How can I prove that $\|Ah\| \le \|A\| \|h\|$ for a linear operator $A$? On http://www.proofwiki.org/wiki/Definition:Norm_(Linear_Transformation) , it is stated that $||Ah|| \leq ||A||||h||$ where $A$ is an operator.
Is this a theorem of some sort? If so, how can it be proved? I've been trying to gather more information about this to no avail.
Help greatly appreciated!

By definition:
$$
\|A\|=\sup_{\|x\|\leq 1}\|Ax||=\sup_{\|x\|= 1}\|Ax||=\sup_{x\neq 0}\frac{\|Ax\|}{\|x\|}.$$
In particular, 
$$
\frac{\|Ax\|}{\|x\|}\leq \|A\|\qquad\forall x\neq 0\quad\Rightarrow\quad\|Ax\|\leq \|A\|\|x\|\qquad \forall x.
$$
Note that $\|A\|$ can alternatively be defined as the least $M\geq 0$ such that $\|Ax\|\leq M\|x\|$ for all $x$. When such an $M$ does not exist, one has $\|A\|=+\infty$ and one says that $A$ is unbounded.
