Prove that $\| A \| = \lvert y \rvert$ 
For all $A \in L(\Bbb R^n, \Bbb R)$ there is a unique $y \in \Bbb R^n$ such that $A\textbf{x} = \textbf{x} \cdot \textbf{y}$. Prove that $\| A \| = \lvert \textbf{y} \rvert$. Hint: Cauchy-Schwarz inequality.

$L(\Bbb R^n, \Bbb R)$ is the space of linear maps from $\Bbb R^n$ to $\Bbb R$. $A \in L(\Bbb R^n, \Bbb R)$ is bounded if there exists a non-negative constant $c$ such that $\|A\textbf{x} \| \leq c \|\textbf{x} \|$. $A$ is bounded if and only if it is continuous.
I believe need to show that $\| A \| \leq \lvert \textbf{y} \rvert$ and $\| A \| \geq \lvert \textbf{y} \rvert$.
First of all, for $A \in L(\Bbb R^n, \Bbb R)$, $\| A \|$ is defined as: $\| A \| = \sup\limits_{\|\textbf{x} \| \leq 1} \| A\textbf{x} \|$. There is also an equivalent definition I found that says $\| A \| = \inf \{c : \| A\textbf{x} \| \leq c \| \textbf{x} \| \}$. For these definitions, I don't understand what $\| A\textbf{x} \|$ means, since $\| A \|$ is defined in terms of $\| A\textbf{x} \|$, and I thought $\| A \|$ and $\| A\textbf{x} \|$ meant the same thing, like $f$ and $f(x)$ are different notations for a function, but their spirit is the same meaning (for example, $f(x) = x^2$ and just $f$ to refer to the function of $x^2$ itself without the input of $x$). Are these two different norms? If so, what is the definition for $\| A\textbf{x} \|$?
I know that since $A\textbf{x} = \textbf{x} \cdot \textbf{y}$, we can write $|A\textbf{x}| = |\textbf{x} \cdot \textbf{y}|$, and using Cauchy-Schwartz we can get $|\textbf{x} \cdot \textbf{y}| \leq |\textbf{x}||\textbf{y}|$. So $|A\textbf{x}| \leq |\textbf{x}||\textbf{y}|$. Can I use this to somehow get $\| A \| \leq \lvert \textbf{y} \rvert$?
For the other direction, it's true that $A\textbf{y} = \textbf{y} \cdot \textbf{y}$, so $|A\textbf{y}| = |\textbf{y} \cdot \textbf{y}| \leq |\textbf{y}|^2$. Can I use this to somehow get $\| A \| \geq \lvert \textbf{y} \rvert$?
 A: You have essentially answered your own question. $||Ax||$ is just different notation for $|Ax|$; it's the norm of $Ax$ taken in $\mathbb{R}$. 
Take your result from Cauchy-Schwartz, $|Ax| \le |x||y|$, and take a supremum over all $|x| \le 1$.
$$||A|| = \sup_{|x| \le 1} |Ax| \le \sup_{|x| \le 1} |x||y| = |y|$$
On the other hand, the supremum is equal to or greater than any achieved value. 
$$||A|| = \sup_{|x| \le 1} |Ax| \ge \left|A \left(\frac{y}{|y|}\right)\right| = |y|$$
A: $\| A \|$ by definiton is the supremum of  $\frac{\| A\textbf{x} \|}{| \textbf{x} |}$ over all $x \in L$. so $|A\textbf{x}| \leq |\textbf{x}||\textbf{y}|$ $\implies$ $\frac{\|A\textbf{x}\|}{|\textbf{x}|} \leq |\textbf{y}|$ for all $x \neq 0$ $\implies$ $\| A \| \leq |\textbf{y}|$
Now $|A\textbf{y}| = |\textbf{y} \cdot \textbf{y}| = |\textbf{y}|^2$, so the supremum is indeed  $|\textbf{y}|$, thus $\| A \| = \lvert \textbf{y} \rvert$
A: By definition of $\| A \| $, and by hypothesis, we have $\| A \| =\sup\limits_{\|\textbf{x} \| \leq 1}  \textbf{y} \cdot \textbf{x} \leq \| \textbf{y}\|$ by Cauchy-Schwarz. 
Since equality is reached with $\textbf{x}={\textbf{y} \over \|\textbf{y}\|}$, of norm $1$, we get $$\| A \| =\sup\limits_{\|\textbf{x} \| \leq 1}  \textbf{y} \cdot \textbf{x}=\textbf{y} \cdot {\textbf{y} \over \|\textbf{y}\|}=\|\textbf{y}\|$$
