Immediate consequence of the definition of Operator Norm. Explain

Question

||Av|| $\leq$ ||A||$_{op}$||v|| for every v $\in$ V

I was wondering why this is true. Wikipedia says it's an immediate consequence of the definition but I just do not follow.

I am using the definition ||A||$_{op}$ =inf { c$\geq$0: ||Av|| $\leq$ c||v|| for all v $\in$V }

Here is the page in case anyone wants to look at it: http://en.wikipedia.org/wiki/Operator_norm

Answer 1

Given two normed vector spaces $V$ and $W$ a linear map $A:V\rightarrow W$ is continuous if and only if it is bounded. In other words if: $$||Av||\leq c||v||$$ The definition of the operator norm is given as: $$||A||_{op} = \inf\{c\geq 0: ||Av||\leq c||v|| \text{for all} v\in V\}.$$ The operator norm is defined by the smallest $c$ so that this is true. From here simply define $||A||_{op}$ as this $c$ and the relationship $$||Av||\leq ||A||_{op}||v||$$ is a direct consequence. This is assuming however that your operator $||A||$ is bounded. There are operators that are not bounded such as the differential operator.

Answer 2

It's a standard procedure which is usually deemed as obvious. Let's give a name to this set: $$ C=\{c\ge 0\ :\ \lVert Av\rVert \le c\lVert v \rVert\}\subset [0, \infty)$$ To say that $C$ is nonempty is the same as to say that $A$ is bounded. In this case, we can take a sequence $c_n \in C$ such that $c_n\to \inf C$. For every $n$ one has, by definition, $$\tag{1} \lVert Av\rVert \le c_n\lVert v \rVert.$$ Letting $n\to \infty$, inequality (1) becomes $$\lVert Av\rVert \le (\inf C)\lVert v\rVert.$$ Now define $\lVert A\rVert_{\mathrm{op}}=\inf C.$