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!
 A: 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.
A: 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.
A: Note, that if $\|h\| = 0$ then the inequality trivially holds. Otherwise,
$$
  \frac{\|Ah\|}{\|h\|}\leq \sup_{\|h\|>0}\frac{\|Ah\|}{\|h\|}
$$
and thus
$$
  \|Ah\|\leq \|h\|\cdot\sup_{\|h\|>0}\frac{\|Ah\|}{\|h\|}
$$
By one of the (equivalent) definitions you linked,
$$
  \|A\| = \sup_{\|h\|>0}\frac{\|Ah\|}{\|h\|}
$$
