$\|L(v)\| \leq \|L\|\cdot\|v\|$ on Banach spaces Let $A,N$ be Banach spaces and let $L: A \rightarrow N$ be a linear transformation.
If $L$ is continuous (which is guaranteed on finite dimensional spaces), the set $ \{ M \geq 0 \ : \|L(v)\| \leq M\|v\|, \ \forall v \in A  \} $ is nonempty (and obviously bounded below), and hence has an infimum, which defines the operator norm of $L$.
Thus $\|L\| = \inf\{ M \geq 0 \ : \|L(v)\| \leq M\|v\|, \ \forall v \in A  \} $ is a norm on the set of all continuous linear maps $L(A,N)$.
Is $\|L\|$ contained in the set $\{ M \geq 0 \ : \|L(v)\| \leq M\|v\|, \ \forall v \in A  \} $ ? In other words, is this set contains its infimum, namely $\|L\|$?
 A: Yes. Fix $v \in A$, and fix $\varepsilon > 0$. Clearly if $\|v\| = 0$ then $\|L(v)\| = 0 = \|L\|\|v\|$, so suppose that $v \not=0$. Since
$$
\|L\| = \inf\{M \ge 0 : \|L(w)\| \le M\|w\|\text{ for all } w \in A\},
$$
there exists $M \ge 0$ such that $\|L(v)\| \le M\|v\|$ for all $v \in A$ and such that $M \le \|L\| + \frac{\varepsilon}{\|v\|}$. In particular,  $\|L(v)\| \le M\|v\| \le \left(\|L\| + \frac{\varepsilon}{\|v\|}\right)\|v\| = \|L\|\|v\| + \varepsilon$. Since this was valid for any $\varepsilon > 0$, we deduce that $\|L(v)\| \le \|L\|\|v\|$, and then since $v \in A$ was arbitrary, we see that $\|L(v)\| \le \|L\|\|v\|$ for all $v \in A$.
This does not mean that there exists a nonzero $v \in A$ such that $\|L(v)\| = \|L\|\|v\|$. To see this, let $c_0$ denote the Banach space of real-valued sequences $a = (a_n)^\infty_{n=1}$ such that $\lim_{n \to \infty} a_n = 0$ under the supremum norm $\|a\|_\infty = \sup_{n \in \mathbb{N}} |a_n|$. There is a bounded linear operator $L : c_0 \to c_0$ given by $L(a)_n := \left(1 - \frac{1}{n}\right) a_n$. Indeed, for $a \in c_0$, we have
$$
\|L(a)\|_\infty = \sup_{n \in \mathbb{N}} \left(1 - \frac{1}{n}\right)|a_n|
   \le \sup_{n \in \mathbb{N}} |a_n| = \|a\|_\infty,
$$
and so $\|L\| \le 1$. For each $n \in \mathbb{N}$, let $\delta_n$ be the sequence $(0,0,\dots, 0, 1, 0, 0, \dots)$ that has a $1$ in the $n^{th}$ term and $0$ in every other term. So $\|\delta_n\|_\infty = 1$ for every $n$. Then $L(\delta_n) = \left(1 - \frac{1}{n}\right)\delta_n$, and in particular $\|L(\delta_n)\|_\infty = (1 - \frac{1}{n}) \to 1$ as $n \to \infty$. So $\|L\| \ge \sup_{n \in \mathbb{N}} \|L(\delta_n)\| = \sup_{n \in \mathbb{N}} 1 - \frac{1}{n} = 1$. Thus, we conclude that $\|L\| = 1$.
For any nonzero $a \in c_0$ we have $|L(a)_n| \to 0$ and so there exists $N$ such that $|L(a)_N| = \|L(a)\| > 0$. So 
$$
\|L(a)\|_\infty = |L(a)_N| = \left(1 - \frac{1}{N}\right) |a_N| < |a_N| \le \|a\|_\infty.
$$
So there is no (nonzero) vector $a \in c_0$ such that $\|L(a)\| = \|L\|\|a\|$.
