Question about definition of bounded linear functionals I'm reading through an analysis textbook, and just working through a section on linear functionals. I have the definition: $f$ is a bounded linear functional if
$$||f|| = \sup\{|f(x)| : x\in X, ||x|| \le 1\} < \infty,$$
and the proposition that 
TFAE:
1. $f$ is bounded;
2. $f$ is continuous;
3. $f$ is continuous at $0$.
Here is my question: it really bothers me, in the definition of a bounded linear functional, that we take just the supremum over all $x$ in the unit ball of the normed space. I know that this is just the definition, but I'm wondering if there is any sort of intuition that would help me to understand why when we find bounds of functions over, say, $R$, we would take the supremum of $|f(x)|$ over all $R$, but with functionals, we take the supremum over just the unit ball?
Another related question that I have is that, in proving $3 \implies 1$ in the proposition above, the book I am reading says:
If $f$ is not bounded, there exists a sequence $x_n \in X$ such that $||x_n|| = 1$ for each $n$, but $|f(x_n)| \to \infty$. 
I see that if $||f|| = \infty$, then there must be some sequence $x_n$ such that $|f(x_n)| \to \infty$, $||x_n|| \le 1$ for each $n$. Why can we find such a sequence such that $||x_n|| = 1$?

The answer to both of your questions is based on the linearity of $f$.  For the first question, notice that, if there is even a single $x$ with $f(x)\neq0$, then by multiplying $x$ by a large positive real number $r$, you get $|f(rx)|=r|f(x)|$, which gets arbitrarily large if you take $r$ large enough.  So the only way $f$ could be bounded everywhere (rather than just on the unit ball $\{x:\Vert x\Vert\leq 1\}$) is to be identically $0$.
For the second question, if you have $x$'s with $\Vert x\Vert\leq 1$ and $|f(x)|$ large, then let $y=x/\Vert x\Vert$ (the denominator isn't $0$ because $|f(0)|$ isn't large), and you have $\Vert y\Vert=1$ and $|f(y)|$ is even larger than $|f(x)|$ because $f(y)=f(x)/\Vert x\Vert$.
