# If $x, y \in X$ with $x \neq y$, then there exists $f \in X^*$ such that $f(x) \neq f(y)$.

Let $X$ be a normed linear space. Prove that if $x, y \in X$ with $x \neq y$, then there exists $f \in X^*$ such that $f(x) \neq f(y)$.

Here $X^*$ denotes the dual space of $X$.

I am getting some smell of using Hahn Banach theorem but not able to prove it. Need some hints.

Hint: Your idea is good. Apply Hahn-Banach on $x-y\neq 0$.

Let $x,y \in X$ such that $x \neq y$.

Assume that $f(x)=f(y) ,\forall f \in X^*\Rightarrow f(x-y)=0$

From the consequences of Hahn-Banach exists $f_0 \in X^*$ such that $||f_0||=1$ and $f_0(y-x)=||x-y||$.

But $$f_0(x-y)=0 \Rightarrow ||x-y||=0 \Rightarrow x=y$$ contradicting our hypothesis that $x \neq y$

Hahn-Banach is the right way. Just notice that the sets $\{x\}$ and $\{y\}$ are convex, nonempty, disjoint and compact.

The Hann-Banach everyone is talking about but not spelling out:

Take $$H:=span \{x-y\}$$ and define a linear functional on $H$ as $$f(t(x-y)):=t \$$ An extension of this $f$ to all of your space will work.

$\forall$ x $\in X$ such that $x \neq 0$ we can find $f \in X^{*}$ such that $f(x)\neq0$ (using Hahn-Banach theorem) and since $x-y\neq0$ you can find such function