The space of all bounded linear functionals is a total set

Question

Prove that for all $X$ normed spaces, $X^*$ is a total set, meaning taht if $f(x)=0$ for every $f$, then $x=0$.

In my class of Functional Analysis, it is simply said that this corollary follows immediately from the following corollary -

For all $x_1\in X$ and for all $x_2\in X$ such that $x_1\neq x_2$, there exists $f\in X^*$ satisfying $f(x_1)\neq f(x_2)$.

I found some difficulty getting the first corollary from the second.

Let $x_0\in X$ such that $f(x_0)=0$ for every $f\in X^*$. Taking some $x\in X$ such that $x\neq x_0$, we get that there exists $f_0\in X^*$ such that $f_0(x)\neq f_0(x_0)=0$. How can I proceed from here to prove that $x_0=0$? I also looked at the following: $f_0(x-x_0)=f_0(x)-f_0(x_0)=f_0(x)\neq 0$, but I could not find any way to continue.

Any hint would be helpful.

Answer 1

Suppose that $p \in X$ is such that for all $f \in X^\ast$, $f(p)=0$. We want to show $p=0$, so assume the contrary.

With $p \neq 0$, now apply the second fact to $x_1= 0$ and $x_2=p$ and note that for all $f \in X^\ast$, $0=f(0) = f(p)=0$ (the first by linearity, the second by the assumption about $p$). So the second fact would fail on this pair. Contradiction!

So $p=0$ and we are done.

Answer 2

It's a simple use the contrapositive. If $x\neq 0$, by second corollary there is an $f$ such that $f(x)\neq 0$.