Let $X$ be a normed space,let $A$ be a non-empty subset of $X$, and let $\lbrace c_x \rbrace_{x \in A}$ be a corresponding set of scalars. Prove the following are equivalent:
a.) There exists $f \in X'$ such that $f(x)=c_x$ for all $x \in A$.
b.) There exists a constant $M \geq 0$ such that $$|\sum_{i=1}^n \alpha_ic_{x_i}| \leq M||\sum_{i=1}^n \alpha_i x_i||$$ for all possible choices $x_1,..., x_n \in A$ and of scalars $\alpha_1,..., \alpha_n \in \mathbb{F}$
I am not really sure how does $\lbrace c_x \rbrace_{x \in A}$ look like?
Assuming (a), we simply have
$$ | \sum\limits_{i=1}^n \alpha_i c_{x_i} | = | \sum\limits_{i=1}^n \alpha_i f(x_i) | = | f(\sum\limits_{i=1}^n \alpha_i x_i) | \leq ||f|| || \sum\limits_{i=1}^n \alpha_i x_i ||$$
Which is (b) when $M = ||f|| $.
Conversely, assume we have (b). Let $A'$ denote the span of $A$. Define a function $f:A' \rightarrow \mathbb{R}$ as in (a). Clearly f is well defined. We need to check its linear and bounded.
$$ | f( \alpha x + y) - \alpha f(x) - f(y) | = | c_{\alpha x +y} - \alpha c_x - c_y| \leq M ||(\alpha x+y)-\alpha x-y|| = 0$$
So $f$ is linear.
$ |f(x)| = |c_x| \leq M ||x|| $ for any $x \in A'$. So $f$ is bounded.
Finally, by Hahn Banach we can extend $f$ to a functional on $X$ with the same norm, agreeing with $f$ on $A'$.