{C\colon \|L(\x)\|\le C\|\x\|\ \mbox{for all}\ \x\in\R^n\}$, lets say it equals to $\alpha$. We wish to prove that $\|L\| = \sup\left\{\frac{\|L(\x)\|}{\|\x\|}\colon \x\ne\0\right\}$ equals to $\beta$. So, it is sufficient to prove that $\alpha = \beta$. From, our definition of $\beta$, for $\x\ne \0$ $$\frac{\|L(\x)\|}{\|\x\|} \le \beta \text{ implies } \|L(\x)\| \le \beta\|\x\|$$ and $\alpha$ is the infimum of all such constant C such that $\|L(\x)\| \le C\|\x\|$. Therefore, $\alpha \le \beta. \hspace{8 pt} (1)$
Now, suppose $D$ is such that $\alpha < D$. By definition there is some $C$ with $\alpha \le C < D$ such that $\|Lx\| \le C \|x\| \le D \|x\|$ for all $x$.
Then $\beta \le D$. Hence $\beta \le \inf\{ D | \alpha < D\} = \alpha. \hspace{8pt} (2)$
By (1) and (2), we arrive at $\alpha = \beta$ so $$\|L\| = \|L\| = \sup\left\{\frac{\|L(\x)\|}{\|\x\|}\colon \x\ne\0\right\}.$$
We can also prove this by letting $$\alpha = \inf\{C\colon \|L(\x)\|\le C\|\x\|\ \mbox{for all}\ \x\in\R^n\} = \inf\{C\colon \frac{\|L(\x)\|}{\|\x\}} \le C, \x\ne \0\}.$$ Therefore $\alpha$ is the infimum of the set of all upper bounds of $$\left\{\frac{\|L(\x)\|}{\|\x\|}\colon \x\ne\0\right\}$$ where $\alpha$ is the supremum of the set $$\left\{\frac{\|L(\x)\|}{\|\x\|}\colon \x\ne\0\right\}.$$ Therefore, $\alpha = \beta$ and we are done.
I didn't follow your logic showing $\beta \le \alpha$.
Suppose $D$ is such that $\alpha < D$. By definition there is some $C$ with $\alpha \le C < D$ such that $\|Lx\| \le C \|x\| \le D \|x\|$ for all $x$.
Then $\beta \le D$. Hence $\beta \le \inf\{ D | \alpha < D\} = \alpha $.