1

I have the following question from a past qualifying exam:

If $X$ is a Banach space and $X^*$ denotes its dual space show that if $x_1\ne x_2\in X$ there is an element $y\in X^*$ such that $y(x_1)\ne y(x_2).$

So i'm trying to construct a linear functional $y:X\rightarrow\mathbb{K}$ with the above property, but I don't seem to be getting anywhere. Perhaps there are elements of the dual space that always exist and which I can take advantage of here? Unfortunately, we didn't talk very much about Banach spaces in my course, so if anyone could point me in the right direction, I'd appreciate it.

zbrads2
  • 1,054
  • 6
  • 13

2 Answers2

3

Let $x_0=x_1-x_2$, this vector is non zero. Let $E$=span$\{x_0\}$. Now we define a function $f:E\to K$ by $f(\lambda x_0)=\lambda$. I'll leave it for you to check that it is a bounded linear functional on $E$ and $f(x_0)=1$. And now by Hahn-Banach theorem we can extend $f$ to a functional $F$ in $X^*$. Since $F(x_0)=f(x_0)\ne 0$ we get $F(x_1)\ne F(x_2)$.

Mark
  • 39,605
  • 1
    @zbrads2 Note that I edited my answer a bit. $x_1$ and $x_2$ might not be in $E$, but the extended functional $F$ still has the property $F(x_0)\ne 0$. – Mark Jun 18 '19 at 20:34
  • I was just about to ask about that. Looks good to me now, thanks. – zbrads2 Jun 18 '19 at 20:40
1

By the Hahn-Banach theorem, there is an element $y\in X^*$ such that $y(x_1-x_2)\neq 0$.

user10354138
  • 33,239