Let $A$ be a bounded linear operator mapping a normed linear space $E_1$ to $E_2$. Suppose $\|A\|$ is defined as the smallest number C such that $\|Af\|$ $\leq$ $C\|f\|$ for all $f\in E_1$.
Prove that $\|A\|$ is also equal to:
$\|A\|$ = $\sup\limits_{x \leq\ 1}$ $\|Ax\|$
AND particularize this for the case of a bounded linear functional on $E_1$
This question I am having a hard time putting my intuition into math terms. Thoughts: To answer this question, this is what I'm thinking (informally): $A$ being bounded means that it maps a bounded subset into another bounded subset all the time. For any $x\in E_1$, A is less than any sphere with radius $C$. $C$ in this case is the lowest upper bound (supremum) as any sphere's lowest upper bound for all points is its radius length. Beyond this, I for some reason lose my train of thought relating the definitions that I'm supposed to equate. As for particularizing this for bounded linear functionals, I am completely lost on it as I've found linear functionals difficult. Any help for this problem would be nice. Thanks!
Let $L_A=\sup_{|x|\le 1} |Ax|$ and $C_A\in\mathbb{R}^+$ be the smallest constant such that $|Ax|\le C_A|x|$ for every $x$, so that we want to show that $L_A=C_A$.
First, fix any $x\not=0$ and let $x'=\frac{x}{|x|}$ so that $|x'|=1$ and $|Ax'|\cdot |x|=|Ax|$. Now $|Ax'|\le L_A$ by definition of $L_A$, and therefore $|Ax|\le L_A|x|$ for any non-zero $x$. This shows that $C_A\le L_A$ (since $C_A$ is the smallest constant that yields the same inequality).
For the other implication, take any $y$ with $|y|\le 1$. By definition of $C_A$ we know that $|Ay|\le C_A|y|$ for every $y$, so that $$L_A:=\sup_{|y|\le 1}|Ay|\le \sup_{|y|\le 1}C_A|y|=C_A$$ as wanted.
The "geometric intuition" here is that since the operator is linear we only need to "know" what happens inside the ball of radius $1$ (or any constant, really) because we can rescale an arbitrary large subset of the normed space to fit inside this unitary ball (that's really what's happened in the first implication).