How to prove the existence of an element of the dual of a Banach space $X$

Question

Let $X$ a Banach space and let $x,y\in X$ such that $x\neq y$. I want to prove that there exists $\Lambda\in X'$ such that $\Lambda x \neq \Lambda y$.

Here $X'$ is the dual space.

How can I prove this? I will apreciate any hint to solve this.

Answer 1

First show there exists $f \in U'$ where $U = \operatorname{span}\{x-y\}$, such that $f(x-y) \ne 0$. Then use Hahn-Banach to extend $f$ to $X$.

Answer 2

You can also use the Hahn-Banach separation theorem (as you often can in lieu of the traditional Hahn-Banach theorem). Let $C = B(x; \|x - y\|)$. The set is open and convex. Using the Hahn-Banach separation theorem, there exists a functional $f \in X^*$ separating $C$ from $\lbrace y \rbrace$. That is, $$f(c) < f(y)$$ for all $c \in C$. In particular, $x \in C$, so $f(x) < f(y)$.