Definitions of norm of operators

Question

Let $A,B$ be normed spaces and $F:A\rightarrow B$ a bounded linear operator.
How do I show that

1. $\inf\{C\geq 0:||Fa||\leq C||a||, \forall a\in A\}=\sup\{||Fa||:||a||\leq 1\}$?
2. $\inf\{C\geq 0:||Fa||\leq C||a||, \forall a\in A\}$ is a minimum?
3. Let $A\neq\{0\}$. Then $$\sup\left\{\frac{||Fa||}{||a||}, a\in A,a\neq0\right\}=\sup\{||Fa||:a\in A,||a||=1\}?$$
4. Why $$\sup\left\{\frac{||Fa||}{||a||}, a\in A,a\neq0\right\}=\sup\{||Fa||:a\in A,||a||=1\}=\inf\{C\geq 0:||Fa||\leq C||a||, \forall a\in A\}=\sup\{||Fa||:||a||\leq 1\}?$$

Unfortunately, I haven't the slightest clue how to begin any of these and I would really appreciate any hint to get me started.

Edit: Using Omnomnom's hints:
1. $$(\|Fa\| \leq C \|a\| \quad \forall a \in A) \iff (\|Fx\| \leq C \quad \forall x : \|x\| = 1)$$ The $\implies$-side seems clear. For the other way, we note that $||Fx||=||F\frac{y}{||y||}||=\frac{||Fy||}{||y||}\leq C$ implies $||Fy||\leq C|y||$.
3. $\sup \{\frac{||Fa||}{||a||}\}=\sup \{||F\frac{a}{||a||}||\}=\sup\{||Fx||:||x||=1\}$

Answer 1

For the first problem, you'll want to show that $$ (\|Fa\| \leq C \|a\| \quad \forall a \in A) \iff (\|Fx\| \leq C \quad \forall x : \|x\| = 1) $$ For the second problem, the infinimum is not necessarily a minimum; perhaps we're missing information. For example, the infimum is necessarily a minimum if $A$ is finite dimensional.

For the third (as well as the first), it helps to note that $\|F\frac{a}{\|a\|}\| = \frac{\|Fa\|}{\|a\|}$

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Let $S_1 = \{C \geq 0 :\|Fa\| \leq C \|a\| \quad \forall a \in A\}$. Let $S_2 = \{\|Fa\| : \|a\| \leq 1\}$. Notably, we may rewrite $S_1 = \{C \geq 0 :\|Fa\| \leq C \text{ whenever } \|a\| \leq 1 \}$.

Note that for every $C \in S_1$ and $M = \|Fa\| \in S_2$ (for some $a$ with $\|a\| \leq 1$), we have $C \geq \|Fa\| = M$. Conclude that $\inf S_1 \geq \sup S_2$.

On the other hand, consider any upper bound $\alpha$ for $S_2$. for every $a$ with $\|a\| \leq 1$, we have $\|Fa\| \in S_2$, so that $\|Fa\| \leq \alpha$. Thus, by definition, $\alpha \in S_1$. So, $S_1$ contains every upper bound of $S_2$. Since the supremum is the least upper bound, $\sup S_2 \geq \inf S_1$.

Thus, $\inf S_1 = \sup S_2$ as desired.

Answer 2

Refer to Sec.2.7 in the book Introductory Functional Analysis With Applications by Erwin Kreyszig.

That section defines what a bounded linear operator is and answers all of your questions.