$F(y) = F(x)$ for aribtrary continuous linear functional $F$, then by Hahn-Banach $y=x$? I am reading an operator theory book and I frequently see something like the following used. For $x$ and $y$ in some Banach space $X$, and $F$ an arbitrary continuous linear functional on $X$, if we have that
$$
F(y) = F(x)
$$
then by the Hahn-Banach theorem $y = x$.
The Hahn-Banach theorem I know of involves extending a linear functional dominated by some sublinear function, which does not seem to come into the above statement at all. So what Hahn-Banach theorem is being used above? 
 A: Often authors in functional analysis books will say "by Hahn-Banach" when they mean "by an easy application of or by one of the standard corollaries of Hahn-Banach" rather than "Directly from the statement of Hahn-Banach".
In this case, the Hahn-Banach theorem (as you state it) gives the result in a very straightforward way. Assume, for a contradiction, that $x \neq y$. Try to construct $F \in \operatorname{span}\{x,y\}^*$ such that $F(x) \neq F(y)$ - it might help to consider separately the cases where $\{x,y\}$ is a linearly independent set and when it is not.
Once you have such a functional, by Hahn-Banach you can extend it to an element of $X^*$ such that $F(x) \neq F(y)$, contradicting your original assumption.
A: There's a slew of theorems, often all referred to as "the Hahn-Banach Theorem", that are all equivalent to, or immediate consequences of, the classic Hahn-Banach theorem involving extending linear functionals dominated by sublinear functionals. Included in this are,


*

*The Hahn-Banach theorem (of course),

*The Hahn-Banach separation theorem: If $C$ and $D$ are convex, disjoint, and $C$ is open, then some functional $f$ exists such that $f(c) < f(d)$ for all $c \in C$ and $d \in D$.

*A corollary of the above using compact sets: if $C$ and $D$ are convex, and $C$ is compact, then some functional $f$ and some real $\alpha$ exist such that $f(c) < \alpha \le f(d)$ for all $c \in C$ and $d \in D$.

*If $Y$ is a subspace of $X$ and $x \in X$, then there exists a functional such that $Y \subseteq \operatorname{ker}(f)$, and $\|f\| = f(x) = d_Y(x)$, where $d_Y(x) = \inf_{y \in Y} \|x - y\|$,


and probably others I'm forgetting. I've always preferred the geometric ones, like $2$ and $3$. Applying $3$ to the compact convex sets $\lbrace x \rbrace$ and $\lbrace y \rbrace$ quickly yields the answer.
A: If $x\neq y$, let $z=x-y$. Then $z\neq0$. Consider the linear map $\varphi\colon\mathbb{R}z\longrightarrow\mathbb R$ defined by $\varphi(\alpha z)=\alpha$. You can extend it to a continuous linear map $F\colon X\longrightarrow\mathbb R$ (by the Hahn-Banach theorem). Then $F(z)=1\neq0$ and therefore $F(x)\neq F(y)$.
