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!

This is defined (in the wiki) for bounded linear operators. This means there exists some $L$ such that $\|Ax\| \le L \|x\|$ for all $x$. The (induced) norm is defined to be the infimum of such constants.
As per Julien's comment below, here is why the $\inf$ of such bounds is itself a bound: Let $\Lambda = \{ L | \|A x \| \le L \|x\| \, \forall x\}$, and suppose $B = \inf  \Lambda$. Then for each $x$ we must have $\|A x \| \le B \|x\|$. To see this, suppose $\|A x_0 \| > B \|x_0\|$ for some $x_0$ (which must be non zero). Then we must have $L \geq \frac{\|A x_0 \|}{\|x_0\|} > B$ for all $L \in \Lambda$, which contradicts the definition of $B$. Hence $B \in \Lambda$.
To illustrate an example of an unbounded operator, let $P$ the be set of polynomials in one variable with norm $\|p\| = \max_{x \in [0,1]} |p(x)|$, and let $\phi: P \to \mathbb{R}$ be the linear operator $\phi(p) = p(2)$ (ie, evaluate the polynomial at $x=2$). Then by taking the sequence $p_n(x) = x^n$, we have a sequence $p_n$ such that $\|p_n\| =1$, but $\phi(p_n) = 2^n$. Hence no constant satisfying the inequality in the question can exist (ie, $\phi$ is an unbounded linear operator).
All linear operators on finite dimensional spaces are bounded.
