TITLE: Equivalent Definitions of the Operator Norm
QUESTION [76 upvotes]: How do you prove that these four definitions of the operator norm are equivalent?
$$\begin{align*}
\lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\
&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\
&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\
&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}.
\end{align*}$$

REPLY [4 votes]: Another equivalent definition, which I can rarely see:
$$\lVert A\rVert=\sup_{\lVert x \rVert < 1} \lVert Ax \rVert$$
It's easy to see that
$$\sup_{\lVert x \rVert < 1} \lVert Ax \rVert \leqslant \sup_{\lVert x \rVert \leqslant  1} \lVert Ax \rVert$$
For the other direction, let $(r_n)$ be a sequence of real numbers, with $0 \leqslant r_n < 1$ and $r_n \to 1$, and $x \in V$ with $\lVert x \rVert \leqslant 1$. Then, we have that $\lVert r_n x\rVert<1$, which implies that $\lVert A(r_n x) \rVert \leqslant \sup_{\lVert x \rVert < 1} \lVert Ax \rVert$, but $\lVert A(r_n x) \rVert \to \lVert A x\rVert$, which implies that $\lVert Ax \rVert \leqslant \sup_{\lVert x \rVert < 1} \lVert Ax \rVert$.