Why do we define $\|L\|=\sup\{\|L x\|\mid \|x\|\leq 1\}$ Let $L:V\to V$ be an linear map in a normed space $V$. Why do we define $$\|L\|=\sup\{\|L x\|\mid \|x\|\leq 1\}\ \ \ ?$$
It mean that it's not defined if $\|x\|>1$, $\|L\|$ is not defined ?
 A: See that $\sup$ in there?  The result does not depend on $x$.  If you like you can define $\|L\|$ as the minimum of all constants $C$ such that
$$
\|Lx\| \le C \|x\|\qquad\text{for all } x \in X
$$
Perhaps you can try to prove this is the same thing as your definition.
A: What all that notation really means is this: take every single vector on the unit sphere (i.e. every vector with $\|x\|=1$), and apply $L$ to them. Now see how long the longest of them became.
It's kind of similar to asking what the biggest eigenvalue of $L$ is (in fact it's the same thing if $L$ is diagonalisable), in that we ask "how much is the most that $L$ stretches vectors?" but this way is more general. It doesn't care that some vectors don't end up a scalar multiple of what it was.
A: No. The norm of an application is defined for the application, not for some values in its domain.
The meaning of the definition is the following: take all the points $x$ whose norm is $1$ or less. If the application is continuous, the set of norms of the images of these points is bounded. The supremum of this set is, by definition, the norm of the application.
There is another equivalent definition that does not mention the points whose norm is $\le 1$:
$$\|L\|=\sup\{k\in\Bbb R: \|Lx\|=k\|x\|\forall x\neq0\}$$
A: The norm of an operator is in fact define by $$\|L\|=\sup\left\{\frac{\|Lx\|}{\|x\|}\mid x\in V\backslash \{0\}\right\} $$ 
But as you can see, if $x\in V\backslash \{0\}$, then
$$\frac{\|Lx\|}{\|x\|}\underset{y=x/\|x\|}{=}\frac{\|Ly\|}{\|y\|},$$
Therefore 
$$\|L\|=\sup\left\{\frac{\|Ly\|}{\|y\|}\mid \|y\|\leq 1\right\}.$$
But as you can also see, 
$$\|Ly\|\underset{u=y/\|y\|}{=}\|Lu\|\|y\|\implies \frac{\|Ly\|}{\|y\|}=\|Lu\|$$
and thus 
$$\|L\|=\sup\left\{\frac{\|Ly\|}{\|y\|}\mid \|y\|\leq 1\right\}=\sup\left\{\|Ly\|\mid \|y\|\leq 1\right\}.$$
