Problem: Let $X$ be a normed vector space where $x,y\in X:x\neq y$. Show that $\exists f\in X^\ast:f(x)\neq f(y)$.

Attempt: Take $z\in X:z=y-x$, where $x\neq y\neq 0$. Then, $\|z\|=\|y-x\|=\delta\gt0$. If $Y=\{0\}$ then $\|z\|=\|x\|=\delta$.

Then we can make use of one of the corollaries to the Hahn-Banach theorem, namely,

"If $X$ a normed vector space and if $x\neq0,\,\exists\, f\in X^*:f(x)=\|x\|$ and $\|f\|=1$"

Using this we have that,


$$\iff f(y)-f(x)=\delta$$

$$\iff f(y)=f(x)+\delta$$

So that then, clearly, $f(y)\ge f(x)$ which is sufficient to say that $f(y)\neq f(x)$ as required.


I want to clarify the reasoning and justification of the above proof.

  1. The corollary to the Hahn Banach theorem we used above is really a corollary of a corollary. It follows from setting $Y=\{0\}$ in this here: "Let $X$ be a normed vector space, $Y\subseteq X$ a linear subspace, and suppose that $\delta=\inf_{y\in Y}\|x-y\|\gt0$. Then $\exists\,f\in X^*:f(x)=\|x\|$ and $\|f\|=1$". Given that we take $z=y-x$, so that it follows, $\|z\|=\|y-x\|=\delta\gt0$, it seems implied that $y\in Y$, so that when we set $Y=\{0\}$ we get that $\|z\|=\|x\|=\delta$. But then, why don't we set $y=0$ in what follows afterwards, namely, in $f(z)=f(y-x)$?

  2. Consulting several different sources, it seems to me that $Y\subseteq X$ in the main Hahn-Banach corollary (that stated in 1. above) should be closed. Is this correct, and should it be stated? Otherwise would it not follow that $\delta=\inf_{y\in Y}\|x-y\|\geq0$?

  • 1
    $\begingroup$ finite-dimensional subspaces of normed spaces are closed. $\endgroup$ – Lord Shark the Unknown May 5 '18 at 14:19
  • $\begingroup$ For question 2, all that's really important is that $Y$ is not dense. $\endgroup$ – Aweygan May 5 '18 at 14:22
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    $\begingroup$ I am a bit confused here. Doesn't your corollary of Hahn-Banach theorem prove your problem, that is, if $y-x\neq 0,$ then by the Corollary, choose $f\in X^*$ such that $\|f\|=1$ and $f(y-x)\neq 0.$ By linearity of $f,$ we have $f(x)\neq f(y).$ $\endgroup$ – Idonknow May 5 '18 at 15:01
  • $\begingroup$ @Idonknow, yes, that is a better proof, and I think I will use that one. The one above was from a set of notes I have, but I didn't see how we could simultaneously let $y$ be arbitrary and then make use of the fact that $y=0$, seemingly, at the same time. $\endgroup$ – Jeremy Jeffrey James May 5 '18 at 15:27

You probably mean this corollary:

Let $X$ be a normed space, $Y \subseteq X$ its subspace, and $x \in X$ such that $\delta = \inf_{y \in Y}\|x-y\| > 0$. Then there exists a bounded linear functional $f \in X^*$ such that $\|f\| = 1$, $f(x) = \delta$ and $f|_Y = 0$.

Notice that by demanding $d(x, Y) = \delta > 0$ you are in fact stating that $x \notin \overline{Y}$. So you can either assume that $Y$ is closed and take $x \notin Y$, or just take $x \notin \overline{Y}.$

Your first corollary

Let $X$ be a normed space and $x \in X, x \ne 0$. Then there exists $f \in X^*$ such that $\|f\| = 1$ and $f(x) = \|x\|$.

is now indeed a consequence of the former one:

Take $x \in X, x \ne 0$. Consider $Y = \{0\}$ and notice $x \notin \{0\} = \overline{Y}$. We have $\delta = d(x, \{0\}) = \|x\|$. The previous corollary implies that there exists $f \in X^*$ such that $\|f\| = 1$, $f(x) = \delta = \|x\|$ and $f|_{\{0\}} = 0$, so $f$ is precisely the desired functional.

Using this, we can show the desired claim, as suggested by @Idontknow:

Take $x, y \in X$, $x \ne y$. We have $x - y \ne 0$ so there exists $f \in X^*$, such that $\|f\| = 1$ and $f(x) - f(y) = f(x - y) = \|x - y\| \ne 0$. It follows that $f(x) \ne f(y)$.

Notice that you cannot claim $f(x) > f(y)$ since those might be complex numbers.


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